What is the integral from 0 to 2 of 6x(2-x) ?

The integral from 0 to 2 of 6x(2-x) is 8

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integral from 0 to 2 of 6x(2-x)

\int_{0}^{2} 6 x (2 - x) d x

8

Solution steps

\int_{0}^{2} 6 x (2 - x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2} x (2 - x) d x

Expand

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

= 6 \cdot \int_{0}^{2} 2 x - 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

= 6 (\int_{0}^{2} 2 x d x - \int_{0}^{2} x^{2} d x)

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

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

= 6 (4 - \frac{8}{3})

Simplify

6 (4 - \frac{8}{3}) : 8

= 8

(2 x - 3) e^{2 - x}

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

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

y = x^{2} - 2, (- 3, 7)

3 x y^{^{'}} = y^{2} - y - 2

What is the integral from 0 to 2 of 6x(2-x) ?
The integral from 0 to 2 of 6x(2-x) is 8