What is the integral from 1 to 4 of (x^2)/(sqrt(x)) ?

The integral from 1 to 4 of (x^2)/(sqrt(x)) is 62/5

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integral from 1 to 4 of (x^2)/(sqrt(x))

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

\frac{62}{5}

Decimal

12.4

Solution steps

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

Simplify

\frac{x ^{2}}{x} : x^{\frac{3}{2}}

= \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_{0}^{1} e^{1 - t} d t

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

\int_{- 2}^{1} 4 d x

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

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

What is the integral from 1 to 4 of (x^2)/(sqrt(x)) ?
The integral from 1 to 4 of (x^2)/(sqrt(x)) is 62/5