What is the integral from 0 to 9 of sqrt(x) ?

The integral from 0 to 9 of sqrt(x) is 18

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integral from 0 to 9 of sqrt(x)

\int_{0}^{9} x d x

18

Solution steps

\int_{0}^{9} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{9}

= [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= 18

f (x) = x^{4} + 2 x^{2}, at x = 1

\frac{d y}{d x} (x^{2} + 3) = x y

\int \frac{16 x ^{2} - 49}{x ^{3}} d x

\frac{( y - 3 )}{y ^{2} - 3 y + 9}

y^{^{''}} + 4 y = t - 2 sin (2 t)

What is the integral from 0 to 9 of sqrt(x) ?
The integral from 0 to 9 of sqrt(x) is 18