Evaluate the definite integral.?

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Answer 1 · 2018-04-24T13:45:57

#0#

This can be solved using the following substitution:

#u=x^2#

Since this is a definite, calculate the new lower and upper bounds using the substitution:

Lower: #u=0^2=0#

Upper: #u=(sqrtpi)^2=pi#

Take the differential:

#du=2xdx#

We see that #xdx# is present in the integral. As a result, we can divide both sides of the differential by #2#:

#1/2du=xdx#

And we get

#1/2int_0^picosudu=1/2(sinu|_0^pi)#

#=1/2(sinpi-sin0)=1/2(0-0)=0#