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

The integral from 0 to 1 of k(1-x^2) is 2/3 k

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integral from 0 to 1 of k(1-x^2)

\int_{0}^{1} k (1 - x^{2}) d x

\frac{2}{3} k

Solution steps

\int_{0}^{1} k (1 - x^{2}) 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^{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

= k (\int_{0}^{1} 1 d x - \int_{0}^{1} x^{2} d x)

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

Join

1 - \frac{1}{3} : \frac{2}{3}

= \frac{2}{3} k

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What is the integral from 0 to 1 of k(1-x^2) ?
The integral from 0 to 1 of k(1-x^2) is 2/3 k