What is the integral from 1 to 7 of 1/(sqrt(8x+8)) ?

The integral from 1 to 7 of 1/(sqrt(8x+8)) is 1

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integral from 1 to 7 of 1/(sqrt(8x+8))

\int_{1}^{7} \frac{1}{8 x + 8} d x

1

Solution steps

\int_{1}^{7} \frac{1}{8 x + 8} d x

Apply u-substitution

= \int_{16}^{64} \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{16}^{64} \frac{1}{u} d u

Apply the Power Rule

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{16}^{64}

Apply radical rule:

= \frac{1}{8} [2 u]_{16}^{64}

Compute the boundaries:

8

= \frac{1}{8} \cdot 8

Simplify

= 1

What is the integral from 1 to 7 of 1/(sqrt(8x+8)) ?
The integral from 1 to 7 of 1/(sqrt(8x+8)) is 1