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

The integral from 0 to 1 of 4x(x^2+3)^3 is 175/2

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

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

\frac{175}{2}

Decimal

87.5

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Simplify

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

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

Compute the boundaries:

\frac{175}{4}

= 2 \cdot \frac{175}{4}

Simplify

= \frac{175}{2}

What is the integral from 0 to 1 of 4x(x^2+3)^3 ?
The integral from 0 to 1 of 4x(x^2+3)^3 is 175/2