What is the integral from 0 to 1 of x(2(x+2))/5 ?

The integral from 0 to 1 of x(2(x+2))/5 is 8/15

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integral from 0 to 1 of x(2(x+2))/5

\int_{0}^{1} x \frac{2 ( x + 2 )}{5} d x

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

\int_{0}^{1} x \frac{2 ( x + 2 )}{5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{5} \cdot \int_{0}^{1} x (x + 2) d x

Expand

x (x + 2) : x^{2} + 2 x

= \frac{2}{5} \cdot \int_{0}^{1} 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

= \frac{2}{5} (\int_{0}^{1} x^{2} d x + \int_{0}^{1} 2 x d x)

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

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

= \frac{2}{5} (\frac{1}{3} + 1)

Simplify

\frac{2}{5} (\frac{1}{3} + 1) : \frac{8}{15}

= \frac{8}{15}

d er i v a t i v e f (x) = x (x - 4)^{3}

\int e^{- s t} e^{- 3 t} d t

a re a y = x, y = 14 x, x = 25

\frac{d y}{d x} = \frac{1}{x} (y^{2} + y)

\int y 10 + 4 y - 4 y^{2} d y

What is the integral from 0 to 1 of x(2(x+2))/5 ?
The integral from 0 to 1 of x(2(x+2))/5 is 8/15