What is the integral from 0 to 1 of cx^2(1-x) ?

The integral from 0 to 1 of cx^2(1-x) is 1/12 c

Popular >

integral from 0 to 1 of cx^2(1-x)

\int_{0}^{1} c x^{2} (1 - x) d x

\frac{1}{12} c

Solution steps

\int_{0}^{1} c x^{2} (1 - 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}^{1} x^{2} (1 - x) d x

Expand

x^{2} (1 - x) : x^{2} - x^{3}

= c \cdot \int_{0}^{1} x^{2} - x^{3} 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}^{1} x^{2} d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= c (\frac{1}{3} - \frac{1}{4})

Join

\frac{1}{3} - \frac{1}{4} : \frac{1}{12}

= \frac{1}{12} c

\int_{0}^{8} 8 e^{- 8 x} d x

\int_{1}^{e} x ln (x) - x d x

\int_{0}^{\infty} \frac{4}{x ^{2}} d x

\int_{3}^{5} \frac{sin ( x - 4 )}{x - 4} d x

\int_{0}^{1} \frac{x y ^{2}}{x ^{2} + 1} d x

What is the integral from 0 to 1 of cx^2(1-x) ?
The integral from 0 to 1 of cx^2(1-x) is 1/12 c