Question

the graph of ( f ) is shown.
evaluate each integral by interpreting it in terms of areas.
(a) ( int_{0}^{6} f(x) d x )
(b) ( int_{0}^{15} f(x) d x )
(c) ( int_{15}^{21} f(x) d x )
(d) ( int_{9}^{21} f(x) d x )
(e) ( int_{9}^{21}|f(x)| d x )
(f) ( int_{6}^{0} f(x) d x )

Explanation:

Step1: Calculate \(\int_{0}^{6}f(x)dx\)

The region from \(x = 0\) to \(x=6\) is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a = 5\), \(b=8\) and \(h = 6\).

$$A=\frac{(5 + 8)\times6}{2}=39$$

Step2: Calculate \(\int_{0}^{15}f(x)dx\)

The region from \(x = 0\) to \(x = 6\) is a trapezoid (\(A_1=39\)), from \(x = 6\) to \(x = 12\) is a rectangle (\(A_2=8\times6 = 48\)), from \(x=12\) to \(x = 15\) is a triangle. The base of the triangle \(b = 3\) and height \(h=8\). The area of the triangle \(A_3=\frac{1}{2}\times3\times8=12\)

$$A=A_1+A_2+A_3=39 + 48+12=99$$

Step3: Calculate \(\int_{15}^{21}f(x)dx\)

The region from \(x = 15\) to \(x = 21\) is a triangle. The base \(b=6\) and height \(h=- 4\) (below the \(x\)-axis). The area \(A=\frac{1}{2}\times6\times(-4)=- 12\)

Step4: Calculate \(\int_{9}^{21}f(x)dx\)

From \(x = 9\) to \(x = 12\): rectangle with \(A_1=8\times3 = 24\). From \(x=12\) to \(x = 15\): triangle with \(A_2=\frac{1}{2}\times3\times8 = 12\). From \(x = 15\) to \(x = 21\): triangle with \(A_3=\frac{1}{2}\times6\times(-4)=-12\)

$$A=24 + 12-12=24$$

Step5: Calculate \(\int_{9}^{21}|f(x)|dx\)

From \(x = 9\) to \(x = 12\): \(A_1=8\times3 = 24\). From \(x=12\) to \(x = 15\): \(A_2=\frac{1}{2}\times3\times8 = 12\). From \(x = 15\) to \(x = 21\): \(\frac{1}{2}\times6\times4 = 12\)

$$A=24+12 + 12=48$$

Step6: Calculate \(\int_{6}^{0}f(x)dx\)

Using the property \(\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx\). Since \(\int_{0}^{6}f(x)dx = 39\), then \(\int_{6}^{0}f(x)dx=-39\)

Answer:

(a) \(39\)
(b) \(99\)
(c) \(-12\)
(d) \(24\)
(e) \(48\)
(f) \(-39\)