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

The integral from 0 to 1 of (x^3-4x^2) is -13/12

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

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

- \frac{13}{12}

Decimal

- 1.08333 \dots

Solution steps

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

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

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

Simplify

\frac{1}{4} - \frac{4}{3} : - \frac{13}{12}

= - \frac{13}{12}

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What is the integral from 0 to 1 of (x^3-4x^2) ?
The integral from 0 to 1 of (x^3-4x^2) is -13/12