Question

the graph of function g is shown below. let $f(x) = \int_{-3}^{x} g(t) dt$. evaluate $f(1)$. $f(1) = \square$

Explanation:

Step1: Define $f(1)$ via integral

$f(1) = \int_{-3}^{1} g(t) dt$

Step2: Split integral into two intervals

$\int_{-3}^{1} g(t) dt = \int_{-3}^{0} g(t) dt + \int_{0}^{1} g(t) dt$

Step3: Calculate $\int_{-3}^{0} g(t) dt$

This is the area of a trapezoid with bases $g(-3)=3$, $g(0)=2$, width $3$.
$\int_{-3}^{0} g(t) dt = \frac{1}{2} \times (3+2) \times 3 = 7.5$

Step4: Calculate $\int_{0}^{1} g(t) dt$

This is the area of a trapezoid with bases $g(0)=3$, $g(1)=5$, width $1$.
$\int_{0}^{1} g(t) dt = \frac{1}{2} \times (3+5) \times 1 = 4$

Step5: Sum the two integrals

$7.5 + 4 = 11.5$

Answer:

$\frac{23}{2}$ or $11.5$