What is the integral from 1 to 4 of sqrt(t) ?

The integral from 1 to 4 of sqrt(t) is 14/3

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integral from 1 to 4 of sqrt(t)

\int_{1}^{4} t d t

\frac{14}{3}

Decimal Notation

4.66666 \dots

Solution steps

\int_{1}^{4} t d t

Apply the Power Rule

: [\frac{2}{3} t^{\frac{3}{2}}]_{1}^{4}

= [\frac{2}{3} t^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{14}{3}

y^{^{'}} = e^{y + 2 x}

\int_{0}^{1} (\frac{3 t ^{2}}{1 + t ^{3}}) d t

\int \frac{1}{6 + x} d x

\int 9 + x^{2} d x

(sec (x^{5} + x))^{^{'}}

What is the integral from 1 to 4 of sqrt(t) ?
The integral from 1 to 4 of sqrt(t) is 14/3