GATE CS 2023 Q.31: Calculus

The value of the definite integral

\int_{-3}^3 \int_{-2}^2 \int_{-1}^1 (4x^2 y - z^3) \, dz \, dy \, dx

is _______. (Rounded off to the nearest integer)

Your answer

Official answer and solution

Solution

Use the property of definite integrals of odd functions over symmetric intervals $[-a, a]$ .

The integral is:

I = \int_{-3}^3 \int_{-2}^2 \int_{-1}^1 (4x^2 y - z^3) \, dz \, dy \, dx

Split the integrand into two terms:

For the first term, $\int_{-3}^3 4x^2 \, dx \cdot \int_{-2}^2 y \, dy \cdot \int_{-1}^1 1 \, dz$ :
Since $y$ is an odd function and the interval $[-2, 2]$ is symmetric about the origin, $\int_{-2}^2 y \, dy = 0$ . Hence the first term is $0$ .
For the second term, $-\int_{-3}^3 1 \, dx \cdot \int_{-2}^2 1 \, dy \cdot \int_{-1}^1 z^3 \, dz$ :
Since $z^3$ is an odd function and the interval $[-1, 1]$ is symmetric about the origin, $\int_{-1}^1 z^3 \, dz = 0$ . Hence the second term is $0$ .

Therefore, $I = 0 - 0 = 0$ .

Final answer

Common trap

Performing tedious nested polynomial integration by hand instead of exploiting odd-function symmetry over symmetric intervals.

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