Evaluate $\displaystyle\int_0^{2}\sqrt{4x + 1} \text{d}x$

Evaluate $\displaystyle\int_0^{2}\sqrt{4x + 1}~\text{d}x$

This becomes:

$\displaystyle\int_0^{2}(4x + 1)^\frac{1}2~\text{d}x$

I am not sure where to go from here, I suspect it might use the chain rule or reverse chain rule.

• Hint: Let $u = 4 x + 1$ – Moo Mar 27 '16 at 13:13

$$\int_{0}^{2}\sqrt{4x+1}\space\text{d}x=$$

Substitute $u=4x+1$ and $\text{d}u=4\space\text{d}x$.

This gives a new lower bound $u=4\cdot0+1=1$ and upper bound $u=4\cdot2+1=9$:

$$\frac{1}{4}\int_{1}^{9}\sqrt{u}\space\text{d}u=\frac{1}{4}\int_{1}^{9}u^{\frac{1}{2}}\space\text{d}u=$$

Use $\int y^{n}\space\text{d}y=\frac{y^{1+n}}{1+n}+\text{C}$

$$\frac{1}{4}\left[\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}\right]_{1}^{9}=\frac{1}{4}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{9}=\frac{1}{4}\cdot\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{1}^{9}=\frac{1}{6}\left[u^{\frac{3}{2}}\right]_{1}^{9}=\frac{1}{6}\left(9^{\frac{3}{2}}-1^{\frac{3}{2}}\right)=\frac{1}{6}(26)=\frac{13}{3}$$

• Instead of giving new bounds, might it not be easier to just replace $u$ with $4x+1$? – Airdish Mar 27 '16 at 13:43
• In this example it is possible, but most integrals get complicated and that's why it's much easier to take the bounds into the new integral – Jan Mar 27 '16 at 13:47
• Ah, I see. You said it can get complicated, but just of curiosity, is the method of replacing $u$ ever actually wrong? – Airdish Mar 27 '16 at 13:48
• @Airdish No, it is right! But notice than that you've to hold your original bounds – Jan Mar 27 '16 at 13:50

A tip with these types of integrations:

Increase the power by one first, and then figure out the coefficient you need.

(Although someone has already provided the solution), this is a helpful way to do it mentally.

Since the derivative of $(4x+1)^{3/2}$ is $6(4x+1)^{\frac{1}{2}}$, $$\int_0^2 \sqrt{4x+1}dx =\left[\frac{1}{6}(4x+1)^{3/2}\right]_0^2.$$