How to tell a definite integral is 0 by the form of integrand?

As given by bskelkar, if the function f (x) is odd, i.e., f (-x) = - f (x), then

∫ (x = -a to a) f (x) = 0.

In the given example, this is not the reason.

Integral, I = ∫ (x = 0 to 1) (1 - 2x) / [1 + cbrt(x - x^2)] dx

= ∫ (x = 0 to 1) [1 - 2(1 - x)] / [1 + cbrt(1 - x - (1 - x)^2)] dx

.........[because ∫ (x = 0 to a) f (x) dx = ∫ (x = 0 to a) f (a - x) dx]

= ∫ (x = 0 to 1) (2x - 1) / [1 + cbrt(1 - x - 1 + 2x - x^2)] dx

= - ∫ (x = 0 to 1) (1 - 2x) / [1 + cbrt(x - x^2)] dx

= - I

=> 2 I = 0

=> I = 0.

Edit:

From the above example, one can infer that

∫ (x = a to b) f (x) dx = 0

if f (a + b - x) = - f (x)

which is the case in the given example where

f (x) = (1 - 2x) / [1 + cbrt(x - x^2)]

and

f (0 + 1 - x) = - (1 - 2x) / [1 + cbrt(x - x^2)]

=> ∫ (x = 0 to 1) f (x) dx = 0.

Edit:

Alternate method hinted by Cool Dude

Let 1 - 2x = t

=> - 2dx = dt

and x - x^2

= x (1 - x)

= (1 - t)/2 * [1 - (1 - t)/2]

= (1/4) (1 - t^2)

Also, x = 0 => t = -1 and x = 1 => t = 1

=> Integral

= ∫ (t = -1 to 1) - (1/2) t dt / [(1/4)(1 - t^2)]^(1/3)

= - (1/2) / (1/4)^(1/3) ∫ (t = - 1 to t = 1) t / (1 - t^2)^(1/3) dt

As the limit of the definite integral is from - 1 to 1 (additive inverse) and the function

t / (1 - t^2)^(1/3) is odd as f (-t) = - f (t),

interal = 0.

Edit:

Oh! that is great Cool Dude.

I wonder why it did not occur to me.

Most elegant substitution, no doubt

x - x^2 = u

=> (1 - 2x) dx = du

=> integral

= ∫ du/u^(1/3) = 0, with integration limits of u from 0 to 0

= (3/2) u^(2/3)

= (3/2) (x - x^2)^(2/3) which is also zero with limits of x from 0 to 1.

Yeah, there's really no single good way of figuring out that a definite integral is equal to 0 just by looking at the integrand. As the first response suggests, an odd function (i.e., one that satisfies f(-x) = -f(x) or, equivalently, one that is rotationally symmetric about the origin) integrated from -c to c (for any constant c) yields 0. Odd functions include, say, polynomials with only odd-degree terms, sin(x) and tan(x). Try integrating some of these guys from -c to c for some constants (appropriately chosen for tan(x) so as to avoid where the function is undefined) to see this in action.

Looking at the graph of your function, you CAN make your problem look like this (integral of odd function from -c to c). You'd have to make the substitution u = x-1/2. (This is what the first response is suggesting in calling the function antisymmetric about x = 1/2.) Now why would you think do this? You probably wouldn't. So let's abandon this line of reasoning, even though it is valid.

Instead, this problem is really begging you to try using a substitution first. Not the one I suggest above, but rather one that is actually "suggested" by the integrand. Something interesting will happen when you do this: you'll end up with an integral where you can IMMEDIATELY tell that it's equal to 0 (and not by looking at the integrand). I leave you to figure out what this substitution is and why it makes it evident that the integral is 0.

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Edit: Just to give a different way of showing that the integral from -c to c of an odd function is 0, consider the following. Let f(x) be an odd function, and let F(x) be an antiderivative of f(x). That is:

(d/dx)F(x) = f(x)

Now I claim F(x) is an even function, i.e. F(-x) = F(x). Indeed:

(d/dx)F(-x) = -f(-x) [chain rule] = f(x) [f(x) is odd]

Since F(x) and F(-x) have the same derivative, they differ by a constant:

F(x) = F(-x) + C

Evaluate the above at 0:

F(0) = F(0) + C => C = 0

So F(x) = F(-x). Now the integral of f(x) from -c to c is:

F(c)-F(-c) = F(c)-F(c) = 0

Neat! This doesn't give you a nice picture like the geometric method suggested above, but it is a lot cleaner than showing the result using definite integrals and substitution. (Really, they're the same, as my use of the chain rule serves the same purpose as the use of substitution in the integral method.)

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Edit 2: Now that I look more closely at Madhukar's method, I'm mostly wrong about my suggestion that his method is like mine. First, he's discussing functions that are not necessarily odd but exhibit the same sort of symmetry about some point on the x-axis. My discussion was simply regarding odd functions. A substitution can change the general case to the odd case (basically, shifting the point of symmetry back to the origin), so with a little more discussion, my method works for what he's doing.

Second, he states:

∫(x=0 to x=a) f(x)dx = ∫(x=0 to x=a) f(a-x)dx

This is true, and I can think of a few ways to show this. One is just using geometry--the region bounded by the integrand in the 1st integral is a horizontal reflection of the region bounded by the integrand in the 2nd. So they have the same area. The other way is substitution, namely letting u = a-x in the 2nd integral. You'll get the first integral back (with u instead of x, but just change the name of the variable and you're cool).

Now what I was suggesting is that, well, substitution is in some sense the "opposite" of the chain rule. The chain rule says a derivative of a composite is the derivative of the inside times the derivative of the outside applied to the inside. Substitution lets us integrate things that look a composite function times the derivative of the inside. And you can prove one from the other. So my using the chain rule serves the same purpose as using substitution to show the fact that Madhukar states.

NOW the last thing in my original post was this: try the substitution u = x-x^2 in the original integral. What do you get?

an integral is finding the area under a curve. if the curve goes below the x axis then you are calculating a negative area. if the area under the curve above the x axis is equal to the area under the curve below the x axis then you have an area plus that same negative area, so its A + -A = 0

with an integral like that one it is very difficult to tell if the integral is zero without actually doing it.

The only case where it is easy to tell is If a function is odd, then the integral from a to -a is always zero

an odd function is one where f(-x) = - f(x), which occurs in functions such as sine and x^3, where their is point symmetry about the origin

another slightly related point is with even functions, where f(-x) = f(x), such as in x^2 and cosine, the integral from -a to a is twice the integral from 0 to a

Integral of and odd function from -a to a is always zero. In this case the function is anti-symmetric about x = 1/2.

when limits are equal & of opposite sign i. e.

if limits are like a to b where b = -a

e. g limits are pi to -pi etc