Thread: 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?

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

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?

I got a different answer:

EDIT: Ah, okay, ecMathGeek to the rescue! (Why do I have to think so complicated all the time?). I'll leave this up though, it might help somebody

Here is another way. If you do not like it then do not use it.