What is the integral from 0 to 9 of sqrt(2x) ?

The integral from 0 to 9 of sqrt(2x) is 18sqrt(2)

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integral from 0 to 9 of sqrt(2x)

\int_{0}^{9} 2 x d x

182

Decimal

25.45584 \dots

Solution steps

\int_{0}^{9} 2 x d x

Apply Integration By Parts

: [2 x x - \int \frac{x}{2} d x]_{0}^{9}

= [2 x x - \int \frac{x}{2} d x]_{0}^{9}

\int \frac{x}{2} d x = \frac{2}{3} x^{\frac{3}{2}}

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

Compute the boundaries:

182

= 182

\int_{0}^{\frac{π}{3}} sec^{2} (x) d x

- 4 x^{2} + 12 x, 2 x^{2} - 5

x \to - 3 lim (10 x - 4)

\frac{d}{d x} (ln (7 - x))

\frac{d}{d t} (t - 1)

What is the integral from 0 to 9 of sqrt(2x) ?
The integral from 0 to 9 of sqrt(2x) is 18sqrt(2)