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

The integral from 0 to 1 of x(4+2x^2)^3 is 65

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

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

65

Solution steps

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

Apply u-substitution

: \int_{4}^{6} \frac{u ^{3}}{4} d u

= \int_{4}^{6} \frac{u ^{3}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{4}^{6} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{4}^{6}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{4}^{6}

Compute the boundaries:

260

= \frac{1}{4} \cdot 260

Simplify

= 65

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What is the integral from 0 to 1 of x(4+2x^2)^3 ?
The integral from 0 to 1 of x(4+2x^2)^3 is 65