Question

What is the value of...

4*integral [ sqrt ( 1 - x^2) ]dx from 0 to 1

Answer 1

this is just area of circle with unit radius.

Answer 2

this expression gives you the area of circle with radius 'a'. Make use of trigonometric substitution \(x = a \cos \theta \) to get the result.
\[ 4\times \int_0^a \sqrt{a^2 - x^2}dx\]

Answer 3

So the expression's value is pi, correct?

Answer 4

How would you solve the integral without the geometry?

Answer 5

From what I remember of trig:

1 - cos^2 (x) = sin^2 (x), so sqrt [ sin^2 ] = sinx

Therefore, we can simplify the expression into 4*integral [ sinx ]dx from 0 to 1

int [ sinx ] = -cosx

From this point, QED