What is the integral from 0 to 18 of x(1/162 x) ?

The integral from 0 to 18 of x(1/162 x) is 12

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integral from 0 to 18 of x(1/162 x)

\int_{0}^{18} x (\frac{1}{162} x) d x

12

Solution steps

\int_{0}^{18} x \frac{1}{162} x d x

Take the constant out:

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

= \frac{1}{162} \cdot \int_{0}^{18} xx d x

Simplify

xx : x^{2}

= \frac{1}{162} \cdot \int_{0}^{18} x^{2} d x

Apply the Power Rule

= \frac{1}{162} [\frac{x ^{3}}{3}]_{0}^{18}

Compute the boundaries:

1944

= \frac{1}{162} \cdot 1944

Simplify

= 12

What is the integral from 0 to 18 of x(1/162 x) ?
The integral from 0 to 18 of x(1/162 x) is 12