current objective find the area of a region bounded above by two different functions question consider the…

Explanation:

Step1: Determine the upper - function

We need to find which of $f(x)=-x^{2}-4x + 5$ and $g(x)=2x + 10$ is the upper - function on the interval $[-5,1]$. We can find the intersection of $f(x)$ and $g(x)$ by setting $f(x)=g(x)$: $-x^{2}-4x + 5=2x + 10$ $x^{2}+6x + 5 = 0$ $(x + 1)(x+5)=0$ The intersection points are $x=-5$ and $x=-1$. We can test a value in the sub - intervals. Let's test $x = - 2$: $f(-2)=-(-2)^{2}-4(-2)+5=-4 + 8 + 5=9$ $g(-2)=2(-2)+10=6$ On the interval $[-5,-1]$, $f(x)\geq g(x)$, and on the interval $[-1,1]$, $g(x)\geq f(x)$.

Step2: Set up the integral for the area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ and the $x$ - axis on the interval $[a,b]$ is given by: $A=\int_{-5}^{-1}f(x)dx+\int_{-1}^{1}g(x)dx$ $\int_{-5}^{-1}(-x^{2}-4x + 5)dx=\left[-\frac{1}{3}x^{3}-2x^{2}+5x\right]{-5}^{-1}$ $=\left(-\frac{1}{3}(-1)^{3}-2(-1)^{2}+5(-1)\right)-\left(-\frac{1}{3}(-5)^{3}-2(-5)^{2}+5(-5)\right)$ $=\left(\frac{1}{3}-2 - 5\right)-\left(\frac{125}{3}-50 - 25\right)$ $=\left(\frac{1 - 6 - 15}{3}\right)-\left(\frac{125-150 - 75}{3}\right)$ $=\frac{-20}{3}-\frac{-100}{3}=\frac{-20 + 100}{3}=\frac{80}{3}$ $\int{-1}^{1}(2x + 10)dx=\left[x^{2}+10x\right]_{-1}^{1}$ $=(1^{2}+10\times1)-((-1)^{2}+10\times(-1))$ $=(1 + 10)-(1 - 10)$ $=11-(-9)=20$

Step3: Calculate the total area

$A=\frac{80}{3}+20=\frac{80 + 60}{3}=\frac{140}{3}$

Answer:

$\frac{140}{3}$