question find the area of the shaded region below. do not round, enter an exact answer. f(x)=x^2 g(x)= - 3x…

Explanation:

Step1: Find intersection points

Set $f(x)=x^{2}$ and $g(x)= - 3x + 4$ equal to each other: $x^{2}=-3x + 4$. Rearrange to $x^{2}+3x - 4=0$. Factor: $(x + 4)(x - 1)=0$. So $x=-4$ or $x = 1$.

Step2: Determine which function is on top

For $x\in[-4,1]$, we can test a value, say $x = 0$. $f(0)=0$ and $g(0)=4$, so $g(x)\geq f(x)$ on $[-4,1]$.

Step3: Use integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}(g(x)-f(x))dx$. Here $a=-4$, $b = 1$, $g(x)=-3x + 4$ and $f(x)=x^{2}$. So $A=\int_{-4}^{1}((-3x + 4)-x^{2})dx=\int_{-4}^{1}(-x^{2}-3x + 4)dx$.

Step4: Integrate term - by - term

$\int(-x^{2}-3x + 4)dx=-\frac{1}{3}x^{3}-\frac{3}{2}x^{2}+4x+C$.

Step5: Evaluate the definite integral

$A=\left(-\frac{1}{3}x^{3}-\frac{3}{2}x^{2}+4x\right)\big|_{-4}^{1}$. $A=\left(-\frac{1}{3}(1)^{3}-\frac{3}{2}(1)^{2}+4(1)\right)-\left(-\frac{1}{3}(-4)^{3}-\frac{3}{2}(-4)^{2}+4(-4)\right)$. $A=\left(-\frac{1}{3}-\frac{3}{2}+4\right)-\left(\frac{64}{3}-24 - 16\right)$. $A=\left(-\frac{2}{6}-\frac{9}{6}+\frac{24}{6}\right)-\left(\frac{128}{6}-\frac{144}{6}-\frac{96}{6}\right)$. $A=\frac{-2 - 9+24}{6}-\frac{128 - 144 - 96}{6}$. $A=\frac{13}{6}-\frac{-112}{6}=\frac{13 + 112}{6}=\frac{125}{6}$.

Answer:

$\frac{125}{6}$