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

The integral from 0 to 2 of (-x^2+2x) is 4/3

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 2 x d x = 4

= - \frac{8}{3} + 4

Simplify

- \frac{8}{3} + 4 : \frac{4}{3}

= \frac{4}{3}

f (x) = 2 x^{- 2}

k = 0 \sum \infty \frac{1}{2 ^{k}}

x \to \frac{π}{3} lim (2 sin (x))

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

\int \frac{1}{5 x + 8} d x

What is the integral from 0 to 2 of (-x^2+2x) ?
The integral from 0 to 2 of (-x^2+2x) is 4/3