Question #73dd0

1 Answer
Astralboy
Apr 15, 2017

#1/3#

Explanation:

Set #u=1-x^2#. Then #du=-2x# We don't have a #-2# though. So we'll multiply by #-2/-2#. Put the #-2# on the top of the numerator inside the integral and leave the #1/-2# outside the integral.

#1/-2 int_0^1 -2xsqrt(1-x^2)dx#

Make the substitution. But note that if you substitute, you have to change the integrand:

For #x=0#: #u=1-0^2=1#

For #x=1#: #u=1-1^2=0#

So we now have

#-1/2 int_1^0 sqrt(u) du#

We can flip the integrand by using the negative on the outside. We can also write #sqrt(u)# as #u^(1/2)# so it's easier to integrate:

#1/2 int_0^1 u^(1/2) du#

Now integrate:

#1/2 (u^(3/2))/(3/2)#

Take out the #1/3/2# and then plug in the integrands:

#(1/2)(1/(3/2)) (1^(3/2)-0^(3/2))#

Simplify:

#(1/3)(1)=1/3#

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