Question

1. use geometry to evaluate the following definite integrals.

a. $$ \int _ { - 3 } ^ { 5 } ( 2 - x ) d x $$
b. $$ \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } d x $$

Explanation:

Part A

Step1: Find the intersection points of \(y = 2 - x\) with the \(x -\)axis

Set \(y=0\), then \(2 - x=0\), so \(x = 2\).

Step2: Divide the integral \(\int_{-3}^{5}(2 - x)dx\) into two parts

\(\int_{-3}^{5}(2 - x)dx=\int_{-3}^{2}(2 - x)dx+\int_{2}^{5}(2 - x)dx\)
The function \(y = 2 - x\) is a straight - line.
For \(y = 2 - x\), when \(x=-3\), \(y=2-(-3)=5\); when \(x = 2\), \(y = 0\); when \(x = 5\), \(y=2 - 5=-3\)
The area of the first triangle (from \(x=-3\) to \(x = 2\)):
The base \(b_1=2-(-3)=5\) and the height \(h_1 = 5\). The area \(A_1=\frac{1}{2}\times b_1\times h_1=\frac{1}{2}\times5\times5=\frac{25}{2}\)
The area of the second triangle (from \(x = 2\) to \(x = 5\)):
The base \(b_2=5 - 2=3\) and the height \(h_2=3\). The area \(A_2=\frac{1}{2}\times b_2\times h_2=\frac{1}{2}\times3\times3=\frac{9}{2}\)
Since the function \(y = 2 - x\) is above the \(x -\)axis for \(x\in[-3,2]\) and below the \(x -\)axis for \(x\in[2,5]\)
\(\int_{-3}^{5}(2 - x)dx=A_1-A_2\)
\(\int_{-3}^{5}(2 - x)dx=\frac{25}{2}-\frac{9}{2}=8\)

Part B

Step1: Recognize the geometric shape

The function \(y=\sqrt{1 - x^{2}}\) (for \(y\geq0\)) is the upper - half of the circle \(x^{2}+y^{2}=1\)
The integral \(\int_{0}^{1}\sqrt{1 - x^{2}}dx\) represents the area of a quarter - circle.
The formula for the area of a circle is \(A=\pi r^{2}\), where \(r = 1\) (since \(x^{2}+y^{2}=1\) has a radius \(r = 1\))

Step2: Calculate the area of the quarter - circle

For a quarter - circle with radius \(r = 1\), the area \(A=\frac{1}{4}\pi r^{2}\)
Substitute \(r = 1\) into the formula, we get \(A=\frac{\pi(1)^{2}}{4}=\frac{\pi}{4}\)

Answer:

A. \(8\)
B. \(\frac{\pi}{4}\)