What is the integral from 0 to 4 of sqrt(t) ?

The integral from 0 to 4 of sqrt(t) is 16/3

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integral from 0 to 4 of sqrt(t)

\int_{0}^{4} t d t

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{4} t d t

Apply the Power Rule

: [\frac{2}{3} t^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} t^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\int_{1}^{2} \frac{5}{x} d x

\int_{3}^{4} 4 x d x

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\int_{1}^{2} x e^{- x^{2}} d x

\int_{0}^{3} (x^{2} - 3 x) d x

What is the integral from 0 to 4 of sqrt(t) ?
The integral from 0 to 4 of sqrt(t) is 16/3