`int_36^81(sqrt x -5)^(3/2)/sqrt x dx=?`

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tiburtius eNotes educator| Certified Educator

`int_36^81(sqrt x -5)^(3/2)/sqrt x dx=|(t=sqrt x-5),(dt=dx/(2 sqrt x) => 2dt=dx/sqrt x),(t_1=sqrt36-5=1),(t_2=sqrt81-5=4)|=`

We have made substitution `t=sqrt x-5` and `t_1` and `t_2` are now new limits of integration for variable `t.` In the following line we use the fact that `int x^(3/2)dx=2/5 x^(5/2)`.

`2int_1^4t^(3/2)dt=2 cdot 2/5t^(5/2)|_1^4=4/5(4^(5/2)-1^5/2)= 4/5(32-1)=124/5` ` `

So your solution is:  `int_36^81(sqrt x -5)^(3/2)/sqrt x dx=124/5`. 

pramodpandey | Student

put  `sqrt(x)-5=t , 1/(2 sqrt(x))dx=dt ,`

`(1/sqrt(x))dx=2dt, `

`put x=81 and x=36 in t ,so we have t=4 and t=1 ,`

`Now int_ (1, 4)2t^(3/2)dt , `

`we get`

`4/5 (t^(5/2))(1,4)`

`(4/5)( 32-1)=124/5`

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