Q. 25 - Calculus

Question

Evaluate the iterated integral :

${\int }_0^1 {\int }_2^4 {\int }_{-1}^5 \left(x+y{z}^2\right)dxdydz$

Step-by-Step Solution

Answer

36

Step 1. Given information.

Given integral is :

${\int }_0^1 {\int }_2^4 {\int }_{-1}^5 \left(x+y{z}^2\right)dxdydz$

We have to evaluate it.

Step 2. Evaluate.

$\begin{array}{r}={\int }_0^1 {\int }_2^4 {\int }_{-1}^5 \left(x+y{z}^{2}dx\right)dydz\text{l}\\ ={\int }_0^1 {\int }_2^4 \left({\int }_{-1}^5 xdx+{\int }_{-1}^5 y{z}^{2}dx\right)dydz\\ ={\int }_0^1 {\int }_2^4 \left({\left(\frac{x^2}{2}\right)}_{-1}^5+y{z}^2(x{)}_{-1}^5\right)dydz\\ ={\int }_0^1 {\int }_2^4 \left(\left(\frac{5^2}{2}-\frac{(-1{)}^2}{2}\right)+y{z}^2(5-(-1\left)\right)dydz\right.\\ ={\int }_0^1 {\int }_2^4 \left(12+6y{z}^2\right)dydz\end{array}$

Step 3. Integrate with respect to y.

Integrate with respect to y 

$\begin{array}{r}={\int }_0^1 \left({\int }_2^4 12dy+6z^2{\int }_2^4 yd\cdot y\right)dz\\ ={\int }_0^1 \left(12(y{)}_2^4+6z^2{\left(\frac{y^2}{2}\right)}_2^4\right)dz\\ ={\int }_0^1 \left(12(4-2)+6z^2\left(\frac{4^2}{2}-\frac{2^2}{2}\right)\right)dz\\ ={\int }_0^1 \left(24+36z^2\right)dz\end{array}$

Integrate with respect to z

$\begin{array}{r}=24{\int }_0^1 dz+36{\int }_0^1 z^{2}dz\text{ Algebra of }\\ =24(z{)}_0^1+36{\left(\frac{z^3}{3}\right)}_0^1\text{ Integ }\\ =24(1-0)+36\left(\frac{1}{3}-0\right)\text{ Apply lin }\\ =24+12\\ =36\\ \therefore {\int }_0^1 {\int }_2^4 {\int }_{-1}^5 \left(x+y{z}^2\right)dxdydz=36\end{array}$