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: Split the integral based on the intersection point

\(\int_{-3}^{5}(2 - x)dx=\int_{-3}^{2}(2 - x)dx+\int_{2}^{5}(2 - x)dx\)
The function \(y = 2 - x\) is a straight - line. The integral \(\int_{a}^{b}(2 - x)dx\) represents the net area between the line \(y = 2 - x\) and the \(x\) - axis from \(x=a\) to \(x = b\).
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 a triangle is \(A=\frac{1}{2}\times base\times height\)
The area of the triangle from \(x=-3\) to \(x = 2\): \(A_1=\frac{1}{2}\times(2 + 3)\times5=\frac{25}{2}\)
The area of the triangle from \(x = 2\) to \(x = 5\): \(A_2=\frac{1}{2}\times(5 - 2)\times3=\frac{9}{2}\)
Since the area below the \(x\) - axis is negative, \(\int_{-3}^{5}(2 - x)dx=A_1-A_2\)
\(\int_{-3}^{5}(2 - x)dx=\frac{25}{2}-\frac{9}{2}=8\)

Part B

Step1: Recognize the geometric shape

The function \(y = \sqrt{1 - x^{2}}\) can be rewritten as \(y^{2}+x^{2}=1,y\geq0\), which is the upper - half of the unit circle \(x^{2}+y^{2}=1\)
The integral \(\int_{0}^{1}\sqrt{1 - x^{2}}dx\) represents the area of a quarter - circle.

Step2: Use the formula for the area of a circle

The area of a full - circle is \(A=\pi r^{2}\), for \(r = 1\), \(A=\pi\)
The area of a quarter - circle with radius \(r = 1\) is \(A=\frac{1}{4}\pi r^{2}\)
Since \(r = 1\), \(\int_{0}^{1}\sqrt{1 - x^{2}}dx=\frac{\pi}{4}\)

Answer:

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