What is the integral from 0 to 8 of k(8-x) ?

The integral from 0 to 8 of k(8-x) is (8-x)*32

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integral from 0 to 8 of k(8-x)

\int_{0}^{8} k (8 - x)

(8 - x) \cdot 32

Solution steps

\int_{0}^{8} k (8 - x) d k

Take the constant out:

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

= (8 - x) \cdot \int_{0}^{8} k d k

\int_{0}^{8} k d k = 32

= (8 - x) \cdot 32

f (x) = - 6 x^{2} + 7 x

f (x) = - 1 - 4 e^{- \frac{x}{3}}, at x = 8

x y^{^{'}} - y = 2 x^{2} - x - 1

\int sin^{2} (\frac{π x}{l}) d x

\frac{d}{d x} (2 (x + 2)^{\frac{3}{2}})

What is the integral from 0 to 8 of k(8-x) ?
The integral from 0 to 8 of k(8-x) is (8-x)*32