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Popular Calculus >

integral from 0 to 1 of (2-t^2)

Solution

\int_{0}^{1} (2 - t^{2}) d t

Solution

\frac{5}{3}

+1

Decimal

1.66666 \dots

Solution steps

\int_{0}^{1} 2 - 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_{0}^{1} 2 d t - \int_{0}^{1} t^{2} d t

\int_{0}^{1} 2 d t = 2

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

= 2 - \frac{1}{3}

Simplify

2 - \frac{1}{3} : \frac{5}{3}

= \frac{5}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of (2-t^2) ?
The integral from 0 to 1 of (2-t^2) is 5/3