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

The integral from 0 to 2 of (4x-2x^2) is 8/3

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integral from 0 to 2 of (4x-2x^2)

\int_{0}^{2} (4 x - 2 x^{2}) d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

= 8 - \frac{16}{3}

Simplify

8 - \frac{16}{3} : \frac{8}{3}

= \frac{8}{3}

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

x \to - \infty lim (\frac{4}{x})

\int \frac{cos ( x )}{sin ^{2} ( x ) - 4} d x

\int 4 sec^{4} (x) d x

y^{^{''}} + 4 y = 8

What is the integral from 0 to 2 of (4x-2x^2) ?
The integral from 0 to 2 of (4x-2x^2) is 8/3