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

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

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

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 1 - \frac{2}{3}

Simplify

1 - \frac{2}{3} : \frac{1}{3}

= \frac{1}{3}

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

\int \frac{2 x + 3}{x} d x

f (x) = 4 - x^{2}, (- 1, 3)

7 x sin (x)

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

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