Question

the following is a graph of the function $f(x)$. find $\int_{-3}^{4} f(x) d x$.
note that the portion of the graph for $-3

Explanation:

Step1: Divide the integral into three parts

$$\int_{-3}^{4}f(x)dx=\int_{-3}^{-1}f(x)dx+\int_{-1}^{0}f(x)dx+\int_{0}^{4}f(x)dx$$

Step2: Calculate $\int_{-3}^{-1}f(x)dx$ (area of trapezoid)

The formula for the area of a trapezoid is $A = \frac{(a + b)h}{2}$. Here, $a = 1$, $b = 0$, $h=2$.
$$\int_{-3}^{-1}f(x)dx=\frac{(1 + 0)\times2}{2}= 1$$

Step3: Calculate $\int_{-1}^{0}f(x)dx$ (area of quarter - circle)

The equation of the circle is $x^{2}+(y - 1)^{2}=9$ (since the center is $(0,1)$ and radius $r = 3$). For $y=f(x)$, we can rewrite it as $y=1+\sqrt{9 - x^{2}}$. The area of a quarter - circle with radius $r = 3$ is $A=\frac{1}{4}\pi r^{2}$.
$$\int_{-1}^{0}f(x)dx=\frac{1}{4}\pi\times3^{2}=\frac{9\pi}{4}$$

Step4: Calculate $\int_{0}^{4}f(x)dx$ (area of trapezoid)

Using the trapezoid area formula $A=\frac{(a + b)h}{2}$, where $a = 4$, $b = 1$, $h = 3$.
$$\int_{0}^{4}f(x)dx=\frac{(4 + 1)\times3}{2}=\frac{15}{2}$$

Step5: Sum up the three parts

$$\int_{-3}^{4}f(x)dx=1+\frac{9\pi}{4}+\frac{15}{2}=\frac{4 + 9\pi+30}{4}=\frac{34 + 9\pi}{4}$$

Answer:

$\frac{34 + 9\pi}{4}$