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

The integral from 0 to 1 of x^3(2x^4+1)^2 is 13/12

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

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

\frac{13}{12}

Decimal Notation

1.08333 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{u ^{2}}{8} d u

= \int_{1}^{3} \frac{u ^{2}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{3} u^{2} d u

Apply the Power Rule

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

= \frac{1}{8} [\frac{u ^{3}}{3}]_{1}^{3}

Compute the boundaries:

\frac{26}{3}

= \frac{1}{8} \cdot \frac{26}{3}

Simplify

= \frac{13}{12}

\int \frac{cos ^{2} ( x )}{x} d x

\frac{d ^{3}}{d x ^{3}} (\frac{2}{x})

\int_{1}^{2} \frac{4}{x ^{3} + x} d x

x \to 0 lim ((1 + 3 x)^{\frac{1}{x}})

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} + 3 z^{2} - 9)

What is the integral from 0 to 1 of x^3(2x^4+1)^2 ?
The integral from 0 to 1 of x^3(2x^4+1)^2 is 13/12