What is the integral from 2 to 4 of k[1-(x-3)^2] ?

The integral from 2 to 4 of k[1-(x-3)^2] is 4/3 k

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integral from 2 to 4 of k[1-(x-3)^2]

\int_{2}^{4} k [1 - (x - 3)^{2}] d x

\frac{4}{3} k

Solution steps

\int_{2}^{4} k [1 - (x - 3)^{2}] d x

Take the constant out:

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

= k \cdot \int_{2}^{4} 1 - (x - 3)^{2} d x

Expand

1 - (x - 3)^{2} : - x^{2} + 6 x - 8

= k \cdot \int_{2}^{4} - x^{2} + 6 x - 8 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_{2}^{4} x^{2} d x + \int_{2}^{4} 6 x d x - \int_{2}^{4} 8 d x)

\int_{2}^{4} x^{2} d x = \frac{56}{3}

\int_{2}^{4} 6 x d x = 36

\int_{2}^{4} 8 d x = 16

= k (- \frac{56}{3} + 36 - 16)

Simplify

k (- \frac{56}{3} + 36 - 16) : \frac{4}{3} k

= \frac{4}{3} k

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What is the integral from 2 to 4 of k[1-(x-3)^2] ?
The integral from 2 to 4 of k[1-(x-3)^2] is 4/3 k