What is the integral from 0 to 7 of 3/(sqrt(x+9)) ?

The integral from 0 to 7 of 3/(sqrt(x+9)) is 6

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integral from 0 to 7 of 3/(sqrt(x+9))

\int_{0}^{7} \frac{3}{x + 9} d x

6

Solution steps

\int_{0}^{7} \frac{3}{x + 9} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{7} \frac{1}{x + 9} d x

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

= 3 [2 u]_{3}^{4}

Compute the boundaries:

2

= 3 \cdot 2

Simplify

= 6

\int_{0}^{\frac{π}{2}} (cos (x) - 1) d x

\int_{- 1}^{1} (\frac{x ^{3} - 8}{x - 2}) d x

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

\int_{0.5}^{0} x d x

\int_{2}^{5} 6 x^{- 2} d x

What is the integral from 0 to 7 of 3/(sqrt(x+9)) ?
The integral from 0 to 7 of 3/(sqrt(x+9)) is 6