find the total area bounded by the x - axis and the curve y = f(x) on the indicated interval. f(x)=x^2 + x…

Explanation:

Step1: Find zeros of the function

Set $f(x)=x^{2}+x - 6 = 0$. Factor: $(x + 3)(x - 2)=0$. Zeros are $x=-3,x = 2$. Only $x=-3$ is in the interval $[-7,-1]$.

Step2: Determine sign of function in sub - intervals

Divide $[-7,-1]$ into $[-7,-3]$ and $[-3,-1]$. Test points: for $x=-5$ in $[-7,-3]$, $f(-5)=25-5 - 6=14>0$; for $x=-2$ in $[-3,-1]$, $f(-2)=4-2 - 6=-4<0$.

Step3: Calculate definite integrals

The area $A=\int_{-7}^{-3}(x^{2}+x - 6)dx-\int_{-3}^{-1}(x^{2}+x - 6)dx$. First integral: $\int(x^{2}+x - 6)dx=\frac{1}{3}x^{3}+\frac{1}{2}x^{2}-6x+C$. $\int_{-7}^{-3}(\frac{1}{3}x^{3}+\frac{1}{2}x^{2}-6x)dx=[\frac{1}{3}(-3)^{3}+\frac{1}{2}(-3)^{2}-6(-3)]-[\frac{1}{3}(-7)^{3}+\frac{1}{2}(-7)^{2}-6(-7)]$ $=(-9+\frac{9}{2}+18)-(-\frac{343}{3}+\frac{49}{2}+42)$ $=(9+\frac{9}{2})-(-\frac{343}{3}+\frac{49}{2}+42)$ $=\frac{18 + 9}{2}-(-\frac{343}{3}+\frac{49}{2}+42)$ $=\frac{27}{2}+\frac{343}{3}-\frac{49}{2}-42$ $=\frac{81 + 686-147 - 252}{6}=\frac{368}{6}=\frac{184}{3}$. Second integral: $\int_{-3}^{-1}(\frac{1}{3}x^{3}+\frac{1}{2}x^{2}-6x)dx=[\frac{1}{3}(-1)^{3}+\frac{1}{2}(-1)^{2}-6(-1)]-[\frac{1}{3}(-3)^{3}+\frac{1}{2}(-3)^{2}-6(-3)]$ $=(-\frac{1}{3}+\frac{1}{2}+6)-(-9+\frac{9}{2}+18)$ $=(-\frac{2 + 3}{6}+6)-(-9+\frac{9}{2}+18)$ $=(-\frac{1}{6}+6)-(-9+\frac{9}{2}+18)$ $=\frac{-1 + 36}{6}-(-9+\frac{9}{2}+18)$ $=\frac{35}{6}-(9+\frac{9}{2})$ $=\frac{35}{6}-\frac{18 + 9}{2}=\frac{35-81}{6}=-\frac{46}{6}=-\frac{23}{3}$. $A=\frac{184}{3}-(-\frac{23}{3})=\frac{184 + 23}{3}=\frac{207}{3}=69$.

Answer:

$69$