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

The integral from 0 to 1 of (x^3)^2 is 1/7

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

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

\frac{1}{7}

Decimal Notation

0.14285 \dots

Solution steps

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

Simplify

(x^{3})^{2} : x^{6}

= \int_{0}^{1} x^{6} d x

Apply the Power Rule

: [\frac{x ^{7}}{7}]_{0}^{1}

= [\frac{x ^{7}}{7}]_{0}^{1}

Compute the boundaries:

\frac{1}{7}

= \frac{1}{7}

\frac{d y}{d x}, e^{x} y = 5

y = \frac{4 x}{x + 2}

\frac{\partial}{\partial x} (e^{s i n (x + c t)})

t \frac{d y}{d t} = t^{3} + 11 t^{3} y

What is the integral from 0 to 1 of (x^3)^2 ?
The integral from 0 to 1 of (x^3)^2 is 1/7