find y' if y=int_1^x2 sqrt(2t+4) dt

Expert Answers
sciencesolve eNotes educator| Certified Educator

You should solve the integral using the change of variable such that:

`2t + 4 = u =gt 2dt = du =gt dt = (du)/2`

You need to evaluate the integral such that:

`int sqrt(2t+4) dt = int sqrt u*(du)/2`

`(1/2)*int u^(1/2) du = (1/2)*(u^(1/2 + 1)/(1/2 + 1))`

`(1/2)*int u^(1/2) du = (1/2)*(u^(3/2))/(3/2)` 

`(1/2)*int u^(1/2) du = (1/3)*(u^(3/2))`

Substitute back `2t + 4`  for u such that:

`int_1^(x^2) sqrt(2t+4) dt =(1/3)*((2t+4)^(3/2))|_1^(x^2)`

`int_1^(x^2) sqrt(2t+4) dt = (1/3)*((2x^2+4)^(3/2) - 6^(3/2))`

`int_1^(x^2) sqrt(2t+4) dt = (1/3)*(sqrt((2x^2+4)^3) - 6sqrt6)`

Since `y = int_1^(x^2) sqrt(2t+4) dt =gt y = (1/3)*(sqrt((2x^2+4)^3) - 6sqrt6)`

You need to find y' such that:

`y' = (1/3)*((2x^2+4)')/(2sqrt(2x^2+4))`

`y' = (1/3)*(4x)/(2sqrt(2x^2+4))`

`y' = x/(3sqrt(2x^2+4))`

Hence, evaluating y' under the given conditions yields `y' = x/(3sqrt(2x^2+4)).`