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

The integral from 0 to 1 of x^2(x^3+1)^4 is 31/15

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

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

\frac{31}{15}

Decimal

2.06666 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{u ^{4}}{3} d u

= \int_{1}^{2} \frac{u ^{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_{1}^{2} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{2}

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

Compute the boundaries:

\frac{31}{5}

= \frac{1}{3} \cdot \frac{31}{5}

Simplify

= \frac{31}{15}

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What is the integral from 0 to 1 of x^2(x^3+1)^4 ?
The integral from 0 to 1 of x^2(x^3+1)^4 is 31/15