 Stoockiltj5

2022-01-23

When do we have to change the limits?
${\int }_{0}^{1}\sqrt{x-{x}^{2}}dx$
completing the square gives me:
$\int \sqrt{-{x}^{2}+x-\frac{1}{4}+\frac{1}{4}}dx$
$\int \sqrt{-\left(x-\frac{1}{2}{\right)}^{2}+\frac{1}{4}}dx$
$\int \sqrt{-{u}^{2}+\frac{1}{4}}du$
$\int \sqrt{\frac{1}{4}-{u}^{2}}du$
Is trig sub the best way to go?
if $u=\frac{1}{2}\mathrm{sin}\theta$ then $du=\frac{1}{2}\mathrm{cos}\theta d\theta$
$=\int \sqrt{\frac{1}{4}-\frac{1}{4}{\mathrm{sin}}^{2}\theta }\frac{1}{2}cos\theta d\theta$
$\int \frac{1}{2}cos\theta \frac{1}{2}cos\theta d\theta$
$\frac{1}{4}\int co{s}^{2}\theta d\theta$
using half angle identity I eventually get:
$\frac{1}{4}\int \frac{1}{2}\left(1+cos\left(2\theta \right)\right)d\theta$
$\frac{1}{8}\int \left(1+cos\left(2\theta \right)d\theta$
$\frac{1}{8}\left(\theta +\frac{1}{2}sin2\theta \right)$
Where do I go from here? Dakota Cunningham

Expert

I would recommend changing the limits every time we substitute. Here, that leads to ${\int }_{-1/2}^{1/2}\sqrt{\frac{1}{4}-{u}^{2}}\phantom{\rule{0.167em}{0ex}}du$ ($u=x-\frac{12}{}$ runs from $0-\frac{12}{}$ to $1-\frac{12}{}$)
$\frac{1}{4}{\int }_{-\pi /2}^{\pi /2}{\mathrm{cos}}^{2}\theta d\theta$ ($u=\frac{12}{\mathrm{sin}\theta }$ so since $\mathrm{sin}\theta$ runs from −1 to 1, $\theta$ runs from )
${\left[\frac{1}{8}\left(\theta +\frac{1}{2}\mathrm{sin}\left(2\theta \right)\right)\right]}_{\theta =-\frac{\pi }{2}}^{\theta =\frac{\pi }{2}}=\left(\frac{\pi }{16}+0\right)-\left(-\frac{\pi }{16}+0\right)=\frac{\pi }{8}$
If you do it that way, there's no need to go back -- we just evaluate the integral right there.
Definite integrals have more tools we can use than indefinite integrals. Even at the basic calculus level you're at, symmetry comes up a lot -- the integral of an odd function from −a to a is zero, for example. If we turn a definite integral problem into an indefinite integral problem and then only convert back at the very end, we lose access to those tools. So then, it makes sense to always bring the limits along and convert them when substituting in an indefinite integral problem.
Is trig sub the best way to go?
No, the best way to go is to not calculate any antiderivatives at all. Once you reach that first line I wrote down (the same point you were at when writing this line, plus the limits), recognize it as the area of a semicircle of radius $\frac{12}{}$ From that, we immediately conclude the result $\frac{1}{2}\pi {\left(\frac{1}{2}\right)}^{2}=\frac{\pi }{8}$
That is, of course, not something we could ever have done with an indefinite integral. Expert

Just so you are aware, this can be handled with the Beta and by extension the Gamma Function.
Here you have:
$I={\int }_{0}^{1}\sqrt{x-{x}^{2}}:dx={\int }_{0}^{1}\sqrt{x\left(1-x\right)}:dx={\int }_{0}^{1}{x}^{\frac{1}{2}}{\left(1-x\right)}^{\frac{1}{2}}:dx$
$=B\left(\frac{1}{2}+1,\frac{1}{2}+1\right)=B\left(\frac{3}{2},\frac{3}{2}\right)$
Using the relationship between the Beta and the Gamma Function this becomes:
$I=B\left(\frac{3}{2},\frac{3}{2}\right)=\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)\cdot \mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{2}+\frac{3}{2}\right)}=\frac{\mathrm{\Gamma }{\left(\frac{3}{2}t\right)}^{2}}{\mathrm{\Gamma }\left(3\right)}=\frac{{\left(\frac{\sqrt{\pi }}{2}\right)}^{2}}{2}=\frac{\pi }{8}$

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