Calculate $\int_{1}^{4}\int_{0}^{2}e^{x^2}\sin[(y-1)^3]\,dy\,dx$

Question

I need to calculate the integral

$$\int_{1}^{4}\int_{0}^{2}e^{x^2}\sin[(y-1)^3]\,dy\,dx$$

using calculation techniques in several variables.

Thanks.

Answer 1

You can split the integral like this $(*)$:$$\int \limits_1^4 e^{x^2}dx \int \limits_{0}^2\sin((y-1)^3) dy$$

Before we go any further with evaluating the first factor, let's have a look at the second. Since sine is an odd function, it calls for some investigation. Let $$I = \int \limits_{0}^2 \sin((y-1)^3)dy$$Since reflection doesn't affect our integral,$$I = \int \limits_0^2\sin(\color{#C00}{2+0-y}-1)^3 dy= \int \limits_0^2\sin(1-y)^3 dy = -\int \limits_0^2 \sin(y-1)^3 dy=-I$$Since $I=-I$, it follows that $I = 0$. Therefore, we can say that the answer to the original problem is also $0$, since anything times zero is zero.

$(*): $ This is because in the inner integral, $e^{x^2}$ acts as a constant, since it does not vary with $y$. Thus$$\int\limits_1^4\int\limits_0^2e^{x^2} \sin((y-1)^3)dydx = \int_1^4e^{x^2}\color{#C00}{\int_0^2 \sin((y-1)^3)dy}dx = \int\limits_1^4\color{#C00}{I}e^{x^2}dx=I\int\limits_0^4e^{x^2}dx$$