Question #e678e

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Question #e678e

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Answer 1 · 2017-12-05T01:07:18

$\int_{0}^{1} x \cdot \sqrt{1 - x^{4}} \cdot d x = \frac{π}{8}$ $int_0^1 x*sqrt(1-x^4)*dx=pi/8$

Explanation:

$\int_{0}^{1} x \cdot \sqrt{1 - x^{4}} \cdot d x$ $int_0^1 x*sqrt(1-x^4)*dx$

= $\frac{1}{2} \cdot \int_{0}^{1} \sqrt{1 - {(x^{2})}^{2}} \cdot 2 x \cdot d x$ $1/2*int_0^1 sqrt[1-(x^2)^2]*2x*dx$

After using $sin u = x^{2}$ $sinu=x^2$ and $cos u \cdot d u = 2 x \cdot d x$ $cosu*du=2x*dx$ substitution, this integral became

$\frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} \sqrt{1 - {(sin u)}^{2}} \cdot cos u \cdot d u$ $1/2*int_0^(pi/2) sqrt[1-(sinu)^2]*cosu*du$

= $\frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} \sqrt{{(cos u)}^{2}} \cdot cos u \cdot d u$ $1/2*int_0^(pi/2) sqrt[(cosu)^2]*cosu*du$

= $\frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} {(cos u)}^{2} \cdot d u$ $1/2*int_0^(pi/2) (cosu)^2*du$

= $\frac{1}{4} \cdot \int_{0}^{\frac{π}{2}} (1 + cos 2 u) \cdot d u$ $1/4*int_0^(pi/2) (1+cos2u)*du$

= ${[\frac{u}{4} + \frac{1}{8} \cdot sin 2 u]}_{0}^{\frac{π}{2}}$ $[u/4+1/8*sin2u]_0^(pi/2)$

= $\frac{π}{8}$ $pi/8$

Answer 2 · 2017-12-05T01:11:02

$\frac{π}{8}$ $pi/8$

Explanation:

.

$\int_{0}^{1} x \sqrt{1 - x^{4}} d x$ $int_0^1xsqrt(1-x^4)dx$

$u = x^{2}$ $u=x^2$
$x = \sqrt{u}$ $x=sqrtu$
$u^{2} = x^{4}$ $u^2=x^4$
$d u = 2 x d x$ $du=2xdx$
$d x = \frac{d u}{2 x} = \frac{d u}{2 \sqrt{u}}$ $dx=(du)/(2x)=(du)/(2sqrtu)$
$\sqrt{1 - x^{4}} = \sqrt{1 - u^{2}}$ $sqrt(1-x^4)=sqrt(1-u^2)$

Now, we substitute the above into the problem integral to turn it into an integral in terms of $u$ $u$ :

$\int (\sqrt{u}) (\sqrt{1 - u^{2}}) (\frac{d u}{2 \sqrt{u}}) =$ $int(sqrtu)(sqrt(1-u^2))((du)/(2sqrtu))=$

$\frac{1}{2} \int \sqrt{1 - u^{2}} d u$ $1/2intsqrt(1-u^2)du$

Now we can use trigonometric substitution to solve this:

$u = sin θ$ $u=sintheta$
$d u = cos θ d (θ)$ $du=costhetad(theta)$

$\frac{1}{2} \int \sqrt{1 - {sin}^{2} θ} cos θ d (θ)$ $1/2intsqrt(1-sin^2theta)costhetad(theta)$

$\frac{1}{2} \int \sqrt{{cos}^{2} θ} cos θ d (θ)$ $1/2intsqrt(cos^2theta)costhetad(theta)$

$\frac{1}{2} \int cos θ cos θ d (θ) = \frac{1}{2} \int {cos}^{2} θ d (θ)$ $1/2intcosthetacosthetad(theta)=1/2intcos^2thetad(theta)$

${cos}^{2} θ = \frac{1 + cos (2 θ)}{2}$ $cos^2theta=(1+cos(2theta))/2$

$\frac{1}{4} \int (1 + cos (2 θ)) d (θ) = \frac{1}{4} \int d (θ) + \frac{1}{4} \int cos (2 θ) d (θ)$ $1/4int(1+cos(2theta))d(theta)=1/4intd(theta)+1/4intcos(2theta)d(theta)$

$\frac{1}{4} θ + \frac{1}{8} sin (2 θ)$ $1/4theta+1/8sin(2theta)$

Now, we can substitute back for $u$ $u$ :

$u = sin θ$ $u=sintheta$ , then $θ = arcsin u$ $theta=arcsinu$
$cos θ = \sqrt{1 - s i x^{2} θ} = \sqrt{1 - u^{2}}$ $costheta=sqrt(1-six^2theta)=sqrt(1-u^2)$
$sin (2 θ) = 2 sin θ cos θ = 2 u \sqrt{1 - u^{2}}$ $sin(2theta)=2sinthetacostheta=2usqrt(1-u^2)$

$\frac{1}{4} θ + \frac{1}{8} sin (2 θ) = \frac{arcsin u + u \sqrt{1 - u^{2}}}{4}$ $1/4theta+1/8sin(2theta)=(arcsinu+usqrt(1-u^2))/4$

Now, we can substitute back for $x$ $x$ :

$\int_{0}^{1} x \sqrt{1 - x^{4}} d x = \frac{{arcsin x}^{2} + x^{2} \sqrt{1 - x^{4}}}{4}$ $int_0^1xsqrt(1-x^4)dx=(arcsinx^2+x^2sqrt(1-x^4))/4$

We will evaluate this from $0$ $0$ to $1$ $1$ :

$\frac{{arcsin 1}^{2} + 1^{2} \sqrt{1 - 1^{4}}}{4} - \frac{{arcsin 0}^{2} + 0^{2} \sqrt{1 - 0^{4}}}{4} =$ $(arcsin1^2+1^2sqrt(1-1^4))/4-(arcsin0^2+0^2sqrt(1-0^4))/4=$

$\frac{\frac{π}{2} + 0}{4} - \frac{0 + 0}{4} = \frac{π}{8}$ $(pi/2+0)/4-(0+0)/4=pi/8$

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