Integrate A Square Root of A function

How would i integrate the following function :

$\int^1_0 \sqrt {2x+1}\ dx$

I'v never seen a problem like this before and my only guess was to use the substitution $u=2x+1$ and $v=u^{1/2}$ but that isn't getting me anywhere

That substitution should work.


$u = 2x+1 \to du = 2\,dx$

Limits:

$x = 1 \to u = 3$
$x = 0 \to u = 1$

Hence:

$\displaystyle I = \int_1^3 u^{\frac{1}{2}} .\, 2\,du = 2\int_1^3 u^{\frac{1}{2}} \, du$

That substitution should work.


$u = 2x+1 \to du = 2\,dx$

Limits:

$x = 1 \to u = 3$
$x = 0 \to u = 1$

Hence:

$\displaystyle I = \int_1^3 u^{\frac{1}{2}} .\, 2\,du = 2\int_1^3 u^{\frac{1}{2}} \, du$

is .2 before du supposed to mean multiply by 2 or a half?

You could aslo use the chain rule

This is integration and NOT differentiation.

Yeah I meant the "chain rule" for integration. See nuodai's post

nuodai's post is the reverse chain rule (I know everyone hates that phrase), which is not the same thing as the chain rule itself. I think the preferred name for it is integration by inspection/recognition. Saying it's the chain rule is a bit misleading.

nuodai's post is the reverse chain rule (I know everyone hates that phrase), which is not the same thing as the chain rule itself. I think the preferred name for it is integration by inspection/recognition. Saying it's the chain rule is a bit misleading.

For some reason, questions like this came up in my C3 before I'd done integration by substitution (which is in C4 on OCR). They can be solved by inspection without the need for a substitution (as such): if you know that $(ax+b)^n$ differentiates to $an(ax+b)^{n-1}$ (by the chain rule), then going backwards reveals that $\displaystyle \int (ax+b)^n\, dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C$. Here, you have $(2x+1)^{\frac{1}{2}}$, so you can integrate it using that result. Note that you can't apply this result to any other type of function; for example you wouldn't be able to integrate $\sqrt{x^2+1}$ using this result.