What is the integral from 0 to 5/2 of c(2x-x^2) ?

The integral from 0 to 5/2 of c(2x-x^2) is 25/24 c

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integral from 0 to 5/2 of c(2x-x^2)

\int_{0}^{\frac{5}{2}} c (2 x - x^{2}) d x

\frac{25}{24} c

Solution steps

\int_{0}^{\frac{5}{2}} c (2 x - x^{2}) d x

Take the constant out:

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

= c \cdot \int_{0}^{\frac{5}{2}} 2 x - x^{2} 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}^{\frac{5}{2}} 2 x d x - \int_{0}^{\frac{5}{2}} x^{2} d x)

\int_{0}^{\frac{5}{2}} 2 x d x = \frac{25}{4}

\int_{0}^{\frac{5}{2}} x^{2} d x = \frac{125}{24}

= c (\frac{25}{4} - \frac{125}{24})

Join

\frac{25}{4} - \frac{125}{24} : \frac{25}{24}

= \frac{25}{24} c

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What is the integral from 0 to 5/2 of c(2x-x^2) ?
The integral from 0 to 5/2 of c(2x-x^2) is 25/24 c