What is #f(x) = int xsqrt(x-2) dx# if #f(2) = 3 #?

1 Answer
NickTheTurtle
Oct 21, 2017

#f(x)=(2x(x-2)^(3/2))/3-(4(x-2)^(5/2))/15+3#

Explanation:

#f(x)=int\ xsqrt(x-2)\ dx#

According to integration by parts,
#int\ u\ dv=uv-int\ v\ du#.

Set #u=x# and #dv=sqrt(x-2)\ dx=(x-2)^(1/2)\ dx#. Then, #du=dx# and #v=(2(x-2)^(3/2))/3#.

Thus,
#f(x)#
#=int\ xsqrt(x-2)\ dx#
#=(2x(x-2)^(3/2))/3-int\ (2(x-2)^(3/2))/3\ dx#
#=(2x(x-2)^(3/2))/3-(4(x-2)^(5/2))/15+C#

Now, since #f(2)=3#,
#(2*2(2-2)^(3/2))/3-(4(2-2)^(5/2))/15+C=3#
#C=3#

Thus, the final answer is #f(x)=(2x(x-2)^(3/2))/3-(4(x-2)^(5/2))/15+3#

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