evaluate the definite integral. use a graphing utility to verify your result.\n int_{0}^{10} left|x^{2}-8x +…

Explanation:

Step1: Find the roots of $x^2 - 8x + 7$

Set $x^2 - 8x + 7=0$, factor it as $(x - 1)(x - 7)=0$. The roots are $x = 1$ and $x = 7$.

Step2: Analyze the sign of $y=x^2 - 8x + 7$

For $x\in[0,1]$, let's take a test - point, say $x = 0$, then $y(0)=7>0$. For $x\in(1,7)$, take $x = 4$, then $y(4)=16-32 + 7=-9<0$. For $x\in[7,10]$, take $x = 8$, then $y(8)=64 - 64+7 = 7>0$. So, $|x^2 - 8x + 7|=\begin{cases}x^2 - 8x + 7, &x\in[0,1]\cup[7,10]\-(x^2 - 8x + 7),&x\in(1,7)\end{cases}$

Step3: Split the integral

$\int_{0}^{10}|x^2 - 8x + 7|dx=\int_{0}^{1}(x^2 - 8x + 7)dx-\int_{1}^{7}(x^2 - 8x + 7)dx+\int_{7}^{10}(x^2 - 8x + 7)dx$

Step4: Integrate term - by - term

The antiderivative of $x^2 - 8x + 7$ is $\frac{1}{3}x^3-4x^2 + 7x$. $\int_{0}^{1}(x^2 - 8x + 7)dx=\left[\frac{1}{3}x^3-4x^2 + 7x\right]{0}^{1}=\frac{1}{3}-4 + 7=\frac{1 - 12+21}{3}=\frac{10}{3}$ $-\int{1}^{7}(x^2 - 8x + 7)dx=-\left[\frac{1}{3}x^3-4x^2 + 7x\right]{1}^{7}=-\left(\frac{7^3}{3}-4\times7^2 + 7\times7-\left(\frac{1}{3}-4 + 7\right)\right)$ $=-\left(\frac{343}{3}-196 + 49-\frac{10}{3}\right)=-\left(\frac{343 - 10}{3}-147\right)=-\left(\frac{333}{3}-147\right)=-(111 - 147)=36$ $\int{7}^{10}(x^2 - 8x + 7)dx=\left[\frac{1}{3}x^3-4x^2 + 7x\right]{7}^{10}=\left(\frac{10^3}{3}-4\times10^2 + 7\times10\right)-\left(\frac{7^3}{3}-4\times7^2 + 7\times7\right)$ $=\left(\frac{1000}{3}-400 + 70\right)-\left(\frac{343}{3}-196 + 49\right)=\frac{1000 - 1200+210}{3}-\frac{343 - 588 + 147}{3}$ $=\frac{1000 - 1200 + 210-(343 - 588 + 147)}{3}=\frac{1010 - 1200-(490 - 588)}{3}=\frac{-190 + 98}{3}=\frac{-92}{3}$ $\int{0}^{10}|x^2 - 8x + 7|dx=\frac{10}{3}+36+\frac{92}{3}=\frac{10 + 108+92}{3}=\frac{210}{3}=70$

Answer:

$70$