How do you find #\int _ { 0} ^ { x } ( 1- 0.1x ) ^ { \frac { 1} { 2} } d x#?

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Answer 1 · 2017-12-03T13:46:45

The answer is #=20/3(1-(1-0.1x)^(3/2))#

We need

#int(x^(1/2)dx)=x^(3/2)/(3/2)+C#

Therefore,

#int_0^x(1-0.1x)^(1/2)dx=[(1-0.1x)^(3/2)/(3/2*-0.1)]_0^x#

#=[-20/3(1-0.1x)^(3/2)]_0^x#

#=-20/3(1-0.1x)^(3/2)+20/3#

#=20/3(1-(1-0.1x)^(3/2))#