What is the integral from 0 to 1 of 2x(8-x^2) ?

The integral from 0 to 1 of 2x(8-x^2) is 15/2

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integral from 0 to 1 of 2x(8-x^2)

\int_{0}^{1} 2 x (8 - x^{2}) d x

\frac{15}{2}

Decimal Notation

7.5

Solution steps

\int_{0}^{1} 2 x (8 - x^{2}) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x (8 - x^{2}) d x

Expand

x (8 - x^{2}) : 8 x - x^{3}

= 2 \cdot \int_{0}^{1} 8 x - 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

= 2 (\int_{0}^{1} 8 x d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 8 x d x = 4

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 2 (4 - \frac{1}{4})

Simplify

2 (4 - \frac{1}{4}) : \frac{15}{2}

= \frac{15}{2}

x \to - 3 lim (- 3 x)

\frac{\partial}{\partial x} (x^{3} - 2 x y^{2} - y^{3})

\int_{0}^{0.5} (200 x) d x

\frac{1}{u}

\frac{2}{s ( s ^{2} + 4 s + 6 )}

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