What is the integral from 0 to 1 of x^2(x^3-3)^4 ?

The integral from 0 to 1 of x^2(x^3-3)^4 is 211/15

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integral from 0 to 1 of x^2(x^3-3)^4

\int_{0}^{1} x^{2} (x^{3} - 3)^{4} d x

\frac{211}{15}

Decimal

14.06666 \dots

Solution steps

\int_{0}^{1} x^{2} (x^{3} - 3)^{4} d x

Apply u-substitution

: \int_{0}^{1} \frac{( u - 3 ) ^{4}}{3} d u

= \int_{0}^{1} \frac{( u - 3 ) ^{4}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} (u - 3)^{4} d u

Apply u-substitution

: \int_{- 3}^{- 2} v^{4} d v

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

Apply the Power Rule

: [\frac{v ^{5}}{5}]_{- 3}^{- 2}

= \frac{1}{3} [\frac{v ^{5}}{5}]_{- 3}^{- 2}

Compute the boundaries:

\frac{211}{5}

= \frac{1}{3} \cdot \frac{211}{5}

Simplify

= \frac{211}{15}

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What is the integral from 0 to 1 of x^2(x^3-3)^4 ?
The integral from 0 to 1 of x^2(x^3-3)^4 is 211/15