What is the integral from 0 to 2 of x^7sqrt(4-x^2) ?

The integral from 0 to 2 of x^7sqrt(4-x^2) is 8192/315

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integral from 0 to 2 of x^7sqrt(4-x^2)

\int_{0}^{2} x^{7} 4 - x^{2} d x

\frac{8192}{315}

Decimal

26.00634 \dots

Solution steps

\int_{0}^{2} x^{7} 4 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 512 sin^{7} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 512 sin^{7} (u) cos^{2} (u) d u

Take the constant out:

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

= 512 \cdot \int_{0}^{\frac{π}{2}} sin^{7} (u) cos^{2} (u) d u

Rewrite using trig identities

= 512 \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (u))^{3} sin (u) cos^{2} (u) d u

Apply u-substitution

: \int_{1}^{0} - v^{2} (1 - v^{2})^{3} d v

= 512 \cdot \int_{1}^{0} - v^{2} (1 - v^{2})^{3} d v

Expand

- v^{2} (1 - v^{2})^{3} : - v^{2} + 3 v^{4} - 3 v^{6} + v^{8}

= 512 \cdot \int_{1}^{0} - v^{2} + 3 v^{4} - 3 v^{6} + v^{8} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 512 (- \int_{0}^{1} - v^{2} + 3 v^{4} - 3 v^{6} + v^{8} 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

= 512 (- (- \int_{0}^{1} v^{2} d v + \int_{0}^{1} 3 v^{4} d v - \int_{0}^{1} 3 v^{6} d v + \int_{0}^{1} v^{8} d v))

\int_{0}^{1} v^{2} d v = \frac{1}{3}

\int_{0}^{1} 3 v^{4} d v = \frac{3}{5}

\int_{0}^{1} 3 v^{6} d v = \frac{3}{7}

\int_{0}^{1} v^{8} d v = \frac{1}{9}

= 512 (- (- \frac{1}{3} + \frac{3}{5} - \frac{3}{7} + \frac{1}{9}))

Simplify

512 (- (- \frac{1}{3} + \frac{3}{5} - \frac{3}{7} + \frac{1}{9})) : \frac{8192}{315}

= \frac{8192}{315}

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What is the integral from 0 to 2 of x^7sqrt(4-x^2) ?
The integral from 0 to 2 of x^7sqrt(4-x^2) is 8192/315