What is the integral from 0 to 4 of (2x+1)^{-1/2} ?

The integral from 0 to 4 of (2x+1)^{-1/2} is 2

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integral from 0 to 4 of (2x+1)^{-1/2}

\int_{0}^{4} (2 x + 1)^{- \frac{1}{2}} d x

2

Solution steps

\int_{0}^{4} (2 x + 1)^{- \frac{1}{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

4

= \frac{1}{2} \cdot 4

Simplify

= 2

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What is the integral from 0 to 4 of (2x+1)^{-1/2} ?
The integral from 0 to 4 of (2x+1)^{-1/2} is 2