What is the integral from 1 to 10 of (2x(x-1))/(81) ?

The integral from 1 to 10 of (2x(x-1))/(81) is 7

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integral from 1 to 10 of (2x(x-1))/(81)

\int_{1}^{10} \frac{2 x ( x - 1 )}{81} d x

7

Solution steps

\int_{1}^{10} \frac{2 x ( x - 1 )}{81} d x

Take the constant out:

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

= \frac{2}{81} \cdot \int_{1}^{10} x (x - 1) d x

Expand

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

= \frac{2}{81} \cdot \int_{1}^{10} x^{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}{81} (\int_{1}^{10} x^{2} d x - \int_{1}^{10} x d x)

\int_{1}^{10} x^{2} d x = 333

\int_{1}^{10} x d x = \frac{99}{2}

= \frac{2}{81} (333 - \frac{99}{2})

Simplify

\frac{2}{81} (333 - \frac{99}{2}) : 7

= 7

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What is the integral from 1 to 10 of (2x(x-1))/(81) ?
The integral from 1 to 10 of (2x(x-1))/(81) is 7