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

The integral from 0 to 1 of 2x(x^2+3)^4 is 781/5

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

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

\frac{781}{5}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{3}^{4} \frac{u ^{4}}{2} d u

= 2 \cdot \int_{3}^{4} \frac{u ^{4}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{3}^{4} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{3}^{4}

= 2 \cdot \frac{1}{2} [\frac{u ^{5}}{5}]_{3}^{4}

Simplify

2 \cdot \frac{1}{2} [\frac{u ^{5}}{5}]_{3}^{4} : [\frac{u ^{5}}{5}]_{3}^{4}

= [\frac{u ^{5}}{5}]_{3}^{4}

Compute the boundaries:

\frac{781}{5}

= \frac{781}{5}

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What is the integral from 0 to 1 of 2x(x^2+3)^4 ?
The integral from 0 to 1 of 2x(x^2+3)^4 is 781/5