Section 5.5
Can somebody verify this solution for me? Thanks!
Find the area of the region between the graphs of $f(x)=x^2+4x+4$ and $g(x)=4x+20$
First thing we need to do is determine the boundaries of integration, we do this by finding the values of $x$ where these functions intersect. So we need to solve when $f(x)=g(x)$, so:
$x^2+4x+4=4x+20$
$\rightarrow x^2-16=0$
$x = \pm 4$.
Cool. So now, we can see between $x=-4$ and $x=4$ that the function $g(x)$ is greater than $f(x)$. Therefore, to find the area between these graphs, we need find the total area under $f(x)$ and subtract it from the total area under $g(x)$. This will give us the area between $g(x)$ and $f(x)$.
Writing this more mathematically, we get:
\begin{align}\int_{-4}^4 (g(x)-f(x))dx&= \int_{-4}^4 4x+20-(x^2+4x+4)dx\\&= \int_{-4}^4 4x+20-x^2-4x-4dx\\&= \int_{-4}^4 -x^2 + 16dx\\&= \left.\frac{-x^3}{3} + 16x\right|_{-4}^4\\&= \left(\frac{-(4)^3}{3} + 16(4)\right) - \left(\frac{-(-4)^3}{3} + 16(-4)\right)\\&= \left(\frac{-64}{3} + 64\right) - \left(\frac{64}{3} - 64\right)\\&= \frac{-64}{3} + 64 - \frac{64}{3} +64\\&=128 - \frac{128}{3}\\&=\frac{384}{3} - \frac{128}{3}\\&=\frac{256}{3}\end{align}