find the area between the curves.\nx = - 5, x = 2, y = 9x, y = x^{2}-10\n\nthe area between the curves is…

Explanation:

Step1: Determine the upper - lower curves

First, find where $9x$ and $x^{2}-10$ intersect. Set $9x=x^{2}-10$, which can be rewritten as $x^{2}-9x - 10=0$. Factoring gives $(x - 10)(x+1)=0$, so the intersection points are $x=-1$ and $x = 10$. In the interval $[-5,2]$, we need to check which function is on top. Let's test a value in the sub - intervals. For $x = 0$, $9x=0$ and $x^{2}-10=-10$, so $y = 9x$ is above $y=x^{2}-10$ in the interval $[-5,2]$.

Step2: Use the area formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $f(x)=9x$, $g(x)=x^{2}-10$, $a=-5$, and $b = 2$. So $A=\int_{-5}^{2}[(9x-(x^{2}-10))]dx=\int_{-5}^{2}(9x - x^{2}+10)dx$.

Step3: Integrate term - by - term

We know that $\int(9x - x^{2}+10)dx=\frac{9}{2}x^{2}-\frac{1}{3}x^{3}+10x+C$.

Step4: Evaluate the definite integral

$A=\left[\frac{9}{2}x^{2}-\frac{1}{3}x^{3}+10x\right]_{-5}^{2}$. First, substitute $x = 2$: $\frac{9}{2}(2)^{2}-\frac{1}{3}(2)^{3}+10(2)=\frac{9}{2}\times4-\frac{8}{3}+20=18-\frac{8}{3}+20=\frac{54 - 8+60}{3}=\frac{106}{3}$. Then substitute $x=-5$: $\frac{9}{2}(-5)^{2}-\frac{1}{3}(-5)^{3}+10(-5)=\frac{9}{2}\times25+\frac{125}{3}-50=\frac{225}{2}+\frac{125}{3}-50=\frac{675 + 250-300}{6}=\frac{625}{6}$. Now, $A=\frac{106}{3}-\left(-\frac{625}{6}\right)=\frac{212 + 625}{6}=\frac{837}{6}=\frac{279}{2}$.

Answer:

$\frac{279}{2}$