What is the integral from 3 to 8 of 1/(sqrt(4x+4)) ?

The integral from 3 to 8 of 1/(sqrt(4x+4)) is 1

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integral from 3 to 8 of 1/(sqrt(4x+4))

\int_{3}^{8} \frac{1}{4 x + 4} d x

1

Solution steps

\int_{3}^{8} \frac{1}{4 x + 4} d x

Apply u-substitution

: \int_{16}^{36} \frac{1}{4 u} d u

= \int_{16}^{36} \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{16}^{36} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{16}^{36}

= \frac{1}{4} [2 u^{\frac{1}{2}}]_{16}^{36}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{4} [2 u]_{16}^{36}

Compute the boundaries:

4

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

Simplify

= 1

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What is the integral from 3 to 8 of 1/(sqrt(4x+4)) ?
The integral from 3 to 8 of 1/(sqrt(4x+4)) is 1