Question

evaluate the integral.
$$ int _ { - 4 } ^ { 4 } f ( x ) d x $$ where

$$ f ( x ) = left{ \begin{array} { l l } { 4 } & { \text { if } - 4 leq x leq 0 } \\ { 16 - x ^ { 2 } } & { \text { if } quad 0 < x leq 4 } end{array} ight. $$

Explanation:

Step1: Split the integral

According to the property \(\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx\) (\(a < c< b\)), we split \(\int_{-4}^{4}f(x)dx\) into \(\int_{-4}^{0}f(x)dx+\int_{0}^{4}f(x)dx\).
Since \(f(x) = 4\) for \(-4\leq x\leq0\) and \(f(x)=16 - x^{2}\) for \(0 < x\leq4\), then \(\int_{-4}^{4}f(x)dx=\int_{-4}^{0}4dx+\int_{0}^{4}(16 - x^{2})dx\).

Step2: Integrate term - by - term

• For \(\int_{-4}^{0}4dx\):

Using the integral formula \(\int kdx=kx + C\) (\(k\) is a constant), we have \(\int_{-4}^{0}4dx=4x\big|_{-4}^0\).
Substitute the upper and lower limits: \(4\times0-4\times(-4)=16\).

• For \(\int_{0}^{4}(16 - x^{2})dx\):

Using the integral formula \(\int(x^{n})dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \(\int kdx=kx + C\), we get \(\int_{0}^{4}(16 - x^{2})dx=\int_{0}^{4}16dx-\int_{0}^{4}x^{2}dx\).
\(\int_{0}^{4}16dx=16x\big|_{0}^4=16\times4-16\times0 = 64\), \(\int_{0}^{4}x^{2}dx=\frac{x^{3}}{3}\big|_{0}^4=\frac{4^{3}}{3}-\frac{0^{3}}{3}=\frac{64}{3}\).
So \(\int_{0}^{4}(16 - x^{2})dx=64-\frac{64}{3}=\frac{192 - 64}{3}=\frac{128}{3}\).

Step3: Sum the results

\(\int_{-4}^{4}f(x)dx = 16+\frac{128}{3}=\frac{48 + 128}{3}=\frac{176}{3}\).

Answer:

\(\frac{176}{3}\)