What is the integral from 0 to 1 of 16x^3sqrt(1-x^2) ?

The integral from 0 to 1 of 16x^3sqrt(1-x^2) is 32/15

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integral from 0 to 1 of 16x^3sqrt(1-x^2)

\int_{0}^{1} 16 x^{3} 1 - x^{2} d x

\frac{32}{15}

Decimal

2.13333 \dots

Solution steps

\int_{0}^{1} 16 x^{3} 1 - x^{2} d x

Take the constant out:

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

= 16 \cdot \int_{0}^{1} x^{3} 1 - x^{2} d x

Apply u-substitution

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

= 16 \cdot \int_{1}^{0} - \frac{( - u + 1 ) u}{2} d u

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

= 16 (- \int_{0}^{1} - \frac{( - u + 1 ) u}{2} d u)

Take the constant out:

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

= 16 (- (- \frac{1}{2} \cdot \int_{0}^{1} (- u + 1) u d u))

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

= 16 (- (- \frac{1}{2} \cdot \int_{0}^{1} - u^{\frac{3}{2}} + u d u))

Apply the Sum Rule:

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

= 16 (- (- \frac{1}{2} (- \int_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u)))

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

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

= 16 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3})))

Simplify

16 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3}))) : \frac{32}{15}

= \frac{32}{15}

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What is the integral from 0 to 1 of 16x^3sqrt(1-x^2) ?
The integral from 0 to 1 of 16x^3sqrt(1-x^2) is 32/15