Question #2e036

2 Answers
Jan 21, 2018

Shown by considering geometry...

Explanation:

I wanted to provide a simple way of evaluating this integral

The first thing we see is that the function is y = sqrt(1-x^2)

=> y^2 = 1 - x^2

=> x^2+y^2 = 1

This is just a circle of radius 1, and centre (0,0)

This is just this shaded region...

When evaluating from x = 0 and x=1

Desmos

What is just a quarter of the area of a circle of radius 1

Full area of the circle = = pir^2 =pi * 1^2 = pi

Hence quarter area = 1/4 * pi

color(red)(= pi/4

Jan 22, 2018

Conventoinal means of substitution...

Explanation:

The conventional way of solving this integral:

We can do a trigonometric substitution...

Let x = sintheta

color(green)(dx = costheta d theta

and we know sin^2x + cos^2x = 1

=> costheta = sqrt(1-sin^2 theta )

=> color(blue)(costheta = sqrt(1-x^2)

Hence our integral becomes:

int color(blue)(costheta) color(green)(costheta d theta

Changing the limits...

theta = sin^(-1) x

x = 1 => theta = sin^(-1)(1) = pi/2

x = 0 => theta = sin^(-1)(0) = 0

=> int_0 ^(pi/2) cos^2 theta d theta

Use cos2theta = cos^2 theta - sin^2 theta

=> cos2theta = 2cos^2 theta -1

cos^2 theta = 1/2( cos2theta + 1 )

=>1/2 int_0 ^(pi/2) cos2theta + 1 d theta

=> 1/2[ 1/2 sin2theta + theta ]_0 ^(pi/2)

=> 1/2[ {0 + pi/2}-{0 + 0 } ]

color(blue)( = pi/4 )