What is the integral from 0 to 1 of k(1-x) ?

The integral from 0 to 1 of k(1-x) is 1/2 k

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integral from 0 to 1 of k(1-x)

\int_{0}^{1} k (1 - x) d x

\frac{1}{2} k

Solution steps

\int_{0}^{1} k (1 - x) d x

Take the constant out:

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

= k \cdot \int_{0}^{1} 1 - x 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

= k (\int_{0}^{1} 1 d x - \int_{0}^{1} x d x)

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} x d x = \frac{1}{2}

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

Join

1 - \frac{1}{2} : \frac{1}{2}

= \frac{1}{2} k

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What is the integral from 0 to 1 of k(1-x) ?
The integral from 0 to 1 of k(1-x) is 1/2 k