each of the regions a, b, and c bounded by the graph of f and the x - axis has area 3. find the value of…

Explanation:

Step1: Use integral property

By the property of definite - integrals $\int_{a}^{b}[f(x)+g(x)]dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$. So, $\int_{-4}^{2}[f(x)+2x + 3]dx=\int_{-4}^{2}f(x)dx+\int_{-4}^{2}2xdx+\int_{-4}^{2}3dx$.

Step2: Calculate $\int_{-4}^{2}f(x)dx$

The area of region $A$ is below the $x$ - axis, so its contribution to the integral of $f(x)$ is $- 3$, the area of region $B$ is above the $x$ - axis, so its contribution is $3$, and the area of region $C$ is below the $x$ - axis, so its contribution is $-3$. Then $\int_{-4}^{2}f(x)dx=-3 + 3-3=-3$.

Step3: Calculate $\int_{-4}^{2}2xdx$

The antiderivative of $2x$ is $x^{2}+C$. Using the fundamental theorem of calculus $\int_{-4}^{2}2xdx=x^{2}\big|_{-4}^{2}=2^{2}-\left(-4\right)^{2}=4 - 16=-12$.

Step4: Calculate $\int_{-4}^{2}3dx$

The antiderivative of $3$ is $3x + C$. Using the fundamental theorem of calculus $\int_{-4}^{2}3dx=3x\big|_{-4}^{2}=3\times(2-( - 4))=3\times6 = 18$.

Step5: Combine the results

$\int_{-4}^{2}[f(x)+2x + 3]dx=-3-12 + 18=3$.

Answer:

$3$