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

The integral from 1 to 10 of x^2*2/81 (x-1) is 107/2

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

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

\frac{107}{2}

Decimal

53.5

Solution steps

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

Expand

x^{2} (x - 1) : x^{3} - x^{2}

= \frac{2}{81} \cdot \int_{1}^{10} x^{3} - 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

= \frac{2}{81} (\int_{1}^{10} x^{3} d x - \int_{1}^{10} x^{2} d x)

\int_{1}^{10} x^{3} d x = \frac{9999}{4}

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

= \frac{2}{81} (\frac{9999}{4} - 333)

Simplify

\frac{2}{81} (\frac{9999}{4} - 333) : \frac{107}{2}

= \frac{107}{2}

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What is the integral from 1 to 10 of x^2*2/81 (x-1) ?
The integral from 1 to 10 of x^2*2/81 (x-1) is 107/2