Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 36

Find the integral $\displaystyle \int^2_{-2} f(x) dx$ where $\displaystyle f(x) = \left \{ \begin{array}{c} 2 & \text{if } & -2 \leq x \leq 0 \\ 4 - x^2 & \text{if } & 0 < x \leq 2 \end{array} \right.$

Using Properties of Integral


$\begin{equation} \begin{aligned} & \int^c_a f(x) dx + \int^b_c f(x) dx = \int^b_a f(x) dx \\ \\ & \int^2_{-2} f(x) dx = \int^0_{-2} f(x) dx + \int^2_0 f(x) dx \\ \\ & \int^2_{-2} f(x) dx = \int^0_{-2} 2 dx + \int^2_0 (4 - x^2) dx \end{aligned} \end{equation}$


Using 2nd Fundamental Theorem of Calculus

$\displaystyle \int^b_a f(x) dx = F(b) - F(a)$, where $F$ is any anti-derivative of $f$.

Evaluating the integral $\displaystyle \int^0_{-2} 2dx,$

Let $\displaystyle f(x) = 2$, then


$\begin{equation} \begin{aligned} & F(x) = 2x + C \\ \\ & \int^0_{-2} 2dx = F(0) - F(-2) \\ \\ & \int^0_{-2} 2dx = 2(0) + C - [2(-2) + C] \\ \\ & \int^0_{-2} 2dx = 0 + C + 4 - C \\ \\ & \int^0_{-2} 2dx = 4 \end{aligned} \end{equation}$


Evaluating the integral $\displaystyle \int^2_0 (4 - x^2) dx$,

Let $f(x) = 4 - x^2$, then


$\begin{equation} \begin{aligned} F(x) =& 4x - \left( \frac{x^{2 + 1}}{2 + 1} \right) + C \\ \\ F(x) =& 4x - \frac{x^3}{3} + C \end{aligned} \end{equation}$




$\begin{equation} \begin{aligned} \int^2_0 (4 - x^2) dx =& F(2) - F(0) \\ \\ \int^2_0 (4 - x^2) dx =& 4(2) - \frac{(2)^3}{3} + C - \left[ 4(0) - \frac{0)^3}{3} + C \right] \\ \\ \int^2_0 (4 - x^2) dx =& 8 - \frac{8}{3} + C - 0 + 0 - C \\ \\ \int^2_0 (4 - x^2) dx =& \frac{24 - 8}{3} \\ \\ \int^2_0 (4 - x^2) dx =& \frac{16}{3} \end{aligned} \end{equation}$


Solving for $\displaystyle \int^2_{-2} f(x) dx$


$\begin{equation} \begin{aligned} \int^2_{-2} f(x) dx =& 4 + \frac{16}{3} \\ \\ \int^2_{-2} f(x) dx =& \frac{12 + 16}{3} \\ \\ \int^2_{-2} f(x) dx =& \frac{28}{3} \end{aligned} \end{equation}$