What is the integral from 1 to 4 of x^{3/2} ?

The integral from 1 to 4 of x^{3/2} is 62/5

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integral from 1 to 4 of x^{3/2}

\int_{1}^{4} x^{\frac{3}{2}} d x

\frac{62}{5}

Decimal Notation

12.4

Solution steps

\int_{1}^{4} x^{\frac{3}{2}} d x

Apply the Power Rule

: [\frac{2}{5} x^{\frac{5}{2}}]_{1}^{4}

= [\frac{2}{5} x^{\frac{5}{2}}]_{1}^{4}

Compute the boundaries:

\frac{62}{5}

= \frac{62}{5}

\int sin (3) (x) d x

\frac{d}{d x} ((x^{2} + y^{2})^{\frac{1}{2}})

x^{6} + y^{5} x^{7} + 4 y^{^{'}} = 0

\frac{d}{d x} (arccos (\frac{7}{x}))

\int x^{4} - x^{3} + x^{2} d x

What is the integral from 1 to 4 of x^{3/2} ?
The integral from 1 to 4 of x^{3/2} is 62/5