Sunday, November 15, 2015

`int_0^3 x/sqrt(36 - x^2) dx` Evaluate the integral

You need to perform the following substitution, such that `36 - x^2 = t => -2xdx = dt => xdx = -(dt)/2`


Replacing the variable yields:


`int_0^3 (xdx)/(sqrt(36 - x^2)) = -int_(t_1)^(t_2) (dt)/(2sqrt t) = -sqrt t|_(t_1)^(t_2)`


Replacing back the variable yields:


`int_0^3 (xdx)/(sqrt(36 - x^2)) = -sqrt(36-x^2)|_0^3`


`int_0^3 (xdx)/(sqrt(36 - x^2)) = -sqrt(36 - 9) + sqrt(36-0)`


`int_0^3 (xdx)/(sqrt(36 - x^2)) = 6 - sqrt 27`


`int_0^3 (xdx)/(sqrt(36 - x^2)) = 6 - 3sqrt 3`


`int_0^3 (xdx)/(sqrt(36 - x^2)) = 3(2 - sqrt3)`


Hence, evaluating the definite integral, yields `int_0^3 (xdx)/(sqrt(36- x^2)) = 3(2 - sqrt3).`

No comments:

Post a Comment

What are hearing tests?

Indications and Procedures Hearing tests are done to establish the presence, type, and sever...