What is the integral from 1 to k of k/(2w^2) ?

The integral from 1 to k of k/(2w^2) is (-1+k)/2

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integral from 1 to k of k/(2w^2)

\int_{1}^{k} \frac{k}{2 w ^{2}} d w

\frac{- 1 + k}{2}

Solution steps

\int_{1}^{k} \frac{k}{2 w ^{2}} d w

Take the constant out:

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

= \frac{k}{2} \cdot \int_{1}^{k} \frac{1}{w ^{2}} d w

Apply the Power Rule

: [- \frac{1}{w}]_{1}^{k}

= \frac{k}{2} [- \frac{1}{w}]_{1}^{k}

Compute the boundaries:

- \frac{1}{k} + 1

= \frac{k}{2} (- \frac{1}{k} + 1)

Simplify

= \frac{- 1 + k}{2}

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What is the integral from 1 to k of k/(2w^2) ?
The integral from 1 to k of k/(2w^2) is (-1+k)/2