What is the integral from 1 to 16 of x/(sqrt(x)) ?

The integral from 1 to 16 of x/(sqrt(x)) is 42

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integral from 1 to 16 of x/(sqrt(x))

\int_{1}^{16} \frac{x}{x} d x

42

Solution steps

\int_{1}^{16} \frac{x}{x} d x

Simplify

\frac{x}{x} : x

= \int_{1}^{16} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{1}^{16}

= [\frac{2}{3} x^{\frac{3}{2}}]_{1}^{16}

Compute the boundaries:

42

= 42

y^{^{'}} = (x - 3) e^{- 2 y}

- \frac{4}{( 5 ) ^{3}}

x \to - 1 lim (x^{2} - x + 7)

f (x) = ln (3 + x)

\frac{d y}{d x} = \frac{y ^{2} + 6 x x ^{2} + y ^{2}}{x y}

What is the integral from 1 to 16 of x/(sqrt(x)) ?
The integral from 1 to 16 of x/(sqrt(x)) is 42