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

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

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integral from 0 to 1 of 2x^2

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{0}^{1} 2 x^{2} 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} 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 \frac{1}{x ^{2} - 2 x - 8} d x

f (x) = x^{2} - 20 x, g (x) = 0

\frac{d}{d x} (ln (\frac{1}{6 x}))

(ln (1 - 7 x))^{^{'}}

6 cos (0.2 t)

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