the graph of the function g on the closed interval 0, 9 consists of four line segments, as shown above. let…

Explanation:

Step1: Recall integral - area relationship

The definite integral $f(x)=\int_{4}^{x}g(t)dt$ represents the net - signed area between the graph of $y = g(t)$ and the $t$ - axis from $t = 4$ to $t=x$.

Step2: Divide the interval $[4,8]$ into sub - intervals

We divide the interval $[4,8]$ into two sub - intervals: $[4,6]$ and $[6,8]$.

Step3: Calculate the area on $[4,6]$

On the interval $[4,6]$, the function $g(t)$ forms a right - triangle with base $b = 2$ and height $h=4$. The area of a triangle is $A=\frac{1}{2}bh$. So, the area of the triangle on $[4,6]$ is $A_1=\frac{1}{2}(2)(4)=4$.

Step4: Calculate the area on $[6,8]$

On the interval $[6,8]$, the function $g(t)$ forms a right - triangle with base $b = 2$ and height $h = 3$. The area of this triangle is $A_2=\frac{1}{2}(2)(3)=3$.

Step5: Calculate $f(8)$

$f(8)=\int_{4}^{8}g(t)dt$. The net - signed area from $t = 4$ to $t = 8$ is the sum of the areas of the two non - overlapping triangles. So, $f(8)=4 + 3=7$. But we made a mistake above.

Let's use the correct approach: The integral $f(8)=\int_{4}^{8}g(t)dt$ is the sum of the areas of the geometric shapes formed by the graph of $y = g(t)$ from $t = 4$ to $t = 8$. The region from $t = 4$ to $t=6$ is a triangle with base $b = 2$ and height $h = 4$, and the region from $t = 6$ to $t = 8$ is a trapezoid. The area of the triangle from $t = 4$ to $t = 6$ is $A_1=\frac{1}{2}(2)(4)=4$. The trapezoid from $t = 6$ to $t = 8$ has bases $b_1 = 4$ and $b_2=1$ and height $h = 2$. The area of a trapezoid is $A=\frac{(b_1 + b_2)h}{2}=\frac{(4 + 1)\times2}{2}=5$. So, $f(8)=\int_{4}^{8}g(t)dt=4+5 = 10$.

Answer:

C. 10