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

The integral from 0 to 1 of x^3x^2 is 1/6

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

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Simplify

x^{3} x^{2} : x^{5}

= \int_{0}^{1} x^{5} d x

Apply the Power Rule

: [\frac{x ^{6}}{6}]_{0}^{1}

= [\frac{x ^{6}}{6}]_{0}^{1}

Compute the boundaries:

\frac{1}{6}

= \frac{1}{6}

\frac{\partial}{\partial x} (e^{- x^{4} - y^{3}})

\int_{1}^{1} x d x

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

\frac{d ^{2}}{d x ^{2}} (ln (2 x + 4))

y = x^{2}, y = 20 - x

What is the integral from 0 to 1 of x^3x^2 ?
The integral from 0 to 1 of x^3x^2 is 1/6