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

The integral from-1 to 1 of k(3-x^2) is 16/3 k

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

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

\frac{16}{3} k

Solution steps

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

\int_{- 1}^{1} 3 d x = 6

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

= k (6 - \frac{2}{3})

Join

6 - \frac{2}{3} : \frac{16}{3}

= \frac{16}{3} k

What is the integral from-1 to 1 of k(3-x^2) ?
The integral from-1 to 1 of k(3-x^2) is 16/3 k