What is the antiderivative of #(-2x+48)^.5#?

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Answer 1 · 2016-03-26T05:06:45

#-(-2x+48)^(3/2)/3+C#

We have:

#int(-2x+48)^.5dx=int(-2x+48)^(1/2)dx#

Let #u=-2x+48#, which implies that #du=-2dx#.

Multiply the inside of the integral by #-2# so that there can be a #-2dx#. Balance this by multiplying the outside by #-1/2#.

#=-1/2int(-2x+48)^(1/2)*-2dx#

Substituting with what we previously defined, this becomes

#=-1/2intu^(1/2)du#

To integrate this, use the rule that #intu^n=u^(n+1)/(n+1)+C#.

#=-1/2(u^(1/2+1)/(1/2+1))+C=-1/2((u^(3/2))/(3/2))+C#

#=-1/2(2/3)u^(3/2)+C=-1/3u^(3/2)+C#

#=-(-2x+48)^(3/2)/3+C#