What is the integral from 0 to 1 of c(1-x^3) ?

The integral from 0 to 1 of c(1-x^3) is 3/4 c

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integral from 0 to 1 of c(1-x^3)

\int_{0}^{1} c (1 - x^{3}) d x

\frac{3}{4} c

Solution steps

\int_{0}^{1} c (1 - x^{3}) 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} 1 - 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} 1 d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 1 d x = 1

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

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

Join

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

= \frac{3}{4} c

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What is the integral from 0 to 1 of c(1-x^3) ?
The integral from 0 to 1 of c(1-x^3) is 3/4 c