Integration Problem: Solving 3/(1+4x)^0.5 from x=0 to x=2

[PhyStan7]

Hi all, I am having a problem with an integration.


Solve the integration 3/(1+4x)^0.5 between x=0 and x=2. Basically find the area under a curve of the equation between x=0 and x=2.

I know the answer is 3 but can't get to it. I tried rewriting it as 3(1+4x)^-0.5 and solving bu got stupid answers like 15. and 21 to lots of dp. I can't find the mark scheme to the past paper i am doing anywhere on the internet so can't find a solution. Help would be much appreciared, thanks

 

[Random Variable]

If you let u=1+4x, then the integral becomes 3/4*u^(-1/2)du for u from 1 to 9.

 

[larstuff]

If you let u=1+4x, then the integral becomes 3/4*u^(-1/2)du for u from 1 to 9.

3/(1+4x)^0.5 = 3 * (1+4x)^(-0,5)

Now you can use the common powers law with the exponent (-0,5).
Adding 1 to this gives (0,5).