Question

5.1/5.2 approximating areas & the definite integral

1. ( y = g ( x ) ) is plotted above. evaluate the following definite integrals.

a. ( int _ { - 1 } ^ { 0 } g ( x ) d x = )
b. ( int _ { 0 } ^ { 6 } g ( x ) d x = )
c. ( int _ { 6 } ^ { 0 } g ( x ) d x = )
d. ( int _ { - 1 } ^ { 6 } g ( x ) d x = )

Explanation:

Step1: Calculate the area of the triangle for $\int_{-1}^{0}g(x)dx$

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

Step2: Calculate the area of the two triangles for \(\int_{0}^{6}g(x)dx\)

For the interval from \(x = 0\) to \(x = 2\), the base \(b_1=2\) and height \(h_1 = 4\). The area of this triangle \(A_1=\frac{1}{2}(2)(4)=4\) (but since the function is below the \(x\) - axis, \(A_1=- 4\)). For the interval from \(x = 2\) to \(x = 6\), the base \(b_2 = 4\) and height \(h_2=4\). The area of this triangle \(A_2=\frac{1}{2}(4)(4)=8\). Then \(\int_{0}^{6}g(x)dx=-4 + 8=4\)

Step3: Use the property \(\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx\) for \(\int_{6}^{0}g(x)dx\)

By the property \(\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx\), so \(\int_{6}^{0}g(x)dx=-\int_{0}^{6}g(x)dx=-4\)

Step4: Use the property \(\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx\) for \(\int_{-1}^{6}g(x)dx\)

We know \(\int_{-1}^{6}g(x)dx=\int_{-1}^{0}g(x)dx+\int_{0}^{6}g(x)dx\). Substitute \(\int_{-1}^{0}g(x)dx=\frac{1}{2}\) and \(\int_{0}^{6}g(x)dx = 4\), then \(\int_{-1}^{6}g(x)dx=\frac{1}{2}+4=\frac{1 + 8}{2}=\frac{9}{2}\)

Answer:

A. \(\frac{1}{2}\)
B. \(4\)
C. \(-4\)
D. \(\frac{9}{2}\)