Chapter 8.2, Problem 53E

To determine

Which of the following gives the area of the region between the graphs of $y=x^2$ and $y=-x$ from $x=0$ to $x=3$

  $\begin{array}{l}\text{(A)2}\\ \text{(B)}\frac{\text{9}}{\text{2}}\\ \text{(C)}\frac{\text{13}}{\text{2}}\\ \text{(D)13}\\ \text{(E)}\frac{\text{27}}{\text{2}}\end{array}$

Answer to Problem 53E

  $\text{(E)}\frac{\text{27}}{\text{2}}$

the area of the region between the graphs of $y=x^2$ and $y=-x$ from $x=0$ to $x=3$ $=\frac{27}{2}$

Explanation of Solution

Given information: the area of the region between the graphs of $y=x^2$ and $y=-x$ from $x=0$ to $x=3$

  

In the interval $\left[0,3\right]$ , the graph only intersect at $0$ and $y=-x$ is always below $y=x^2$ in the interval $\left[0,3\right]$

Therefore, the area is

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{3}{\int }}\left[x^2+x\right]}dx\\ A={\left[\frac{x^3}{3}+\frac{x^2}{2}\right]}_0^3\\ A=9+\frac{9}{2}\\ A=\frac{27}{2}\end{array}$