question let r be the region bounded by the functions f(x)=4x² + 2x and g(x)= -x as shown in the diagram…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $4x^{2}+2x=-x$. Rearranging gives $4x^{2}+3x = 0$. Factoring out $x$ we get $x(4x + 3)=0$. Solving for $x$ gives $x = 0$ and $x=-\frac{3}{4}$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)-g(x)=4x^{2}+2x-(-x)=4x^{2}+3x$, and $a =-\frac{3}{4}$, $b = 0$. So $A=\int_{-\frac{3}{4}}^{0}(4x^{2}+3x)dx$.

Step3: Integrate

Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(4x^{2}+3x)dx=4\times\frac{x^{3}}{3}+3\times\frac{x^{2}}{2}+C=\frac{4x^{3}}{3}+\frac{3x^{2}}{2}+C$.

Step4: Evaluate definite integral

$A=\left[\frac{4x^{3}}{3}+\frac{3x^{2}}{2}\right]_{-\frac{3}{4}}^{0}$. First, substitute $x = 0$: $\frac{4(0)^{3}}{3}+\frac{3(0)^{2}}{2}=0$. Then substitute $x=-\frac{3}{4}$: $\frac{4}{3}\times\left(-\frac{3}{4}\right)^{3}+\frac{3}{2}\times\left(-\frac{3}{4}\right)^{2}=\frac{4}{3}\times\left(-\frac{27}{64}\right)+\frac{3}{2}\times\frac{9}{16}=-\frac{9}{16}+\frac{27}{32}=\frac{-18 + 27}{32}=\frac{9}{32}=0.28125\approx0.281$.

Answer:

$0.281$