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

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

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integral from 0 to 1 of x(2x)

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{0}^{1} x \cdot 2 x 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} xx d x

Simplify

xx : x^{2}

= 2 \cdot \int_{0}^{1} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{0}^{1}

= 2 [\frac{x ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= 2 \cdot \frac{1}{3}

Simplify

= \frac{2}{3}

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

\int_{0}^{1} \frac{5 x ^{4}}{x ^{5} + 6} d x

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

\int_{0}^{1} \frac{10}{1 + p ^{2}} d p

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

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