What is the integral from 0 to 6 of 1/18 x ?

The integral from 0 to 6 of 1/18 x is 1

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integral from 0 to 6 of 1/18 x

\int_{0}^{6} \frac{1}{18} x d x

1

Solution steps

\int_{0}^{6} \frac{1}{18} x d x

Take the constant out:

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

= \frac{1}{18} \cdot \int_{0}^{6} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{6}

= \frac{1}{18} [\frac{x ^{2}}{2}]_{0}^{6}

Compute the boundaries:

18

= \frac{1}{18} \cdot 18

Simplify

= 1

\frac{d}{d x} (x π)

\frac{1}{( s + 3 ) ^{2}}

\frac{5 x}{4 - x ^{2}}

8 x^{2} cos (x) cot (x)

y^{^{'}} = \frac{6 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

What is the integral from 0 to 6 of 1/18 x ?
The integral from 0 to 6 of 1/18 x is 1