I would say you need to integrate the expression you arrived at twice, first at:

[math]\int_{-1}^{0}\mathrm x^2-2x\,\mathrm{d}x[/math]

and then at:

[math]\int_{0}^{2}\mathrm x^2-2x\,\mathrm{d}x[/math]

Once you have evaluated both expressions, add the absolute values. If you try to evaluate the expression over the entire interval you will come up with 0, which the following figure shows cannot be the case. The space between the two curves is the area you're looking for.

The answer should be [math]\frac {8}{3}[/math]

[math]\large\displaystyle\star[/math] A2A

Original Question:

From where do you solve tricky integration problems?

I try to solve them by constant use of a pen and paper with lots of mess(rough work) until I get the result.

Though probability of getting result is not very high as you might come across with some integrals like,

[math]\rightarrow\large\displaystyle\int \frac{x}{\ln x} \, dx[/math]

[math]\rightarrow\large\displaystyle\int e^{x^{2}} \, dx[/math]

[math]\rightarrow\large\displaystyle\int \frac{\sin x}{x} \, dx[/math]

which looks easier but actually are non integrable (using elementary functions).

I will give you a glimpse of various ways which could be followed(in the given order) to solve(attempt) the problems based on indefinite or definite integrals:

• Integration is the reverse process of differentiation. So, first of all make sure, you know the derivatives of all elementary functions so that it can be applied whenever required.

Some of the examples:

[math]\rightarrow\large\displaystyle\frac{d}{dx}(\tan^{-1} x) = \large\displaystyle\frac{1}{1 + x^2}[/math]

[math]\rightarrow\large\displaystyle\frac{d}{dx}(\ln x) = \large\displaystyle\frac{1}{x}[/math]

[math]\rightarrow\large\displaystyle\frac{d}{dx}(\cot x) = \large\displaystyle - \csc^2 x[/math]

[math]\large\displaystyle\ldots[/math] and so on.

• The next way is to check if there is any possibility of “substitution”. This could make a complex problem simple.

For example:

If you were asked to integrate,

[math]\large\displaystyle I = \large\displaystyle\int \frac{1 + \displaystyle\frac{1}{x} - \displaystyle\frac{1}{x^2}}{x + \ln x + \displaystyle\frac{1}{x}} \, dx[/math]

It seems a bit complicated right!

But you don’t need to think about it more. Just substitute the denominator as [math]\large\displaystyle t[/math] and you have its derivative in the numerator. So, the integral would be simply,

[math]\large\displaystyle I = \large\displaystyle\boxed{\large\displaystyle\ln \left|x + \ln x + \frac{1}{x} \right| + c}[/math]

• The other important way of tackling these problems is using basic algebraic simplifications like “rationalisation”.

Let me illustrate this using a simple example:

[math]\large\displaystyle I = \large\displaystyle\int \frac{1}{1 + \sin x} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \frac{1 - \sin x}{(1 + \sin x) (1 - \sin x)} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \frac{1 - \sin x}{(1 - \sin^2 x)} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \frac{1 - \sin x}{\cos^2 x} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \sec^2 x \, dx - \int \tan x \sec x \, dx[/math]

[math]\implies\large\displaystyle\boxed{\large\displaystyle I = \large\displaystyle \tan x - \sec x + C}[/math]

• This one would be useful when you come across the integrals involving various trigonometrical functions. You need to remember some trigonometric identities like:

[math]\rightarrow\large\displaystyle\cos 2 x = \large\displaystyle\cos^2 x - \sin^2 x[/math]

[math]\rightarrow\large\displaystyle 1 + \sin 2x = \large\displaystyle (\sin x + \cos x)^2[/math]

[math]\rightarrow\large\displaystyle\sin 2 x = \large\displaystyle 2 \sin x \cos x[/math]

Consider an example:

[math]\large\displaystyle I = \large\displaystyle\int \tan^{-1} \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2x}} \, dx[/math]

So, you just apply the trigonometric identity and get the result.

[math]\implies\large\displaystyle I = \large\displaystyle\int \tan^{-1} \sqrt{\frac{2 \sin^2 x}{2 \cos^2 x}} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \tan^{-1} \sqrt{\tan^2 x} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \tan^{-1} (\tan x) \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int x \, dx[/math]

[math]\implies\large\displaystyle\boxed{\large\displaystyle I = \large\displaystyle\frac{x^2}{2} + c} [/math]

• This method is slightly underrated but it could turn out as a crucial way to solve an integral. I’m talking about solving by adding and subtracting something which could make the expression easy.

Example:

[math]\large\displaystyle I = \large\displaystyle\int \frac{x^2}{x + 1} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \frac{x^2 + 1 - 1}{x + 1} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int \frac{x^2 - 1}{x + 1} \, dx + \int\frac{1}{x + 1} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\int x \, dx - \int 1 \, dx + \int\frac{1}{x + 1} \, dx[/math]

[math]\implies\large\displaystyle I = \large\displaystyle\frac{x^2}{2} - x + \ln\left|x + 1 \right| + C[/math]

[math]\implies\large\displaystyle\boxed{\large\displaystyle I =\large\displaystyle\frac{x^2 - 2x}{2} + \ln\left|x + 1 \right| + C}[/math]

• Then there are certain standard forms and properties of definite and indefinite integrals which could be useful if you keep it in your memory.

Few examples of this kind:

[math]\rightarrow\large\displaystyle\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c[/math]

[math]\rightarrow\large\displaystyle\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left|\frac{x - a}{x + a} \right| + c[/math]

[math]\rightarrow\large\displaystyle\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx[/math]

[math]\large\displaystyle\ldots[/math] and many more.

• Now, The ultimate method which you will surely going to need in order to solve any problems(not just restricted to calculus). It is simply

[math]\hspace{55 mm}\large\displaystyle\boxed{\large\displaystyle\boxed{\large\displaystyle\textbf{PRACTICE}}}[/math]

Any other method won’t work if you don’t implement this.

P.S. - Now, Try to solve a simple integral using any of the above approaches and comment the solution.

[math]\large\displaystyle\boxed{I = \large \displaystyle \int e^x \left( \sin^{-1} x - \frac{3x(3 + 2x^2)}{(1 - x^2)^{\frac{7}{2}}} \right)\,dx}[/math]

It can be solved in just three steps.

Thanks!

Image source: - My rough notebook.

Edit: I have corrected the integral (mentioned in P.S.) now.

Edit 2: Please see the solution to the integral :

[math]\large\displaystyle I = \large \displaystyle \int e^x \left( \sin^{-1} x - \frac{3x(3 + 2x^2)}{(1 - x^2)^{\frac{7}{2}}} \right)\,dx[/math]

Step 1 : Adding something and subtracting something.

[math]\large\displaystyle I = \large \displaystyle \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}} + \frac{x}{(1 - x^2)^{\frac{3}{2}}} - \frac{x}{(1 - x^2)^{\frac{3}{2}}} + \frac{2x^2 + 1}{{(1 - x^2)}^{\frac{5}{2}}} - \frac{2x^2 + 1}{{(1 - x^2)}^{\frac{5}{2}}} - \frac{3x(3 + 2x^2)}{(1 - x^2)^{\frac{7}{2}}} \right)\,dx[/math]

Step 2 : Little bit of rearranging.

[math]\implies\large\displaystyle I = \large \displaystyle \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1 - x^2}} \right)\,dx - \int e^x \left( \frac{1}{\sqrt{1 - x^2}} + \frac{x}{(1 - x^2)^{\frac{3}{2}}} \right)\,dx + \int e^x \left( \frac{x}{(1 - x^2)^{\frac{3}{2}}} + \frac{2x^2 + 1}{{(1 - x^2)}^{\frac{5}{2}}} \right)\,dx - \int e^x \left( \frac{2x^2 + 1}{{(1 - x^2)}^{\frac{5}{2}}} + \frac{3x(3 + 2x^2)}{(1 - x^2)^{\frac{7}{2}}} \right)\,dx[/math]

Step 3 : Now just apply the standard integral,

[math]\rightarrow\large\displaystyle\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c[/math]

[math]\implies\large\displaystyle I = \large \displaystyle e^x \left( \sin^{-1} x \right) - e^x \left( \frac{1}{\sqrt{1 - x^2}} \right) + e^x \left( \frac{x}{(1 - x^2)^{\frac{3}{2}}} \right) - e^x \left( \frac{2x^2 + 1}{{(1 - x^2)}^{\frac{5}{2}}} \right) + C[/math]

Finally, rearranging it to get a better expression.

[math]\implies\large\displaystyle\boxed{I = \large \displaystyle e^x \left[ \sin^{-1} x - \frac{1}{\sqrt{1 - x^2}} + \frac{x}{(1 - x^2)^{\frac{3}{2}}} - \frac{2x^2 + 1}{{(1 - x^2)}^{\frac{5}{2}}} \right] + C}[/math]

This (integral) was something I constructed just as a brain racking exercise.

Hope! you tried well before checking the solution.

P.P.S. - Pardon for the delay in posting the solution (got a “little” busy)!

P.P.P.S. - Feel free to ask in case you have any doubt in any step!

[math]\large\displaystyle\boxed{\huge{\huge{\displaystyle\ddot\smile}}}[/math]

Thanks for the A2A. If you are a student learning integration by parts, then the other answers address this question already. If you are not a student, then what you are doing wrong is trying to solve this using integration by parts in the first place.

It is EXTREMELY useful to begin to see exponential functions and sines and cosines as just the same kind of function. When you do, you realize that it is often beneficial to treat them as the same kind of function in a problem. This is just such an example.

Instead of your integral, let’s think about a related integral.

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx[/math]

Of course this integral doesn’t look like yours, nor does it give the same answer as yours, but using Euler's formula and some algebra, we can see how this integral is related to the one you care about.

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx=\int_0^\pi e^{6x}(\cos(10x)+i\sin(10x))\ dx=\int_0^\pi e^{6x}\cos(10x)\ dx+i\int_0^\pi e^{6x}\sin(10x)\ dx[/math]

Notice that the answer to our new integral has a real part, and an imaginary part. The real part is a similar integral to the one you care about but with the sine function replaced by a cosine. The imaginary part of the original integral is exactly that answer that you care about. So instead of solving your problem using integration by parts twice and then doing some algebra, why not solve this related integral which ONLY has exponential functions in it and then take the imaginary part as the answer to your question?

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx= \int_0^\pi e^{x(6+10i)}\ dx[/math]

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx=\frac{1}{6+10i} e^{x(6+10i)}|_0^\pi[/math]

We are now DONE integrating. Not nearly as much work as integrating by parts twice. All we have left is to do some complex arithmetic to sort out the real and imaginary part of this answer.

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx=\frac{1}{6+10i}\left( e^{\pi(6+10i)}-1\right)[/math]

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx=\frac{1}{6+10i}\left( e^{6\pi}e^{10\pi i}-1\right)[/math]

Now [math]e^{10\pi i}=\left(e^{2\pi i}\right)^5=1^5=1[/math]. Furthermore, [math]\frac{1}{6+10i}=\frac{6-10i}{6^2+10^2}=\frac{3}{68}-i\frac{5}{68}[/math].

So we have:

[math]\int_0^\pi e^{6x}e^{i\cdot 10x}\ dx=\left(\frac{3}{68}-i\frac{5}{68}\right)\left( e^{6\pi}-1\right)[/math]

And finally, we know that the answer to your problem is just the imaginary part of this result:

[math]\int_0^\pi e^{6x}\sin(10x)\ dx=- \frac{5}{68} \left( e^{6\pi}-1\right)[/math]

A line integral is the generalization of simple integral.

A surface integral is generalization of double integral.

A volume integral is generalization of triple integral.

A multiple integral is any type of integral.

Let us go a little deeper. For simplicity, we will restrict our discussion to only Cartesian coordinates, but the same argument holds for other coordinates as well.

1. Simple Integral Vs Line Integral: A simple integral is evaluated along the x axis (or y axis) only). [math]\int_a^b\ f(x)\,dx[/math] denotes integration of f(x) along the x axis. The points which go into evaluation of this integral - the domain of integration- come only from x axis (their y coordinates are identically zero).

But a line integral generalizes the idea of a simple integral. In a line integral, the curve along which the integral is evaluated is not necessarily a x (or y) axis, or even a straight line. It can be any curve lying in higher dimensional space; though the curve itself is a 2 D entity, by definition.

A simple integral:

Here the integration is along the x axis only, which necessitates that the y coordinate is zero along the path of integration.

A line integration:

Here the path of integration is along the curve, which is not an x axis or y axis in general. The picture shows a closed curve but the concept of line integrals is valid for open curves as well.

This idea, that a line integral is a generalization of simple integral, is critical to our evaluation of a line integral.

We evaluate the line integral by expressing the curve C in parametric form, in terms of a parameter “t”, say. Geometrically, this is equivalent to stretching the curve into a straight line on the “t” axis.

This reduces the line integral into a simple integral in term of variable t.

(the line integral of a vector valued function F has been expressed as simple integral in terms of t, c(t) is a parametric representation of curve C. A similar expression for scalar function is shown below:

2. Surface Integral Vs Double Integral : Just as a line integral extends the idea of a simple integral to general curves, a surface integral extends the idea of double integral to a general surface.

In a double integral, the points which go into the evaluation of the integration come from a 2 D planar surface.

The domain of integration, region D in the expression above is a 2 D planar region over which the integration is calculated. This is shown in the figure below:

The region D lies on the xy plane and is bounded by [math]a<x<b; y=g_2(x); y=g_1(x)[/math]

However, a plane is only one kind of surface, there are other surfaces as well, like a paraboloid, an ellipsoid, a sphere. What should we do if we have to calculate integrations over these , non-planar surfaces? For eg. the flux of an elctric field through a sphere?

This is what leads to the concept of a surface integral.

The surface S can be any surface in space.

Even in multidimensional space, a surface is a 2D entity and can be expressed completely and uniquely by two parameters, u and v (say) . This helps us to convert a surface integral into a double integral. Geometrically, it is akin to stretching, and flattening out a surface into a plane on u and v axes.

Here D is a region on u v plane whose limits are based on the fact that

is a parametric representation of surface S, which makes D a projection of S on u v plane.

For example , if S is a cylindrical surface given by:

[math]x^2+y^2=9 , 0<z<1[/math]

then

[math]x=3 cos v[/math]

[math]y=3 sin v[/math]

[math]z= u[/math]

[math]0< v < 2 pi; 0<u<1[/math]

will be the parametric representation of S through u and v. This means that D will have the following representation on u v plane:

Though this is not asked in the question, but for the sake of completion, let me add that it is possible to convert a line integral into a double integral using Green’s theorem , and line integral into surface integral using Stoke’s theorem.

To summarize:

[math]\int \sin \left(e^x+1\right)\;dx[/math]

Substitute: [math]u = e^x[/math]; [math]dx=\frac{du}{u}[/math]

[math]=\int \frac{\sin (u+1)}{u}\;du[/math]

Sum of angles formula:

[math]=\int \frac{\sin (u)\cos(1)+\cos(u)\sin(1)}{u}\;du[/math]

Break apart the integral:

[math]=\int \frac{\sin (u)\cos(1)}{u}\;dx+\int\frac{\cos(u)\sin(1)}{u}\;du[/math]

Pull out the constants:

[math]=\cos(1)\int \frac{\sin (u)}{u}\;dx+\sin(1)\int\frac{\cos(u)}{u}\;du[/math]

Special functions: Sine Integral and Cosine Integral:

[math]=\cos(1)\operatorname{Si}(u)+\sin(1)\operatorname{Ci}(u)+C[/math]

Undo the substitution:

[math]=\cos(1)\operatorname{Si}(e^x+1)+\sin(1)\operatorname{Ci}(e^x+1)+C[/math]

Edited:

I went to check my answer on Symbolab, and to my surprise, there is an elementary integral!

It involves the substitution [math]u=e^x+1[/math], followed by a Weierstrass Substitution, followed by a Partial Fraction decomposition, followed by some Trig substitutions. So that’s the strategy, if you want to go for it.

It gives new nuance to the term “elementary”.

Edited again:

I believe the Symbolab derivation has a bug, and the answer it gives is certainly not correct. It applies a Weierstrass Substitution [math]t=\tan\frac{x}{2}[/math]:

[math]\int\frac{\sin u}{u-1}\;du\Rightarrow\int\frac{4t\;dt}{(1+t^2)^2(t-1)}[/math]

This is incorrect; the [math]u-1[/math] is magically turned into [math]t-1[/math].

So it looks like there might not be an elementary integral after all.

Line Integral vs Single Integral : A line integral is an integral where the function is integrated or evaluated along a curve which lies on higher dimensional space thus it is also called path integral.We all know about simple integral ,in geometric view, we find area under a curve in 2D region similarly in line integral we will find a area under curve which in higher dimension(more than 2),in physical view,we find total work done both line & normal integral but difference is work will be evaluated along the path or curve given. for better understand please go these two links:

Surface Integral vs Double Integral : A Surface integral is an integral where the function is integrated or evaluated along a surface which lies on higher dimensional space.A (two dimensional) surface integral is taken on a shape embedded in a higher-dimensional space.

But in double integral it can only integrals a function which is bounded by 2D region with respect to infinitesimal area

That is, we can take the surface integral of a sphere, say, in three dimensions. We can map the sphere's surface to a plane, and then take the integral.

Another example would be that of a cube in 3D. Clearly, the surface of the cube is 2D in nature, but the cube itself is embedded in 3D space. We can take the integral over this surface.

You can think about surface integrals this way: if we can somehow unfold, stretch, rotate, cut, and bend the surface of some shape to make it flat, then we can take the surface integral over the boundary of the shape. But the shape itself is not necessarily flat and certainly not two-dimensional.

A double integral (in its main meaning as an iterated integral) can only be taken on a two-dimensional space. That is, we can only take it over a region of 2D space. Like a square, or a circle, or any other shape with an inside.

So, a surface integral can lead to an iterated integral if we can map (stretch, rotate, etc) the surface to a two-dimensional space, and, conversely, if we can map the two dimensional space into a higher dimensional surface, then we can take the surface integral! It's a nice symmetry between the two for nice enough surfaces and shapes (though the surface integral is more general if we consider exceptional cases).

Surface integral turns into double integral when the surface is projected onto arbitrary plane region.

Volume Integral vs Triple Integral : A Volume integral is an integral where the function is integrated or evaluated along a volume which lies on higher dimensional space.A (three dimensional) volume integral is taken on a shape embedded in a higher-dimensional space.

But in triple integral it can only integrals a function which is bounded by 3D region with respect to infinitesimal volume.A volume integral is a specific type of triple integral.

There are several issues with attempting to answer that question.

In order for something to be classified as “an NP problem” or not, it needs to be a decision problem with a specific size parameter (usually called “[math]n[/math]"), so that we can examine whether instances of that decision problem can be verified in time polynomial in [math]n[/math].

The term “solving integrals” is not in any obvious way such a thing. Presumably, you mean something like this: a program that receives as input an elementary function described in some formal way, and produces as output an elementary function which is an antiderivative of the input. We can assume that the size parameter [math]n[/math] is some indication of the length of the input expression.

So first of all, that's not a decision problem. Decision problems are only supposed to answer “yes” or “no” on a given input. This can be fixed, by changing the question to be about the class FNP instead of NP. FNP is the analog of NP for function problems, whereas NP applies to decision problems only.

Naively, you may argue that verifying an answer to the problem of computing an antiderivative can be done very efficiently, since it merely involves the computation of a derivative. This can certainly be done in polynomial time. So it seems that symbolic integration of elementary functions is indeed in FNP.

However, you are then left with the problem of deciding whether two explicit elementary functions are the same: the input, and the computed derivative of the output. That problem is undecidable. It's not a matter of efficiency: there's no algorithm to do this at all. So the problem of computing symbolic integrals has bigger issues than being in FNP or not: it's not algorithmically tractable.

In practice, there are good algorithms which handle symbolic integration fairly efficiently, and return the correct answer in most cases. However, it's meaningless to ask whether such algorithms are or aren't in NP or FNP, since they're not algorithms at all: they are semi-algorithms which only successfully work on a subset of their inputs, and that subset isn't recognizable by any algorithm.

The classic “hard to integrate” function is the Dirichlet Function which is discontinuous everywhere. This needed a whole new field in analysis Measure (mathematics) - Wikipedia

This is the foundation of Lebesgue integration - Wikipedia. The Lebesgue integral is equivalent to the Riemann integral - Wikipedia where the Riemann integral is defined, but of course can do much more. Like evaluate the integral of the Dirichlet function. It is also the foundation of probability.

Note that the fundamental theorem of calculus only applies to the Riemann. There are tons of cases in Lebesgue integration where “anti derivatives” don't exist.

Likewise , Itô calculus - Wikipedia can integrate stochastic processes, which are random values discontinuous-everywhere functions. This is huge in financial mathematics.

IBP works when one function goes to a constant, while the other stays the same or similar when derived infinitely. In this equation, neither function reduces to a constant after any amount of deriving or integrating.

So here’s what you do. You do IBP twice and set that equal to the original. At that point, you should have the integral on both sides, albeit with some constant being multiplied by one of the integrals. Take both integrals on one side, then simplify.

Here’s an example of this done with a simpler problem:

[math]\int_0^\pi e^x \sin(x) dx[/math]

By LIPET, you usually take [math]u = e^x[/math], but the way you do it works as well. To make it more applicable to your problem, I’ll take [math]u = \sin(x)[/math].

[math]\int_0^\pi e^x \sin(x) = uv - \int_0^\pi v du[/math]

[math]= e^x \sin(x) - \int_0^\pi e^x[/math][math] \cos(x) dx[/math]

Take IBP once more

[math]u = \cos(x), [/math][math]dv = e^x dx [/math], [math]v [/math][math]= e^x [/math][math], du = -\sin(x) dx[/math]

[math]\int_0^\pi e^x \sin(x) = e^x \sin(x) - ( e^x \cos(x) + \int_0^\pi e^x \sin(x) dx )[/math]

[math]=[/math] [math]e^x \sin(x) - e^x \cos(x) - \int_0^\pi e^x \sin(x) dx[/math]

Now you have [math]\int_0^\pi e^x \sin(x) dx [/math]on both sides. It is positive on the left of the equal sign, and negative on the right. Add [math]\int_0^\pi e^x \sin(x) dx[/math] to both sides to get:

[math]2 \int_0^\pi e^x \sin(x) = e^x \sin(x) - e^x \cos(x)[/math]

Now divide both sides by 2

[math]\int_0^\pi e^x \sin(x) = \dfrac {e^x \sin(x) - e^x \cos(x)}{2}[/math]

Your problem will be more complicated, but those are the basics.

I hope that helped! And good luck in your future endeavors.

I wouldn’t try it by hand. I’m sure it’s possible, but it’s a lot of steps and you’d get one wrong. In the old days I sometimes had to do stuff like this, but now we have Wolfram Computational Knowledge Engine.

I am legitimately shocked no one mentioned the mean value theorem. It’s only the reason why the fundamental theorem of calculus even works in the first place.

I assume you’re asking this question not out of confusion about sums of rectangles, surely you understand that. Rather, I’m assuming you want to know why Area is computed by f(b) - f(a). And that is an outstanding question.

The reason is pretty simple. It’s the mean value theorem, which says that for any continuous interval [a, b], say, there exists a ‘c’ in [a, b] s.t. if f(x) is differentiable on the open interval (a, b),

then, f’(c) = (f(b) - f(a))/(b-a).

Okay, now at first glance you might want to call bullshit. That’s fair. It does look suspect upon first glance.

But consider this, it is certainly true that for a smooth curve there exists f(c) st. for c in [a, b] f(a) <\ f(c) <\ f(b). Easy enough.

Now, what the mean value theorem says is something different. What it says is that if you drove 60 miles in one hour, you must have driven exactly 60mph at least once during your trip.

Now if you don’t believe me that’s fine. Just think about it, if you drove less than 60mph the entire way you would not have driven 60 miles in one hour; not possible.

You could drive a variety of speeds and have it come out to 60 miles over 1 hour; that’s fine too. But here’s the thing, if you always drove less than 60mph, you would not have gone 60 miles in 1 hour, since this is not possible as we said. But if you drove sometimes slower than 60mph and sometimes faster than 60mph, but never stopped your spedometer at exactly 60mph, you still drove 60mph at least once, since you have to have gone 60mph when accelerating to a speed faster than 60mph. After all, a speedometer can only get to 70mph if the needle first touches and passes 60mph, and to slow down from 70mph to, say 50mph, the needle must go back down and touch and cross 60mph again, therefore you must have gone 60mph at least once.

The only other option is to go 60mph the entire way, which hammers home the point nicely.

Now, doings a bit of algebra, we see that f’(c)(b-a) = f(b) - f(a), and well what do you know? That looks riemann-sum-ish, except it isn’t since we’re using f’(c) instead of f(c). Riemann used the actual function (which makes more sense, mind you, to me anyway), but the smart-ass who figured out that the mean value theorem applies here is actually more accurate in getting the area right.

Turns out that f(b) - f(a) = the infinite sum of (f(Xn) - f(Xn-1)) = f’(c)(b - a) = the infinite sum f’(Cn)(Xn - Xn-1).

Altogether then, the integral of g(x) from a to b = f’(c)(b-a) = f(b) - f(a), where f’(c) = g(c).

And that’s why you find the anti-derivative before calculating the area of the function from a to b. It all seems like a lucky coincidence doesn’t it! I bet that’s not too far from the truth.

I’d still rather take Riemann sums manually myself. They make more sense, ¯\_(ツ)_/¯.

But wait! We want the integral for f(x) not g(x)!

Yeah, so find an anti-derivative for f(x). Call that r(x), (isn’t ‘r’ such a cool letter for a function? I like it.) then the integral of f(x) from a to b = r’(c)(b-a) = r(b) - r(a).

So, I mean, yeah it’s slightly weird. What the hell does r(x) evaluated at a and b have to do with f(x)?! You might protest.

Well, it just happens that f’(x) isn’t what we want. We want r(x) so that r’(x) = f(x) not f’(x).

Perhaps, less perversely, then, the integral of f(x) from a to b = f(c)(b -a) = r(b) - r(a). Since r’(c) = f(c). And that is the relationship going on here. And why the definite integral gives area.

Note, that car example I have was not my original example but I lost the link it came from. I’ll add it here if I an find it again. Here’s a pretty good alternative link.

Note that this is not the definition of the definite integral. Integrable functions do not need to have continuous domains. The fundamental theorem of calculus only works if the function you want to integrate is both continuous on the interval you want to integrate over and differentiable over the interior of that interval.

A function is Riemann integrable, in general, when the sup{L(f, P) : P is a partition of [a, b]} = inf{U(f, P) : P is a partition of [a, b]}. This, is the real criterion for differentiability with respect to Darboux integrals, at least. There are other kinds of integrals, but these are the ones studied in a calculus class.

Don’t forget this as it’s very important to understand conceptually. People often think the integral is defined as F(b) - F(a), no it is not.

To find area under a curve from one value of x to another value of x, or from one point on the curve to another, in terms of x, we take the integral at the first point and subtract it from the integral at the second point. This is the area under the curve, between the curve and the x-axis between from some vertical line, x = a, to another one, x = b.

Now the general form of the integral of the function f(x) is required to find the definite integral. The general integral will be in the form

F(x) = ʃ f(x) * dx = F(x) + c where c is some constant. You can ignore the c when taking the definite integral.

To find the definite integral we take the integral evaluated at x = a and subtract it from the integral evaluated at x = b.

So, if you have a curve like y = f(x) =[math] 3x^2[/math] + 4 and you want to know the are underneath it from x = 1 to x = 2,

ʃʃʃ

If you don’t know about integrals yet, the integral is the limit as n → ꝏ of the summation of the product of f(x) and dx, where f(x) is a function and considered over an interval from x=a to x=b, which is divided into n vertical cross sections, each cross section has a width of dx = (b - a)/n. The greater the number of cross sections taken from a to b, the greater the accuracy of the area found by the sum, until if n = ꝏ, the summation equals the precise value of the area.

Integration and accumulation of change

[math]\int_{x = a}^{b} f(x) dx[/math] is a big sum (that's why there's a big S there). What is it adding up? Lots and lots of little products: [math]f(x) \times dx[/math], as we run through [math]x[/math] values from [math]a[/math] to [math]b[/math] in tiny steps of size [math]dx[/math].

Now, this doesn't have to be interpreted in terms of area, but that's one application it has: [math]f(x)[/math] describes the "height" of a function's graph at a point, and [math]dx[/math] describes an infinitesimal width around that point, so that this gives us the area of a thin rectangle "under" the function's graph at a particular point. Adding a bunch of these pieces up, we end up with the total area "under" the function's graph throughout some interval.

Note: It's not a great idea to try to understand all integration in terms of area, in the same way that it's not a great idea to try to understand all addition in terms of area calculation. Sometimes, calculating area is why you're carrying out a sum, and sometimes, area has nothing to do with it.

Also, keep in mind, this uses a "signed" notion of area: when [math]f(x)[/math] goes negative, the area "under" it is considered negative. Also, more subtly, the sign of [math]dx[/math] plays into it, so that [math]\int_{x = a}^{b} f(x) dx = - \int_{x = b}^{a} f(x) dx[/math]. If you really wanted the area between the function's graph and the horizontal axis, no funny business about signs, you'd use [math]\int_{x = a}^{b} |f(x) dx|[/math].

There is good news and there is bad news. The good news is that there is indeed a closed form antiderivative. The bad news is that it’s a bitch to work it through manually. You would definitely not want to run in to this on a college integration bee. I’ll get it going.

Start by using the following reduction rule:

[math]\displaystyle\Large\int\frac{dx}{(a+bx^n)^p}\ \rightarrow \frac{x}{(n(p-1)a)(a+bx^n)^{p-1}}-\frac{1-np+n}{n(p-1)a}\int\frac{dx}{(a+bx^n)^{p-1}}\tag*{}[/math]

Applying this reduction rule twice and simplifying reduces the integral to

[math]\displaystyle\Large\frac{x}{8(x^4+1)^2}+\frac{7}{8}\left(\frac{x}{4(x^4+1)}+\frac{3}{4}\int\frac{dx}{x^4+1}\right)\tag*{}[/math]

We still have an integral in the expression, and this is where it gets hairy. To be continued after research…

Ok, I’m back.

Like I said, it’s a bitch. The above is from Engineering Mathematics Handbook, (3rd ed., McGraw Hill, auth. Jan J. Tuma)

So if we carefully use this antiderivative rule, we have [math]p=1, m=2, a=1.[/math]

Working on it, back momentarily…

So making the substitutions and simplifying, we have

[math]\displaystyle\large\frac{21}{128}\sqrt{2}\arctan{(\sqrt{2}x-1)}+\frac{21}{128}\sqrt{2}\arctan{(\sqrt{2}x+1)}+\frac{x(7x^4+11)}{32(x^4+1)^2}-\frac{21}{256}\sqrt{2}\ln{\left(\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)}+C[/math]

Somebody please check this against Wolfram and / or point out the errors…

There is obviously no single hardest integration problem, but one standard integral you might have some trouble with is the [math]\displaystyle \int \limits_0^1 \dfrac{\ln{(1-x)}}{x} \mathrm{d}x[/math]

Taking “integration problem” to be any sort of math question involving integrals, I daresay that finding the average distance between two randomly-selected points in a circle (with radius [math]R[/math]) is fairly challenging. This, however, is more a matter of solid conceptual understanding than raw ability to antidifferentiate.

Then again, I have never taken a class beyond an introductory calculus, and don't have a solid understanding of anything beyond differential equations, and so there are other things like contour integrals, that I have just heard of, but haven't really looked into yet, and couldn't tell you anything about their difficulty.

Edit: Solutions are in the comments. If you noticed that the first integral is the [math]-\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^2}[/math], try to do it a different way to prove what that sum converges to.

Substitute [math]x={cos^2}(t)[/math]
now [math]1-cos(t)={sin^2}(\frac{t}{2})[/math]
and [math]1+cos(t)={cos^2}(\frac{t}{2})[/math]

Thus the whole factor along with power [math]\frac{1}{2}[/math] becomes [math]tan (\frac{t}{2})[/math]
also there is a (cost)^2 in the denominator and -2*cost*sint in the numerator

thus finally the questions is

integral of -2*tan(t/2)*tan(t).dt where x=(cost)^2
from now on i will not see th constants and you can deal with them on your own.
convert tan(t) = 2tan(t/2)/[1-tan(t/2)^2]
now with tan(t/2)^2 in numerator convert it into 1-tan(t/2)^2 i.e. denominator
Now the integral left is 1/1-tan(t/2)^2
convert into cos and sin and use cosx= cos(x/2)^2-sin(x/2)^2 = 2cos(x/2)^2 - 1 and finish the integral

IMPORTANT : take care of constants and dont forget to substitute t with x in the end.

I would strongly recommend to try to resist thinking about it this way.

I can understand why you would have this impression, because this is usually how students in a senior high school or introductory college calculus course will first meet integration. The first calculus course a student will usually take, the integration of a function of one variable over an interval is usually given the geometric interpretation of the area between the curve and the axis of the variable.

This is OK as a way to visualize the sum in the integral. It helps make things concrete. Which is fine if you're just learning. But if you just understand integration as finding the area under a curve, you are going to get misconceptions that may make it harder when you go to take higher level calculus courses as part of a math, physics, or engineering degree.

In higher level calculus and science/engineering courses (e.g. Cal 3, vector/advanced calculus, complex valued functions), you will encounter a whole range of other types of integrals, none of which can be interpreted as an area under a curve. For example, multiple integrals, line integrals, surface integrals, complex contour integration, integration of probability distributions.

It's better to think of integration in the following slightly more abstract way:

Suppose you have a function defined on some set (the domain), with the function taking on some value at each, or at least most points in the set. The set could be different things. It could be:

• real numbers on an interval on a number line.
• a 2D region in a flat x-y plane.
• a region in 3D space
• a curved path in space
• a curved surface in space
• or even more abstract sets like the high dimensional “phase space” of a system in advanced classical mechanics or statistical mechanics or the sample space in a probability problem

Suppose as well that you have some “measure” of the “size” of a small, infinitesimal subset of the region. This is what the differential dx does in a single variable integration. In that case, dx is an infinitesimal length element along the number line where the function is defined. In a 2d integral, the differential could represent an area element. In a 3d integral, a volume element.

Then the integral becomes a sum of all the values the function takes in points in the domain set, weighted by the “size” of each little infinitesimal chunk of the domain set.

So to go back to the 1d integral.

Instead of thinking of it as an area under the curve sketched out in the x-y plane, think of it like this. You have the 1d number line, and on each point on the line, you attach a number representing the value of the function. If you want a concrete visualization, picture a metal rod, and at each point along the rod you have a different temperature. Then the integral is adding up all the values along the line.

The answer using concept of gamma functions

EDIT : problems with math typing

Rachel Britt-Busler's comments on latex are good enough for me. I'll try to do this in words.

If a rectangle has long side, y, and short side, delta_x, then the rectangle's area = y*delta_x. The integral of y*delta_x is literally the sum of a lot of rectangles. Each rectangle has a long side equal to y, which is the height of function above the x-axis. Each rectangle has a short side, delta_x, which is a short distance along the x-axis. So the sum of the areas of these rectangles is the total area under the curve, from x_1, the left side of the first rectangle, to x_2, the right side of the last rectangle. In numerical analysis you actually estimate definite integrals this way, using more and more rectangles with smaller and smaller values of delta_x to get more and more accurate estimates.

I don't know the history of the invention of calculus. Newton used it to solve problems of planetary motion, so he might have invented differentiation to get slopes and integration to get areas. I don't know if Leibniz's approach was more formal.

Some of my friends who were math majors really, really hated infinitesimals. They completely rejected "deriving" (their quote marks) differential and integral calculus by multiplying and dividing by "numbers" that were "smaller than any numbers." So an alternative approach is to formally define differentiation and integration, then apply those definitions to a wide range of functions, and then later to show that for some functions with "nice" properties, integration can be used to find areas, volumes, etc.

Sorry if that's way more than you wanted.

if you don't memorize the general rules and don't have a keen eye. If you are solving sums for MCQ purposes then you should just try and differentiate the options instead of tearing your hair off thinking about the possible substitutions that can work. However if its a easy one you should rather try to integrate it. Differentiating the options is bound to give you the results. It may take a minute more but you'll get your answer.

Now if you are solving it for subjective purposes, you need to first learn all the basic substitutions in specific cases. You need to memorize them literally. Unless you are fully thorough with them you cant solve a single sum. Don't get worried if you cant solve some of them. There are sums which even the best teachers have to think hard before solving. And of course you need to go through the examples given in various books. They help a lot.

Last but not the least. Practice yourself.

Hope that was useful.

Area= Area1 + Area2 -Area3

• Area1 =∫ x dx = [(x^2)/2] from x=0 to x=b
• Area1 = (b^2)/2 -0 =(b^2)/2
• Area2 = ∫ b dx =[ bx ] from x=b to x=b+d
• Area2 = b*(b+d)-b*b = bd
• Area3 = ∫ x-d dx =[(x^2)/2 -dx] from x=d to x=b+d
• Area3 = [((b+d)^2)/2 -d(b+d)]-( (d^2)/2 -d*d)
• Area3 = [((b^2+2bd+d^2)/2 -(db+d^2)]-( (d^2)/2 -d^2)
• Area3 = [((b^2+d^2)/2 -(d^2)]-(-(d^2)/2)
• Area3 = [((b^2-d^2)/2]+((d^2)/2)
• Area3 = (b^2)/2
• Area= Area1 + Area2 -Area3
• Area=(b^2)/2 + bd -(b^2)/2
• Area= bd

This is sort of a broad question, but it really depends on the problem. For future references, you can look at this website Calculus III - Line Integrals

First of all, we need to look at what a line integral. They are of multiple types, so I’ll treat them one by one.

The first one I will explain is a line integral with respect to arc length. Well, this integral, just like the Riemann integral represents area. What sort of area? Well imagine you have a string(that’s your curve). And somewhere, let’s say above it, you have the surface of a function f(x,y)(sort of a bowl). What you do is project each point from the curve to the bowl. Now you have a curtain. Well, on this curtain you have small arc elements, dS. It starts to smell a bit of Riemann. You now have a nice 2D surface. How do you translate this into a Riemann integral? Well, you take the curtain which is sort of bunched up and you stretch it until it becomes as smooth as possible. Now you have a basic Riemann integral that you can solve by calculating the area under the curve.

The second line integral is the integral of a vector field. These integrals are giving you the work done by a particle travelling across that line, through that vector field. Well, the most straightforward way to calculate these ones is by parametrizing the curve(Line Integrals of Vector Fields) and then apply the formula that I have provided through that website(I still have to learn how to type in Latex, sorry). Now, you sometimes might have some ugly curves or vector fields? What do you do? Well, here comes into play mister Stokes(or his 2 friends, Riemann and Green). What does Stokes say? He says that if you want to find the circulation of a vector field across a loop, you can calculate the integral of the rotor of the vector field, on a surface that sits on the given loop that you have. Notice that I said loop: this means only closed curves(also no self-intersections). Well, this is pretty straightforward to calculate so I’ll just explain some intuition behind this theorem. Imagine that you have a closed loop and you chop up into small pieces the space enclosed by it. Well, now think about the microscopic circulation inside each small square formed by the cuts. What Stokes says is that the sum of those microscopic circulations is equal to the circulation across the loop itself. Of course that you can choose any surface of your liking, it will be the same answer.

Hope that this helped!

Since definite (Riemann) integrals are effectively a limiting Riemann sum, they can always be interpreted as the area under some function (in the univariate case).

If you look up disc integration, you will see the univariate definite integral in that case is better interpreted as representing the volume of a solid of revolution.

If you look up the definition of the expected value of an absolutely continuous random variable, it can help to simply think about the integral as a continuous version of a weighted average.

More generally, it often suffices to simply think about any definite integral as a “continuous” sum (without invoking any pictures of area), where the thing you are summing depends on your application of interest.