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

1 Answer
Dec 3, 2017

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

Explanation:

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))#