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}
$