What is the integral from 0 to 1 of 9x^2(1-5x^3)^2 ?

The integral from 0 to 1 of 9x^2(1-5x^3)^2 is 13

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integral from 0 to 1 of 9x^2(1-5x^3)^2

\int_{0}^{1} 9 x^{2} (1 - 5 x^{3})^{2} d x

13

Solution steps

\int_{0}^{1} 9 x^{2} (1 - 5 x^{3})^{2} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{1} x^{2} (1 - 5 x^{3})^{2} d x

Apply u-substitution

: \int_{0}^{1} \frac{( 1 - 5 u ) ^{2}}{3} d u

= 9 \cdot \int_{0}^{1} \frac{( 1 - 5 u ) ^{2}}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int_{0}^{1} (1 - 5 u)^{2} d u

Apply u-substitution

: \int_{1}^{- 4} - \frac{v ^{2}}{5} d v

= 9 \cdot \frac{1}{3} \cdot \int_{1}^{- 4} - \frac{v ^{2}}{5} d v

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

= 9 \cdot \frac{1}{3} (- \int_{- 4}^{1} - \frac{v ^{2}}{5} d v)

Take the constant out:

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

= 9 \cdot \frac{1}{3} (- (- \frac{1}{5} \cdot \int_{- 4}^{1} v^{2} d v))

Apply the Power Rule

: [\frac{v ^{3}}{3}]_{- 4}^{1}

= 9 \cdot \frac{1}{3} (- (- \frac{1}{5} [\frac{v ^{3}}{3}]_{- 4}^{1}))

Simplify

9 \cdot \frac{1}{3} (- (- \frac{1}{5} [\frac{v ^{3}}{3}]_{- 4}^{1})) : \frac{3}{5} [\frac{v ^{3}}{3}]_{- 4}^{1}

= \frac{3}{5} [\frac{v ^{3}}{3}]_{- 4}^{1}

Compute the boundaries:

\frac{65}{3}

= \frac{3}{5} \cdot \frac{65}{3}

Simplify

= 13

n = 1 \sum \infty \frac{n + 1}{2 n - 3}

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What is the integral from 0 to 1 of 9x^2(1-5x^3)^2 ?
The integral from 0 to 1 of 9x^2(1-5x^3)^2 is 13