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

The integral from 0 to 1 of-x^3+4x^2-3x is -5/12

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

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

- \frac{5}{12}

Decimal Notation

- 0.41666 \dots

Solution steps

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

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

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

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

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

Simplify

- \frac{1}{4} + \frac{4}{3} - \frac{3}{2} : - \frac{5}{12}

= - \frac{5}{12}

\int_{- 5}^{0} (5 + 25 - x^{2}) d x

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\int_{1}^{8} (x^{2} - 4 x - 12) d x

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\int_{2}^{4} 16 - r^{2} r d r

What is the integral from 0 to 1 of-x^3+4x^2-3x ?
The integral from 0 to 1 of-x^3+4x^2-3x is -5/12