What is the integral from 0 to 1 of xsqrt(1-x^4) ?

The integral from 0 to 1 of xsqrt(1-x^4) is pi/8

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integral from 0 to 1 of xsqrt(1-x^4)

\int_{0}^{1} x 1 - x^{4} d x

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{0}^{1} x 1 - x^{4} d x

Simplify

x^{4} : (x^{2})^{2}

= \int_{0}^{1} x 1 - (x^{2})^{2} d x

Apply u-substitution

: \int_{0}^{1} \frac{1 - u ^{2}}{2} d u

= \int_{0}^{1} \frac{1 - u ^{2}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{1} 1 - u^{2} d u

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d v + \int_{0}^{\frac{π}{2}} cos (2 v) d v)

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= \frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

\frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{8}

= \frac{π}{8}

\int x x^{\frac{3}{2}} - 1 d x

\frac{d}{d x} (lo g_{5} (2 x^{2} + 1))

(2 x + \frac{2}{y}) + (\frac{( 2 x ^{2} )}{y} + \frac{2 x}{y ^{2}}) y^{^{'}} = 0

f (x) = x^{4} - 10 x^{2} + 9

\int - \frac{14}{x ^{15}} d x

What is the integral from 0 to 1 of xsqrt(1-x^4) ?
The integral from 0 to 1 of xsqrt(1-x^4) is pi/8