Question

(7) 9. for the function f graphed below evaluate the definite integrals
a) $$ \int _ { - 5 } ^ { 0 } f ( x ) d x = $$
b) $$ \int _ { 0 } ^ { 4 } f ( x ) d x = $$
c) $$ \int _ { - 6 } ^ { 6 } f ( x ) d x = $$

Explanation:

Step1: Calculate \(\int_{-5}^{0}f(x)dx\)

The area from \(x = - 5\) to \(x=-3\) is a rectangle with height \(h = 3\) and width \(w=2\). The area \(A_1=3\times2 = 6\).
The area from \(x=-3\) to \(x = 0\) is a triangle with base \(b = 3\) and height \(h = 3\). The area \(A_2=\frac{1}{2}\times3\times3=\frac{9}{2}\).
Since the function is above the \(x\) - axis in this interval, \(\int_{-5}^{0}f(x)dx=A_1 + A_2\).

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Step2: Calculate \(\int_{0}^{4}f(x)dx\)

The area from \(x = 0\) to \(x=4\) is a triangle. The base \(b = 4\) and the height \(h=- 2\) (below the \(x\) - axis). The area of the triangle \(A=\frac{1}{2}\times4\times2 = 4\). Since the function is below the \(x\) - axis in this interval, \(\int_{0}^{4}f(x)dx=-4\)

Step3: Calculate \(\int_{-6}^{6}f(x)dx\)

The area from \(x=-6\) to \(x=-5\) is a rectangle with height \(h = 3\) and width \(w = 1\), so \(A_3=3\times1=3\).
We already know \(\int_{-5}^{0}f(x)dx = 10.5\) and \(\int_{0}^{4}f(x)dx=-4\).
The area from \(x = 4\) to \(x=6\) is a triangle with base \(b = 2\) and height \(h = 1\), so \(A_4=\frac{1}{2}\times2\times1 = 1\)

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Answer:

a) \(\frac{21}{2}\)
b) \(-4\)
c) \(\frac{21}{2}\)