∫[1,9] (x-1)/√x dx
let u^2 = x
2u du = dx
Then you have
∫[1,3] (u^2-1)/u * 2u du = 2 ∫[1,3] u^2-1 du = 2 (1/3 u^3 - u)[1,3] = 40/3
But you don't really need a u substitution, since
(x-1)/√x is just x^(1/2) - x^(-1/2)
so the integral is 2/3 x^(3/2) - 2x^(1/2)
integrals
evaluate the definite integral from 1 to 9 of ((x-1) / (sqrt x))
dx ? Would this one be a u sub?
evaluate the integral from 0 to 10 of abs value (x-5) dx? I think this one would be split but not sure how or why?
1 answer
2 answers