How to integrate $\sqrt{1+(2/3)x}$?

Question

How would you solve the following (step by step please!): $$\int^6_5\sqrt{1+\frac23x}\ dx$$

I started with $u=1+\frac23x$, $du=\frac23\,dx$, now what?

Answer 1

You have $$\int^6_5\sqrt{1+\frac23x}\ dx$$

Now let $u = 1 + \frac{2}{3}x$, and $du = \frac{2}{3}\ dx$, so that means $dx = \frac{3}{2}du$, so make the subsitution:

$$\int^{6 = x}_{5=x}\sqrt{u}\frac{3}{2}\ du = \frac{3}{2} \int^{6 = x}_{5 = x}u^{\frac{1}{2}}\ du$$

Now we need to find the bounds for $u$, and then we will use the power rule for antiderivatives: Since $ u = 1 + \frac{2}{3}x$, when $x = 6$, $u =5$, and when $x = 5$, $u =\frac{13}{3}$, so those are our new bounds. So the integral is: $$\frac{3}{2} \int^{5}_{\frac{13}{3}}u^{\frac{1}{2}} du = \frac{3}{2}(\frac{2}{3}(5^\frac{3}{2}) - \frac{2}{3}(\frac{13}{3})^\frac{3}{2}) = 5^\frac{3}{2} - (\frac{13}{3})^\frac{3}{2} = \sqrt{125} - \sqrt{(\frac{13}{3})^3}$$$$ = 5\sqrt{5} - \frac{13\sqrt{\frac{13}{3}}}{3}$$

Alternately, instead of finding the new bounds for $u$, you can just put the antiderivative back in terms of $x$ and solve from there.