What is the integral from-k to k of (x-1/k)^3 ?

The integral from-k to k of (x-1/k)^3 is -2k^2-2/(k^2)

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integral from-k to k of (x-1/k)^3

\int_{- k}^{k} (x - \frac{1}{k})^{3} d x

- 2 k^{2} - \frac{2}{k ^{2}}

Solution steps

\int_{- k}^{k} (x - \frac{1}{k})^{3} d x

Expand

(x - \frac{1}{k})^{3} : x^{3} - \frac{3 x ^{2}}{k} + \frac{3 x}{k ^{2}} - \frac{1}{k ^{3}}

= \int_{- k}^{k} x^{3} - \frac{3 x ^{2}}{k} + \frac{3 x}{k ^{2}} - \frac{1}{k ^{3}} 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

= \int_{- k}^{k} x^{3} d x - \int_{- k}^{k} \frac{3 x ^{2}}{k} d x + \int_{- k}^{k} \frac{3 x}{k ^{2}} d x - \int_{- k}^{k} \frac{1}{k ^{3}} d x

\int_{- k}^{k} x^{3} d x = 0

\int_{- k}^{k} \frac{3 x ^{2}}{k} d x = 2 k^{2}

\int_{- k}^{k} \frac{3 x}{k ^{2}} d x = 0

\int_{- k}^{k} \frac{1}{k ^{3}} d x = \frac{2}{k ^{2}}

= 0 - 2 k^{2} + 0 - \frac{2}{k ^{2}}

Simplify

= - 2 k^{2} - \frac{2}{k ^{2}}

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What is the integral from-k to k of (x-1/k)^3 ?
The integral from-k to k of (x-1/k)^3 is -2k^2-2/(k^2)