How do you find the definite integral for: #(9-x^2)^(1/2)# for the intervals #[-3, 3]#?

2 Answers
Nov 5, 2016

The answer is #=(9pi)/2#

Explanation:

First, we calculate #intsqrt(9-x^2)dx#
let #x=3sinu#, then #dx=3cosudu#
#sqrt(9-9sin^2u)=3cosu#
#:.intsqrt(9-x^2)dx=int3cosu*3cosudu#
#=9intcos^2udu#
#cos2u=2cos^2u-1##=>##cos^2u=(cos2u+1)/2#
#9intcos^2udu=9/2int(cos2u+1)du#
#=9/2((sin2u)/2+u)= 9/4((sin2u)+2u)#
#sin2u=2sinucosu=2*x/3*sqrt(9-x^2)/3=2/9x(sqrt(9-x^2))#
#9/4((sin2u)+2u)=9/4*2/9*(xsqrt(9-x^2))+9/2arcsin(x/3)#
#int_-3^3sqrt(9-x^2)dx=(1/2(xsqrt(9-x^2))+9/2arcsin(x/3))_-3^3#
#0+9/2*arcsin1-0-9/2arcsin(-1)#
#0+9pi/4--9pi/4=9pi/2#

Nov 6, 2016

#(9pi)/2#

Explanation:

We can also think about this geometrically. #y=sqrt(9-x^2)# is the upper half of the circle #x^2+y^2=9#, a circle with radius #3# and center at #(0,0)#.

Since the bounds span the length of the semicircle, the integral will be equal to #1//2# the circle's area, or #(9pi)/2#.