How to evaluate the definite integral using the rule of substitution.

Question

$\int_1^2 x\sqrt {x-1} \:dx$

I have so far that $u = x-1$, when $x=1$, $u=0$ and when $x=2$, $u=1$.

I do not know how to get a $dx$ that will eliminate the other $x$ from the equation. I tired doing $dx=du$ and then $\int_0^1 (u+1)\sqrt {u} \:du$ but I could not get the right answer. The correct answer is $\frac {16}{15} $.

Answer 1

Your substitution is correct, your further calculations are where the problem is.

Let $x-1=u \implies \mathrm{d}x=\mathrm{d}u$ and the bounds will be : $u=0$, $u=1$. The new integral becomes:

\begin{align} \int_0^1\sqrt{u}(u+1)\,\mathrm{d}u &= \int_0^1 (u^{3/2} + \sqrt{u})\,\mathrm{d}u \\ &=\int_o^1 u^{3/2}\,\mathrm{d}u + \int_0^1\sqrt{u}\,\mathrm{d}u \\ &= \bigg[\frac{2u^{5/2}}{5}\bigg]_0^1+\bigg[\frac{2u^{3/2}}{3}\bigg]_0^1 \\ & =\frac{16}{15}. \end{align}

Answer 2

You're on the right track! Since that $\sqrt{u}$ in your second integral is nothing more than $u^{1/2}$, you can distribute it through the rest of the terms in the integral to get $\int_1^2 u^{3/2}+u^{1/2} \:du$, which you can then evaluate with the power rule. And don't forget to change your bounds of integration once you make the u-substitution: $\int_0^1 u^{3/2}+u^{1/2} \:du$

Answer 3

Probably simpler would have been $$\sqrt{x-1}=u \implies x=u^2+1\implies dx=2u \,du$$ making $$\int_1^2 x\sqrt {x-1} \,dx=2\int_0^1 (u^4+u^2)\,du=2\left(\frac 15+\frac 13 \right)=\frac {16}{15}$$