Ask a New Question
Search
Evaluate the integral from 0 to 24 of the square root of 9+3x dx.
1 answer
if you let u = 9+3x, du = 3dx and you have
∫√(9+3x) dx = ∫√u (du/3) = 1/3 ∫u^(1/2) du
take it from there.
Similar Questions
evaluate the following integral using trigonometric substitution.
integral (dx)/(square root 289-x^2)) Rewrite given interval
1 answer
Evaluate the integral.
The integral from 0 to the square root of 2 over 2 of the function 8/sq. rt.(1-t^2)
1 answer
Evaluate the integral.
The integral from the square root of three over three to the square root of three of the function
1 answer
how do you solve the integral of 1/[(square root of x)(lnx)] from 2 to infinity?
i did the p- integral theorem with 1/square root
1 answer
more similar questions