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

The integral from 0 to 1 of x^2(3+4x^3)^3 is 145/3

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

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

\frac{145}{3}

Decimal

48.33333 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int_{3}^{7} \frac{v ^{3}}{4} d v

= \frac{1}{3} \cdot \int_{3}^{7} \frac{v ^{3}}{4} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{4} \cdot \int_{3}^{7} v^{3} d v

Apply the Power Rule

: [\frac{v ^{4}}{4}]_{3}^{7}

= \frac{1}{3} \cdot \frac{1}{4} [\frac{v ^{4}}{4}]_{3}^{7}

Simplify

\frac{1}{3} \cdot \frac{1}{4} [\frac{v ^{4}}{4}]_{3}^{7} : \frac{1}{12} [\frac{v ^{4}}{4}]_{3}^{7}

= \frac{1}{12} [\frac{v ^{4}}{4}]_{3}^{7}

Compute the boundaries:

580

= \frac{1}{12} \cdot 580

Simplify

= \frac{145}{3}

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What is the integral from 0 to 1 of x^2(3+4x^3)^3 ?
The integral from 0 to 1 of x^2(3+4x^3)^3 is 145/3