Question

The value of \( \int_1^3 \int_2^5 x^2 y \, dx \, dy \) is ________(answer in integer).

Hint:

For double integrals, first evaluate the inner integral and then the outer integral. Always simplify the integrand first when possible.

Answer 1

Step 1: Evaluate the inner integral. The inner integral is with respect to \( x \), so we compute: \[ \int_2^5 x^2 y \, dx. \] Since \( y \) is a constant with respect to \( x \), we can factor it out: \[ y \int_2^5 x^2 \, dx. \] Now, integrate \( x^2 \): \[ \int x^2 \, dx = \frac{x^3}{3}. \] Evaluating from 2 to 5: \[ \left[\frac{x^3}{3}\right]_2^5 = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39. \] Thus, the inner integral becomes: \[ y \times 39 = 39y. \] Step 2: Evaluate the outer integral. Now we evaluate the outer integral: \[ \int_1^3 39y \, dy. \] Integrating \( 39y \): \[ \int 39y \, dy = \frac{39y^2}{2}. \] Evaluating from 1 to 3: \[ \left[\frac{39y^2}{2}\right]_1^3 = \frac{39 \times 3^2}{2} - \frac{39 \times 1^2}{2} = \frac{39 \times 9}{2} - \frac{39 \times 1}{2} = \frac{351}{2} - \frac{39}{2} = \frac{312}{2} = 156. \] Thus, the value of the integral is: \[ \boxed{156}. \]