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Popular Calculus >

integral from-1 to 2 of (1-t^2)t

Solution

\int_{- 1}^{2} (1 - t^{2}) t d t

Solution

- \frac{9}{4}

+1

Decimal

- 2.25

Solution steps

\int_{- 1}^{2} (1 - t^{2}) t d t

Expand

(1 - t^{2}) t : t - t^{3}

= \int_{- 1}^{2} t - t^{3} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{2} t d t - \int_{- 1}^{2} t^{3} d t

\int_{- 1}^{2} t d t = \frac{3}{2}

\int_{- 1}^{2} t^{3} d t = \frac{15}{4}

= \frac{3}{2} - \frac{15}{4}

Simplify

\frac{3}{2} - \frac{15}{4} : - \frac{9}{4}

= - \frac{9}{4}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from-1 to 2 of (1-t^2)t ?
The integral from-1 to 2 of (1-t^2)t is -9/4