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

The integral from 0 to 1/2 of t^2 is 1/24

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integral from 0 to 1/2 of t^2

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

\frac{1}{24}

Decimal Notation

0.04166 \dots

Solution steps

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

Apply the Power Rule

: [\frac{t ^{3}}{3}]_{0}^{\frac{1}{2}}

= [\frac{t ^{3}}{3}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{24}

= \frac{1}{24}

\frac{\partial}{\partial x} (8 + e^{- x y} cos (y))

f (x) = \frac{1}{2 + 3 x}, (2, \frac{1}{8})

y = 1 - 2 x

g (x) = (x^{2} + 1)^{3} (x^{2} + 2)^{6}

\frac{s + 3}{s ^{2} + 2 s + 1}

What is the integral from 0 to 1/2 of t^2 ?
The integral from 0 to 1/2 of t^2 is 1/24