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

The integral from 0 to 1 of 3x(x^2+2)^3 is 195/8

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

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

\frac{195}{8}

Decimal

24.375

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2}^{3}

= 3 \cdot \frac{1}{2} [\frac{u ^{4}}{4}]_{2}^{3}

Simplify

3 \cdot \frac{1}{2} [\frac{u ^{4}}{4}]_{2}^{3} : \frac{3}{2} [\frac{u ^{4}}{4}]_{2}^{3}

= \frac{3}{2} [\frac{u ^{4}}{4}]_{2}^{3}

Compute the boundaries:

\frac{65}{4}

= \frac{3}{2} \cdot \frac{65}{4}

Simplify

= \frac{195}{8}

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What is the integral from 0 to 1 of 3x(x^2+2)^3 ?
The integral from 0 to 1 of 3x(x^2+2)^3 is 195/8