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

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

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integral from 1 to 2 of k+(1-k)(x-1)^2

\int_{1}^{2} k + (1 - k) (x - 1)^{2} d x

\frac{2}{3} k + \frac{1}{3}

Solution steps

\int_{1}^{2} k + (1 - k) (x - 1)^{2} 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_{1}^{2} k d x + \int_{1}^{2} (1 - k) (x - 1)^{2} d x

\int_{1}^{2} k d x = k

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

= k + (1 - k) \frac{1}{3}

Simplify

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

= \frac{2}{3} k + \frac{1}{3}

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