What is the integral from 0 to 1 of 4x-3x^2-x^3 ?

The integral from 0 to 1 of 4x-3x^2-x^3 is 3/4

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integral from 0 to 1 of 4x-3x^2-x^3

\int_{0}^{1} 4 x - 3 x^{2} - x^{3} d x

\frac{3}{4}

Decimal Notation

0.75

Solution steps

\int_{0}^{1} 4 x - 3 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

= \int_{0}^{1} 4 x d x - \int_{0}^{1} 3 x^{2} d x - \int_{0}^{1} x^{3} d x

\int_{0}^{1} 4 x d x = 2

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

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

= 2 - 1 - \frac{1}{4}

Simplify

2 - 1 - \frac{1}{4} : \frac{3}{4}

= \frac{3}{4}

\int_{0}^{2} 2 x + x d x

\int_{- 5}^{0} x^{2} + 5 x d x

\int_{0}^{\frac{1}{2}} \frac{arctan ( x )}{x} d x

\int_{- 3}^{0} x^{2} + 6 x + 9 d x

\int_{0}^{5} \frac{3 x + 2}{x ^{2} + 9} d x

What is the integral from 0 to 1 of 4x-3x^2-x^3 ?
The integral from 0 to 1 of 4x-3x^2-x^3 is 3/4