What is the integral from 0 to 1 of-t^2 ?

The integral from 0 to 1 of-t^2 is -1/3

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integral from 0 to 1 of-t^2

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

- \frac{1}{3}

Decimal

- 0.33333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: [\frac{t ^{3}}{3}]_{0}^{1}

= - [\frac{t ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= - \frac{1}{3}

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