Question

the graph of the function g on the closed interval 0, 9 consists of four line segments, as shown above. let f be the function defined by f(x)=(int_{4}^{x}g(t)dt). what is the value of f(8)? a -1 b 2 c 10 d 14

Explanation:

Step1: Recall integral - area relationship

The value of $f(8)=\int_{4}^{8}g(t)dt$ is equal to the net - signed area between the graph of $y = g(t)$ and the $t$ - axis from $t = 4$ to $t = 8$.

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

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

Step3: Calculate area of the first sub - interval $[4,6]$

The graph of $g(t)$ on $[4,6]$ forms a right - triangle with base $b_1=2$ and height $h_1 = 4$. The area of this triangle $A_1=\frac{1}{2}\times b_1\times h_1=\frac{1}{2}\times2\times4 = 4$.

Step4: Calculate area of the second sub - interval $[6,8]$

The graph of $g(t)$ on $[6,8]$ forms a right - triangle with base $b_2 = 2$ and height $h_2=2$. The area of this triangle $A_2=\frac{1}{2}\times b_2\times h_2=\frac{1}{2}\times2\times2=2$.

Step5: Calculate the net - signed area

Since the function $g(t)$ is above the $t$ - axis on the interval $[4,8]$, the net - signed area $\int_{4}^{8}g(t)dt=A_1 + A_2$. So, $f(8)=4 + 2=6$. But we made a mistake above.
Let's calculate the area in another way.
The area under the curve from $t = 4$ to $t=6$: The line segment from $(4,0)$ to $(6,4)$ has the area of a trapezoid with bases $b_1 = 0$ and $b_2=4$ and height $h = 2$. Using the trapezoid area formula $A=\frac{(b_1 + b_2)h}{2}=\frac{(0 + 4)\times2}{2}=4$.
The area from $t = 6$ to $t = 8$: The line segment from $(6,4)$ to $(8,2)$ also forms a trapezoid with bases $b_1=2$ and $b_2 = 4$ and height $h = 2$. Using the trapezoid area formula $A=\frac{(b_1 + b_2)h}{2}=\frac{(2 + 4)\times2}{2}=6$.
The sum of the areas of the two trapezoids: $\int_{4}^{8}g(t)dt=4 + 6=10$.

Answer:

C. 10