What is the integral from 0 to 9 of k(9-x) ?

The integral from 0 to 9 of k(9-x) is (9-x) 81/2

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integral from 0 to 9 of k(9-x)

\int_{0}^{9} k (9 - x)

(9 - x) \frac{81}{2}

Solution steps

\int_{0}^{9} k (9 - x) d k

Take the constant out:

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

= (9 - x) \cdot \int_{0}^{9} k d k

\int_{0}^{9} k d k = \frac{81}{2}

= (9 - x) \frac{81}{2}

\frac{d}{d x} (arctan (\frac{a ^{2} - x ^{3}}{b ^{2}}))

\frac{d}{d x} (x \cdot (x^{5} + x^{4} + 3 x + 2))

\frac{\partial ^{2}}{\partial K ( b ) \partial b} (a K (b) (b)^{a} L^{b})

\frac{d y}{d x} - 6 x y = sin (x^{2})

\frac{d}{d x} (\frac{4 x}{( 3 x ^{2} - 1 ) ^{2}})

What is the integral from 0 to 9 of k(9-x) ?
The integral from 0 to 9 of k(9-x) is (9-x) 81/2