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

The integral from 0 to 1 of 3x^2(1+x^3)^5 is 21/2

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

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

\frac{21}{2}

Decimal Notation

10.5

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{1}^{2}

= 3 \cdot \frac{1}{3} [\frac{u ^{6}}{6}]_{1}^{2}

Simplify

3 \cdot \frac{1}{3} [\frac{u ^{6}}{6}]_{1}^{2} : [\frac{u ^{6}}{6}]_{1}^{2}

= [\frac{u ^{6}}{6}]_{1}^{2}

Compute the boundaries:

\frac{21}{2}

= \frac{21}{2}

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What is the integral from 0 to 1 of 3x^2(1+x^3)^5 ?
The integral from 0 to 1 of 3x^2(1+x^3)^5 is 21/2