What is the integral from 1/2 to 5 of sqrt(2x-1) ?

The integral from 1/2 to 5 of sqrt(2x-1) is 9

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integral from 1/2 to 5 of sqrt(2x-1)

\int_{\frac{1}{2}}^{5} 2 x - 1 d x

9

Solution steps

\int_{\frac{1}{2}}^{5} 2 x - 1 d x

Apply u-substitution

: \int_{0}^{9} \frac{u}{2} d u

= \int_{0}^{9} \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{9} u d u

Apply the Power Rule

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

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= \frac{1}{2} \cdot 18

Simplify

= 9

\int_{- 6}^{1} (2 x + 4) d x

\int_{0}^{4} 3 x 25 - x^{2} d x

\int_{1}^{4} \frac{3 x + 5}{x} d x

\int_{0}^{1} 3 x + 6 d x

\int_{5}^{7} \frac{1}{x ^{2}} d x

What is the integral from 1/2 to 5 of sqrt(2x-1) ?
The integral from 1/2 to 5 of sqrt(2x-1) is 9