What is the integral from 0 to 6 of k(6-x) ?

The integral from 0 to 6 of k(6-x) is (6-x)*18

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integral from 0 to 6 of k(6-x)

\int_{0}^{6} k (6 - x)

(6 - x) \cdot 18

Solution steps

\int_{0}^{6} k (6 - x) d k

Take the constant out:

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

= (6 - x) \cdot \int_{0}^{6} k d k

\int_{0}^{6} k d k = 18

= (6 - x) \cdot 18

\frac{5 ( 1 - s ^{2} )}{( s + 2 ) ^{3}}

\int \frac{( 2 x + 3 )}{( x + 1 ) ^{2}} d x

(e^{y} + 1)^{2} \cdot e^{- y} d x + (e^{x} + 1)^{3} \cdot e^{- x} d y = 0

\int_{- 1}^{1} π (6 - 6 x^{2})^{2} d x

\int_{0}^{2} (4 - x^{2}) d x

What is the integral from 0 to 6 of k(6-x) ?
The integral from 0 to 6 of k(6-x) is (6-x)*18