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

The integral from 0 to 1 of x^2(1+2x^3)^5 is 182/9

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

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

\frac{182}{9}

Decimal Notation

20.22222 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{u ^{5}}{6} d u

= \int_{1}^{3} \frac{u ^{5}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{1}^{3} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{1}^{3}

= \frac{1}{6} [\frac{u ^{6}}{6}]_{1}^{3}

Compute the boundaries:

\frac{364}{3}

= \frac{1}{6} \cdot \frac{364}{3}

Simplify

= \frac{182}{9}

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What is the integral from 0 to 1 of x^2(1+2x^3)^5 ?
The integral from 0 to 1 of x^2(1+2x^3)^5 is 182/9