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Popular Calculus >

integral from 3 to 2 of (2t-t^2)

Solution

\int_{3}^{2} (2 t - t^{2}) d t

Solution

\frac{4}{3}

+1

Decimal

1.33333 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{3} 2 t - t^{2} 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_{2}^{3} 2 t d t - \int_{2}^{3} t^{2} d t)

\int_{2}^{3} 2 t d t = 5

\int_{2}^{3} t^{2} d t = \frac{19}{3}

= - (5 - \frac{19}{3})

Simplify

- (5 - \frac{19}{3}) : \frac{4}{3}

= \frac{4}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 3 to 2 of (2t-t^2) ?
The integral from 3 to 2 of (2t-t^2) is 4/3