What is the integral from 0 to 1 of (-x^3-2x^2+3x) ?

The integral from 0 to 1 of (-x^3-2x^2+3x) is 7/12

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integral from 0 to 1 of (-x^3-2x^2+3x)

\int_{0}^{1} (- x^{3} - 2 x^{2} + 3 x) d x

\frac{7}{12}

Decimal

0.58333 \dots

Solution steps

\int_{0}^{1} - x^{3} - 2 x^{2} + 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

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

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

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

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

= - \frac{1}{4} - \frac{2}{3} + \frac{3}{2}

Simplify

- \frac{1}{4} - \frac{2}{3} + \frac{3}{2} : \frac{7}{12}

= \frac{7}{12}

\int_{1}^{3 x} (x^{3} e^{x y}) d y

\int_{1}^{4} 1 + \frac{1}{u} + \frac{1}{u ^{2}} d u

\int_{x}^{x^{3}} e^{t^{2}} d t

\int_{0}^{2} 4 - y^{2} d y

\int_{0}^{1} 3^{5 x} d x

What is the integral from 0 to 1 of (-x^3-2x^2+3x) ?
The integral from 0 to 1 of (-x^3-2x^2+3x) is 7/12