What is the integral from-1 to 4 of sqrt(2x+1) ?

The integral from-1 to 4 of sqrt(2x+1) is 9+i 1/3

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integral from-1 to 4 of sqrt(2x+1)

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

9 + i \frac{1}{3}

Solution steps

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

Apply u-substitution

: \int_{- 1}^{9} \frac{u}{2} d u

= \int_{- 1}^{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_{- 1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{- 1}^{9}

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

Compute the boundaries:

18 + i \frac{2}{3}

= \frac{1}{2} (18 + i \frac{2}{3})

Simplify

= 9 + i \frac{1}{3}

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What is the integral from-1 to 4 of sqrt(2x+1) ?
The integral from-1 to 4 of sqrt(2x+1) is 9+i 1/3