find the area of the shaded region on the right.

Explanation:

Step1: Find the intersection points

Set $5x^{2}=40 - 10x$. Rearrange to $5x^{2}+10x - 40=0$, then $x^{2}+2x - 8=0$. Factor: $(x + 4)(x - 2)=0$. So $x=-4,2$. We are interested in $x\in[0,4]$.

Step2: Use the integral formula for area between curves

The area $A=\int_{0}^{2}((40 - 10x)-5x^{2})dx+\int_{2}^{4}(5x^{2}-(40 - 10x))dx$.

First integral:

$\int_{0}^{2}(40 - 10x-5x^{2})dx=\left[40x-5x^{2}-\frac{5}{3}x^{3}\right]_{0}^{2}=40\times2-5\times2^{2}-\frac{5}{3}\times2^{3}=80 - 20-\frac{40}{3}=\frac{240 - 60 - 40}{3}=\frac{140}{3}$.

Second integral:

$\int_{2}^{4}(5x^{2}+10x - 40)dx=\left[\frac{5}{3}x^{3}+5x^{2}-40x\right]_{2}^{4}=\left(\frac{5}{3}\times4^{3}+5\times4^{2}-40\times4\right)-\left(\frac{5}{3}\times2^{3}+5\times2^{2}-40\times2\right)$ $=\left(\frac{320}{3}+80 - 160\right)-\left(\frac{40}{3}+20 - 80\right)=\left(\frac{320}{3}-80\right)-\left(\frac{40}{3}-60\right)=\frac{320 - 240}{3}-\frac{40 - 180}{3}=\frac{80}{3}+\frac{140}{3}=\frac{220}{3}$.

Step3: Sum the two - integral results

$A=\frac{140}{3}+\frac{220}{3}=\frac{360}{3}=120$.

Answer:

$120$