Evaluate the triple integral of 4x^2*z^3*y with x from 2 to 3, y from 1 to 2, and z from 0 to 1.

Step-by-step solution

1. 1

   Notice that the integrand separates. $4x^2z^3y$ is a product $\left(4x^2\right)\cdot(y)\cdot\left(z^3\right)$ of one factor per variable, and the box has constant limits in every variable. Under those two conditions the triple integral factors into a product of three single integrals.

2. 2

   Integrate in $x$ from 2 to 3. $\displaystyle\int_2^3 4x^2\,dx=\left[\frac{4x^3}{3}\right]_2^3=\frac{4(27-8)}{3}=\frac{76}{3}$.

3. 3

   Integrate in $y$ from 1 to 2. $\displaystyle\int_1^2 y\,dy=\left[\frac{y^2}{2}\right]_1^2=\frac{4-1}{2}=\frac32$.

4. 4

   Integrate in $z$ from 0 to 1. $\displaystyle\int_0^1 z^3\,dz=\left[\frac{z^4}{4}\right]_0^1=\frac14$.

5. 5

   Multiply the three results. $\frac{76}{3}\cdot\frac32\cdot\frac14=\frac{76}{8}=\frac{19}{2}=9.5$.

6. 6

   Verify numerically. A nested composite Simpson evaluation of the same iterated integral returns $9.4999999\ldots$, agreeing with the exact $\tfrac{19}{2}$ to seven digits.