What is the integral from 0 to 2 of x^2(1-0.5x) ?

The integral from 0 to 2 of x^2(1-0.5x) is 2/3

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integral from 0 to 2 of x^2(1-0.5x)

\int_{0}^{2} x^{2} (1 - 0.5 x) d x

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{0}^{2} x^{2} (1 - 0.5 x) d x

Expand

x^{2} (1 - 0.5 x) : x^{2} - 0.5 x^{3}

= \int_{0}^{2} x^{2} - 0.5 x^{3} 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} 0.5 x^{3} d x

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

\int_{0}^{2} 0.5 x^{3} d x = 2

= \frac{8}{3} - 2

Simplify

\frac{8}{3} - 2 : \frac{2}{3}

= \frac{2}{3}

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

n = 1 \sum \infty \frac{1}{3 ^{n} + 2}

t cos (2 t)

k = 0 \sum \infty (\frac{e}{10})^{k}

x \to 1 - lim (- x^{2} + 3 x - 4)

What is the integral from 0 to 2 of x^2(1-0.5x) ?
The integral from 0 to 2 of x^2(1-0.5x) is 2/3