`int_0^1 r^3 / sqrt(4 + r^2) dr` Evaluate the integral

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gsarora17 eNotes educator| Certified Educator

`int_0^1r^3/sqrt(4+r^2)dr`

Let's first evaluate the indefinite integral using the method of substitution,

Substitute `x=4+r^2, =>r^2=x-4`

`=>dx=2rdr` 

`intr^3/sqrt(4+r^2)dr=int(x-4)/(2sqrt(x))dx`

`=1/2int(x/sqrt(x)-4/sqrt(x))dx`

`=1/2int(sqrt(x)-4/sqrt(x))dx`

`=1/2((x^(1/2+1)/(1/2+1))-4(x^(-1/2+1)/(-1/2+1)))`

`=1/2((x^(3/2)/(3/2))-4(x^(1/2)/(1/2)))`

`=x^(3/2)/3-4x^(1/2)`

substitute back `x=r^2+4`  and add constant C to the solution,

`=(r^2+4)^(3/2)/3-4(r^2+4)^(1/2)+C`

Now let's evaluate the definite integral,

`int_0^1r^3/sqrt(4+r^2)dr=[(r^2+4)^(3/2)/3-4(r^2+4)^(1/2)]_0^1`

`=[(1^2+4)^(3/2)/3-4(1^2+4)^(1/2)]-[(0^2+4)^(3/2)/3-4(0^2+4)^(1/2)]`

`=[(5)^(3/2)/3-4(5)^(1/2)]-[4^(3/2)/3-4(4^(1/2)]` 

`=[(5^(3/2)-12(5)^(1/2))/3]-[(2^2)^(3/2)/3-4*2]`

`=[5^(1/2)((5-12)/3)]-[2^3/3-8]`

`=[-7/3sqrt(5)]-[8/3-8]`

`=(-7/3sqrt(5))-((8-24)/3)`

`=-7/3sqrt(5)-(-16/3)`

`=-7/3sqrt(5)+16/3`

`=1/3(16-7sqrt(5))`

 

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