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

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

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

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

1

Solution steps

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

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

\int_{1}^{10} 1 d x = 9

= \frac{2}{81} (\frac{99}{2} - 9)

Simplify

\frac{2}{81} (\frac{99}{2} - 9) : 1

= 1

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