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) dx = )
b. ( int_{0}^{6} g(x) dx = )
c. ( int_{6}^{0} g(x) dx = )
d. ( int_{-1}^{6} g(x) dx = )

Explanation:

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

The formula for the area of a triangle is \(A=\frac{1}{2}bh\). For the region from \(x = - 1\) to \(x=0\), the base \(b = 1\) and the height \(h = 1\). 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 trapezoid and triangle for \(\int_{0}^{6}g(x)dx\)

First, for the region from \(x = 0\) to \(x = 2\): it is a triangle with base \(b = 2\) and height \(h=- 4\) (below the \(x\) - axis). The area \(A_1=\frac{1}{2}(2)(4)=4\), and since it is below the \(x\) - axis, its contribution to the integral is \(-4\).
For the region from \(x = 2\) to \(x = 6\): it is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a=-4\), \(b = 0\), and \(h = 4\). \(A_2=\frac{(-4+0)\times4}{2}=- 8\). But wait, we can also calculate it as two parts: a triangle and a rectangle. Alternatively, using the integral property.
The integral \(\int_{0}^{6}g(x)dx\):
The area of the triangle from \(x = 0\) to \(x = 2\) (negative) and the area of the trapezoid (or decomposed shapes) from \(x=2\) to \(x = 6\).
The area of the part from \(x = 2\) to \(x = 6\): The base of the trapezoid (or equivalent shapes) gives us. The integral \(\int_{0}^{6}g(x)dx=-4+\frac{1}{2}(4)(4 - 0)=-4 + 8=4\)

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 \(\int_{-1}^{6}g(x)dx=\int_{-1}^{0}g(x)dx+\int_{0}^{6}g(x)dx\). Substituting the values we found: \(\frac{1}{2}+4=\frac{1 + 8}{2}=\frac{9}{2}\)

Answer:

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