Integral of square root $x-a$
I had just started learning Calculus recently and I encountered an Integral that was not talked about on my textbook, for example: $$\int_1^0x\sqrt{x-1}\,dx$$
How do I solve this integral step by step? I thought it would turn into something like $$\frac{x^2}{2}\times\frac{(x-1)^\frac{3}{2}}{\frac{3}{2}}$$ but that does not seem to be the right way to solve it.
$\endgroup$ 36 Answers
$\begingroup$I think you might have the wrong limits: the integrand is not defined on $[0,1]$.
Here's the work for the indefinite integral.
Let $u = x-1$. Then $du = dx$, but also $u+1=x$. Thus, $$ \begin{aligned}[t] \int x \sqrt{x-1} \, dx = \int(u+1) \sqrt{u} \, du &= \int (u+1) u^{1/2} \, du \\ &= \int (u^{3/2} + u^{1/2} ) \, du \\ &= \dfrac{u^{5/2}}{5/2} + \dfrac{u^{3/2}}{3/2} + C \\[1ex] &= \tfrac{2}{5} (x-1)^{5/2} + \tfrac{2}{3} (x-1)^{3/2} + C. \end{aligned} $$ (I'll leave it to you to take care of the definite integral.)
$\endgroup$ $\begingroup$For example by parts:
$$\begin{cases}&u=x,&u'=1\\{}\\ &v'=\sqrt{x-1},&v=\frac23(x-1)^{3/2}\end{cases}\implies \int x\sqrt{x-1}\,dx=$$$${}$$
$$=\frac23(x-1)^{3/2}-\frac23\int(x-1)^{3/2}\,dx$$
Can you end now? Observe that as commented, $\;\sqrt{x-1}\;$ isn't real for $\;x\in[0,1]\;$ ...
$\endgroup$ $\begingroup$Hint: $$x \sqrt{1-x}=(x-1) \sqrt{1-x}+\sqrt{1-x}=-(1-x)^{3/2}+(1-x)^{1/2}$$ This trick of adding and substracting is verry common in integration. It is not true in general that $$ \int_a^b f(x)g(x)dx=\int_a^b f(x) dx\int_a^b g(x)dx.$$
$\endgroup$ $\begingroup$\begin{align} u & = \sqrt{x-1} \\[10pt] u^2 & = x-1 \\[10pt] 2u\,du & = dx \\[10pt] x & = u^2 + 1 \end{align} Therefore $$ \int x\sqrt{x-1} \,dx = \int (u^2+1)(2u\,du) = \int (2u^3 + 2u)\,du = \cdots $$ This is a "rationalizing substitution". "Rationalizing" means getting rid of the radical.
$\endgroup$ $\begingroup$I would use the traditional substitution in this case: $$t=\sqrt{x-1}\iff t^2=x-1,\enspace t\ge0,\qquad\mathrm dx=2t\,\mathrm dt.$$ You end up in the integral $$\int2t^2(t^2+1)\,\mathrm dt=\frac{2t^5}{5}+\frac{2t^3}{3}=\frac{2\sqrt{x-1}((x-1)(3x+2)}{15}.$$
$\endgroup$ $\begingroup$$\small\bf\text{The definite integral's formula is:}$
$\normalsize\color{green}{\int_{a}^{b} f(x)dx=|_{a}^{b}F(x)=F(b)-F(a)}$
FYI: The definite integrals are from minimum to maximum, so instead of $\int_{1}^{0}$ use $\int_{0}^{1}$
$\small\bf\text{So lets get started how to solve your example:}$
$\int_{0}^{1}x\sqrt{x-1}\,dx$
Let use $u=x-1$, so then $\frac{du}{dx}=1$ since a derivative of $x$ is $1$.
Now we have:
$u=x-1\longrightarrow x=u+1$
$dx=1\cdot du$
Then:
$\int_{0}^{1}(u+1)\sqrt{u}\,du=\int_{0}^{1}u\sqrt{u}+\sqrt{u}\,du\\=\int_{0}^{1}u^{\frac{3}{2}}+u^{\frac{1}{2}}\,du= \int_{0}^{1}u^{\frac{3}{2}}\,du+\int_{0}^{1}u^{\frac{1}{2}}\,du$
$\normalsize\color{green}{\int x^n dx=\frac{1}{n+1}x^{n+1}+C}$
$=|_{0}^{1}\frac{1}{\frac{3}{2}+1}u^{\frac{3}{2}+1}+|_{0}^{1}\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}\\=|_{0}^{1}\frac{2}{5}u^{\frac{5}{2}}+|_{0}^{1}\frac{2}{3}u^{\frac{3}{2}}=|_{0}^{1}\frac{2}{5}u^2\sqrt{u}+|_{0}^{1}\frac{2}{3}u\sqrt{u}\\=|_{0}^{1}\frac{2}{5}u^2\sqrt{u}+\frac{2}{3}u\sqrt{u}$
$\normalsize\bf\text{Method 1:}$
$u=x-1\Longrightarrow u=0\rightarrow u=0-1=-1\\ \hspace{29mm}u=1\rightarrow u=1-1=0$
$=|_{-1}^{0}\frac{2}{5}u^2\sqrt{u}+\frac{2}{3}u\sqrt{u}=(\frac{2}{5}0^2\sqrt{0}+\frac{2}{3}0\sqrt{0})-(\frac{2}{5}(-1)^2\sqrt{-1}+\frac{2}{3}(-1)\sqrt{-1})\\=0-(\frac{2}{5}\sqrt{-1}-\frac{2}{3}\sqrt{-1})\\=-\frac{2}{5}\sqrt{-1}+\frac{2}{3}\sqrt{-1}=-\frac{2}{5}i+\frac{2}{3}i\\=-\frac{6}{15}i+\frac{10}{15}i=\frac{4}{15}i$
$\normalsize\bf\text{Method 2:}$
$=|_{0}^{1}\frac{2}{5}u^2\sqrt{u}+\frac{2}{3}u\sqrt{u}\\=|_{0}^{1}\frac{2}{5}(x-1)^2\sqrt{x-1}+\frac{2}{3}(x-1)\sqrt{x-1}\\=(\frac{2}{5}(1-1)^2\sqrt{1-1}+\frac{2}{3}(1-1)\sqrt{1-1})-(\frac{2}{5}(0-1)^2\sqrt{0-1}+\frac{2}{3}(0-1)\sqrt{0-1})\\=0-(\frac{2}{5}\sqrt{-1}-\frac{2}{3}\sqrt{-1})\\=-\frac{2}{5}\sqrt{-1}+\frac{2}{3}\sqrt{-1}=-\frac{2}{5}i+\frac{2}{3}i\\=-\frac{6}{15}i+\frac{10}{15}i=\frac{4}{15}i$
$\endgroup$
