0
$\begingroup$

Evaluate the definite integral $\int_\frac{1}{2}^1 \frac{x^2}{\sqrt{2x-x^2}}$ by first "completing the square" in the denominator and then making a suitable trigonometric change of variable by estimating the integral directly. Show that it is less than $\frac{1}{2}$.

$\endgroup$
  • 3
    $\begingroup$ So, what have you tried? You have been given nice hints already. It remains to see if you can apply them properly. $\endgroup$ – Pedro Tamaroff Aug 16 '13 at 4:45
  • 2
    $\begingroup$ We have $2x-x^2=1-(1-x)^2$. Now is it clear what to do? $\endgroup$ – André Nicolas Aug 16 '13 at 4:46
2
$\begingroup$

As Andre Nicolas wrote, $\sqrt{2x-x^2}=\sqrt{1-(1-x)^2}\Rightarrow\displaystyle \int_\frac 1 2^1\frac{x}{\sqrt{2x-x^2}}dx=\int_\frac 1 2^1\frac {x}{\sqrt{1-(1-x)^2}}dx$. We can substitue simply $a=(1-x)$ (and as follows $x=1-a,dx=-da$) so: $$\displaystyle\int_\frac 1 2^1 f(x)dx=\int_0 ^\frac 1 2\frac{(1-a)^2 da}{\sqrt{1-a^2}}=\int\frac 1 {\sqrt{1-a^2}}da+\int\frac{-2a}{\sqrt{1-a^2}}da+\int_0^{0.5}\frac{a^2}{\sqrt{1-a^2}}da$$ Here the calculating for each integral (in the last simply integrate by parts) is simple: $$=\arcsin(a)\mid_0^\frac 1 2+\sqrt{1-a^2}\mid_0^\frac 1 2+(\frac{\arcsin(a)-a\sqrt{1-a^2}}{2})\mid_0^\frac 1 2=0.5235987756+\frac{\sqrt 3}{2}-1+0.04529303685=0.4349172162<0.5$$ as required.

$\endgroup$
2
$\begingroup$

The integration procedure has been done in detail by Coargu Aliquis. We tackle the estimate only. If you have a graphing calculator, or know how to access a graphing program (there are free ones on the web), you might start by graphing $y=\frac{x^2}{\sqrt{1-x^2}}$, to see what is going on. You may notice that the function seems to be increasing in our interval, and reaches $1$ at $x=1$. We will show that indeed $$\frac{x^2}{\sqrt{2x-x^2}}\lt 1$$ for all $x$ in the interval $\frac{1}{2}\le x\lt 1$. Equivalently, we show that $\frac{x^4}{2x-x^2}\lt 1$.

Equivalently, we show that $\frac{x^3}{2-x}\lt 1$, that is, that $2-x-x^2\gt 0$. But this is clear, since both $x^3$ and $x$ are less than $1$ in the interval $[\frac{1}{2},1)$.

Thus our integrand is $\lt 1$ in the interval $[\frac{1}{2},1)$. Since the interval has length $\frac{1}{2}$, it follows that the integral is $\lt \frac{1}{2}$.

$\endgroup$
0
$\begingroup$

$$I=\int_\frac12^1 \frac{x^2}{\sqrt{2x-x^2}} =\int_\frac12^1 \frac{x^2}{\sqrt{1-(x-1)^2}}$$

Putting $x-1=\sin\theta,dx=\cos\theta d\theta$

When $x=\frac12, \sin\theta=-\frac12\implies \theta=-\frac\pi6$

$$I=\displaystyle\int_{-\frac\pi6}^0\frac{(1+\sin\theta)^2}{\cos\theta}\cos\theta d\theta$$

$$=\displaystyle\int_{-\frac\pi6}^0\frac{3+2\sin\theta-\cos2\theta}2d\theta$$

$$=\displaystyle\frac{6\theta-4\cos\theta-\sin2\theta}4\big|_{-\frac\pi6}^0$$

$$=\displaystyle\frac{-4-\left(-\pi-4\frac{\sqrt3}2+\frac{\sqrt3}2\right)}4$$

$$=\frac{3\sqrt3+2\pi-8}8=0.43491721623$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.