What is the integral of (k/(2018))x^2 ?

The integral of (k/(2018))x^2 is (kx^3)/(6054)+C

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integral of (k/(2018))x^2

\int (\frac{k}{2018}) x^{2} d x

\frac{k x ^{3}}{6054} + C

Solution steps

\int \frac{k}{2018} x^{2} d x

Take the constant out:

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

= \frac{k}{2018} \cdot \int x^{2} d x

Apply the Power Rule

: \frac{x ^{3}}{3}

= \frac{k}{2018} \cdot \frac{x ^{3}}{3}

Simplify

\frac{k}{2018} \cdot \frac{x ^{3}}{3} : \frac{k x ^{3}}{6054}

= \frac{k x ^{3}}{6054}

Add a constant to the solution

= \frac{k x ^{3}}{6054} + C

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What is the integral of (k/(2018))x^2 ?
The integral of (k/(2018))x^2 is (kx^3)/(6054)+C