the graph of g is shown below. evaluate the definite integral of ∫ - 9 9 g(x)dx.

Explanation:

Step1: Divide the integral into sub - integrals

$\int_{-9}^{9}g(x)dx=\int_{-9}^{-4}g(x)dx+\int_{-4}^{0}g(x)dx+\int_{0}^{4}g(x)dx+\int_{4}^{9}g(x)dx$

Step2: Calculate area of each part using geometric shapes

• For $\int_{-9}^{-4}g(x)dx$: The region is a trapezoid with bases $b_1 = 2$ and $b_2=4$ and height $h = 5$. The area $A_1=\frac{(2 + 4)\times5}{2}=15$.
• For $\int_{-4}^{0}g(x)dx$: The region is a triangle with base $b = 4$ and height $h = 4$. The area $A_2=\frac{4\times4}{2}=8$.
• For $\int_{0}^{4}g(x)dx$: The region is a triangle with base $b = 4$ and height $h = 4$. The area $A_3=-\frac{4\times4}{2}=- 8$ (negative since below x - axis).
• For $\int_{4}^{9}g(x)dx$: The region is a trapezoid with bases $b_1 = 4$ and $b_2 = 0$ and height $h = 5$. The area $A_4=\frac{(4+0)\times5}{2}=10$.

Step3: Sum up the areas of sub - integrals

$\int_{-9}^{9}g(x)dx=A_1 + A_2+A_3+A_4=15 + 8-8 + 10=25$

Answer:

$25$