# Evaluate the integral? :  int_0^2 xsqrt(2x-x^2) dx

Jan 20, 2018

${\int}_{0}^{2} \setminus x \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx} = \frac{\pi}{2}$

#### Explanation:

Consider the indefinite integral:

$I = \int \setminus x \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx}$

Which we can write as:

$I = \int \setminus \left(x - 1\right) \sqrt{2 x - {x}^{2}} + \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \left(x - 1\right) \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx} + \int \setminus \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx}$

For the first integral, we have:

${I}_{1} = \int \setminus \left(x - 1\right) \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx}$

we can use a substitution, Let:

$u = 2 x - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 - 2 x = - 2 \left(x - 1\right)$

And if we perform this substitution then we get:

${I}_{1} = \int \setminus \left(- \frac{1}{2}\right) \sqrt{u} \setminus \mathrm{du}$
$\setminus \setminus \setminus = - \frac{1}{2} {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)$
$\setminus \setminus \setminus = - \frac{1}{3} {u}^{\frac{3}{2}}$

And restoring the substitution we get:

${I}_{1} = - \frac{1}{3} {\left(2 x - {x}^{2}\right)}^{\frac{3}{2}}$

Next we consider the second integral,

${I}_{2} = \int \setminus \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \sqrt{1 - {\left(x - 1\right)}^{2}} \setminus \mathrm{dx}$

And here we can perform a substitution; Let

$\sin \theta = x - 1 \implies \frac{d \theta}{\mathrm{dx}} \cos \theta = 1$

And if we perform this substitution then we get:

${I}_{2} = \int \setminus \sqrt{1 - {\sin}^{2} \theta} \setminus \left(\cos \theta\right) \setminus d \theta$
$\setminus \setminus \setminus = \int \setminus {\cos}^{2} \theta \setminus d \theta$
$\setminus \setminus \setminus = \int \setminus \frac{\cos \left(2 \theta\right) + 1}{2} \setminus d \theta$
$\setminus \setminus \setminus = \frac{\sin 2 \theta + \theta}{2}$

And restoring the substitution we find that $\theta = \arcsin \left(x - 1\right)$ and $\sin \left(2 \theta\right) = 2 \left(x - 1\right) \sqrt{1 - {\left(x - 1\right)}^{2}}$ and so:

${I}_{2} = \frac{\arcsin \left(x - 1\right) + \left(x - 1\right) \sqrt{\left(x - 1\right) - {\left(x - 1\right)}^{2}}}{2}$
$\setminus \setminus \setminus = \frac{\arcsin \left(x - 1\right) + \left(x - 1\right) \sqrt{2 x - {x}^{2}}}{2}$

Combining our two results we then gave:

$I = \frac{\arcsin \left(x - 1\right) + \left(x - 1\right) \sqrt{2 x - {x}^{2}}}{2} - \frac{1}{3} {\left(2 x - {x}^{2}\right)}^{\frac{3}{2}}$

Given this result we then have:

${\int}_{0}^{2} \setminus x \sqrt{2 x - {x}^{2}} \setminus \mathrm{dx} = {\left[\frac{\arcsin \left(x - 1\right) + \left(x - 1\right) \sqrt{2 x - {x}^{2}}}{2} - \frac{1}{3} {\left(2 x - {x}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{2}$

 " " = 1/2((arcsin1 - arcsin(-1))

$\text{ } = \frac{1}{2} \left(\frac{\pi}{2} - \left(- \frac{\pi}{2}\right)\right)$

$\text{ } = \frac{\pi}{2}$

Jan 20, 2018

${\int}_{0}^{2} x \sqrt{2 x - {x}^{2}} \cdot \mathrm{dx} = \frac{\pi}{2}$

#### Explanation:

${\int}_{0}^{2} x \sqrt{2 x - {x}^{2}} \cdot \mathrm{dx}$

=${\int}_{0}^{2} x \sqrt{{1}^{2} - {\left(x - 1\right)}^{2}} \cdot \mathrm{dx}$

After using $x - 1 = \sin u$, $x = 1 + \sin u$ and $\mathrm{dx} = \cos u \cdot \mathrm{du}$ transforms, this integral became

${\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \sin u\right) \cdot \cos u \cdot \cos u \cdot \mathrm{du}$

=${\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \sin u\right) \cdot {\left(\cos u\right)}^{2} \cdot \mathrm{du}$

=$\frac{1}{2} {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \sin u\right) \cdot \left(1 + \cos 2 u\right) \cdot \mathrm{du}$

=$\frac{1}{2} {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left(1 + \cos 2 u + \sin u + \cos 2 u \cdot \sin u\right) \cdot \mathrm{du}$

=$\frac{1}{2} {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left[1 + \cos 2 u + \sin u + \frac{1}{2} \cdot \left(\sin 3 u - \sin u\right)\right] \cdot \mathrm{du}$

=$\frac{1}{4} {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \left(2 + 2 \cos 2 u + \sin 3 u + \sin u\right) \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\left[2 u + \sin 2 u - \frac{1}{3} \cdot \cos 3 u - \cos u\right]}_{- \frac{\pi}{2}}^{\frac{\pi}{2}}$

=$\frac{1}{4} \cdot 2 \pi$

=$\frac{\pi}{2}$