What is the integral from 0 to 2pi of (t)^2 ?

The integral from 0 to 2pi of (t)^2 is (8pi^3)/3

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integral from 0 to 2pi of (t)^2

\int_{0}^{2 π} (t)^{2} d t

\frac{8 π ^{3}}{3}

Decimal Notation

82.68340 \dots

Solution steps

\int_{0}^{2 π} t^{2} d t

Apply the Power Rule

: [\frac{t ^{3}}{3}]_{0}^{2 π}

= [\frac{t ^{3}}{3}]_{0}^{2 π}

Compute the boundaries:

\frac{8 π ^{3}}{3}

= \frac{8 π ^{3}}{3}

\int_{1}^{2} 4 r^{2} + 1 \cdot r d r

\int_{0}^{π} \frac{2 - x}{x ^{2}} d x

\int_{0}^{1} (7 x - 4) d x

\int_{0}^{t} sin^{2} (wt) d t

\int_{0}^{π} \frac{1}{aπ + b x} d x

What is the integral from 0 to 2pi of (t)^2 ?
The integral from 0 to 2pi of (t)^2 is (8pi^3)/3