# Math Help - Evaluating Integrals

1. ## Evaluating Integrals

Evaluate the following integral by interpeting it in terms of areas:

The definite integral of f(x) from 1 to 0

f(x) = sqrt(1 - x^2)dx = (1/4)3.14x^2 = (1/4)3.14(1)^2 = 3.14/4

This is an example that was given in my text book. I don't undersatnd the second step (that part in bold type). Can someone please exlpain why the anti-derivative of sqrt(1 - x^2)dx is (1/4)3.14x^2?

2. Originally Posted by zachb
Evaluate the following integral by interpeting it in terms of areas:

The definite integral of f(x) from 1 to 0

f(x) = sqrt(1 - x^2)dx = (1/4)3.14x^2 = (1/4)3.14(1)^2 = 3.14/4

This is an example that was given in my text book. I don't undersatnd the second step (that part in bold type). Can someone please exlpain why the anti-derivative of sqrt(1 - x^2)dx is (1/4)3.14x^2?
They cheated! But it works.

Notice that f(x) = sqrt(1 - x^2) is the equation of the upper half of a circle with radius 1. Notice that if you integrate from x = 0 to x = 1, you are finding the area of (1/4) of the area of that circle. Let's cheat and do this the easy way:

(1/4)*A(rea of a circle) = (1/4)*pi*r^2 = (1/4)*3.14*1 = 3.14/4

3. Originally Posted by zachb
Evaluate the following integral by interpeting it in terms of areas:

The definite integral of f(x) from 1 to 0

f(x) = sqrt(1 - x^2)dx = (1/4)3.14x^2 = (1/4)3.14(1)^2 = 3.14/4

This is an example that was given in my text book. I don't undersatnd the second step (that part in bold type). Can someone please exlpain why the anti-derivative of sqrt(1 - x^2)dx is (1/4)3.14x^2?