What is the integral from 0 to 3 of c(3-x) ?

The integral from 0 to 3 of c(3-x) is 9/2 c

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integral from 0 to 3 of c(3-x)

\int_{0}^{3} c (3 - x) d x

\frac{9}{2} c

Solution steps

\int_{0}^{3} c (3 - x) d x

Take the constant out:

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

= c \cdot \int_{0}^{3} 3 - x 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

= c (\int_{0}^{3} 3 d x - \int_{0}^{3} x d x)

\int_{0}^{3} 3 d x = 9

\int_{0}^{3} x d x = \frac{9}{2}

= c (9 - \frac{9}{2})

Join

9 - \frac{9}{2} : \frac{9}{2}

= \frac{9}{2} c

\int \frac{5 x - 3}{x + 1} d x

f (x) = \frac{4}{3 x ^{2}}

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

\int - \frac{18}{x ^{19}} d x

\frac{d x}{d t} - \frac{x}{t - 1} = t^{2} + 2

What is the integral from 0 to 3 of c(3-x) ?
The integral from 0 to 3 of c(3-x) is 9/2 c