Question

the graph of f is shown in the figure. let $a(x)=int_{-2}^{x}f(t)dt$ and $f(x)=int_{4}^{x}f(t)dt$ be two area - functions for f. evaluate the following area functions: a. $a(-2)$ b. $f(8)$ c. $a(4)$ d. $f(4)$ e. $a(8)$ a. $a(-2)=0$ (simplify your answer.) b. $f(8)=18$ (simplify your answer.)

Explanation:

Step1: Recall the definition of definite - integral as area

The definite integral $\int_{a}^{x}f(t)dt$ represents the net - signed area between the curve $y = f(t)$ and the $t$ - axis from $t=a$ to $t = x$. For $A(x)=\int_{-2}^{x}f(t)dt$ and $F(x)=\int_{4}^{x}f(t)dt$.

Step2: Evaluate $A(-2)$

By the fundamental theorem of calculus, if $A(x)=\int_{-2}^{x}f(t)dt$, when $x=-2$, $A(-2)=\int_{-2}^{-2}f(t)dt = 0$.

Step3: Evaluate $F(8)$

Since $F(x)=\int_{4}^{x}f(t)dt$, when $x = 8$, we consider the area between $t = 4$ and $t = 8$. The area between $t = 4$ and $t = 8$ is the area above the $t$ - axis minus the area below the $t$ - axis. The area above the $t$ - axis from $t = 4$ to $t$ - value where the curve crosses the $t$ - axis is $18$ and the area below the $t$ - axis from that point to $t = 8$ is $9$. So $F(8)=\int_{4}^{8}f(t)dt=18 - 9=9$.

Step4: Evaluate $A(4)$

$A(4)=\int_{-2}^{4}f(t)dt$. The area from $t=-2$ to $t = 4$ is the sum of the area from $t=-2$ to $t = 0$ and from $t = 0$ to $t = 4$. The area from $t=-2$ to $t = 0$ is $7$ and from $t = 0$ to $t = 4$ is $18$. So $A(4)=7 + 18=25$.

Step5: Evaluate $F(4)$

$F(4)=\int_{4}^{4}f(t)dt = 0$.

Step6: Evaluate $A(8)$

$A(8)=\int_{-2}^{8}f(t)dt$. We know that the area from $t=-2$ to $t = 0$ is $7$, from $t = 0$ to $t = 4$ is $18$, and from $t = 4$ to $t = 8$ is $18-9 = 9$. So $A(8)=7+18 + 9=34$.

Answer:

a. $A(-2)=0$
b. $F(8)=9$
c. $A(4)=25$
d. $F(4)=0$
e. $A(8)=34$