What is the integral from 2 to 9 of 1/(2sqrt(x+7)) ?

The integral from 2 to 9 of 1/(2sqrt(x+7)) is 1

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integral from 2 to 9 of 1/(2sqrt(x+7))

\int_{2}^{9} \frac{1}{2 x + 7} d x

1

Solution steps

\int_{2}^{9} \frac{1}{2 x + 7} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{9} \frac{1}{x + 7} d x

Apply u-substitution

: \int_{3}^{4} 2 d u

= \frac{1}{2} \cdot \int_{3}^{4} 2 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} [2 u]_{3}^{4}

Compute the boundaries:

2

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

Simplify

= 1

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What is the integral from 2 to 9 of 1/(2sqrt(x+7)) ?
The integral from 2 to 9 of 1/(2sqrt(x+7)) is 1