What is the integral from 0 to 12 of 9/(sqrt(12-x)) ?

The integral from 0 to 12 of 9/(sqrt(12-x)) is 36sqrt(3)

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integral from 0 to 12 of 9/(sqrt(12-x))

\int_{0}^{12} \frac{9}{12 - x} d x

363

Decimal

62.35382 \dots

Solution steps

\int_{0}^{12} \frac{9}{12 - x} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{12} \frac{1}{12 - x} d x

Apply u-substitution

= 9 \cdot \int_{12}^{0} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 9 (- \int_{0}^{12} - \frac{1}{u} d u)

Take the constant out:

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

= 9 (- (- \int_{0}^{12} \frac{1}{u} d u))

Apply the Power Rule

= 9 (- (- [2 u^{\frac{1}{2}}]_{0}^{12}))

Simplify

9 (- (- [2 u^{\frac{1}{2}}]_{0}^{12})) : 9 [2 u]_{0}^{12}

= 9 [2 u]_{0}^{12}

Compute the boundaries:

43

= 9 \cdot 43

Simplify

= 363

What is the integral from 0 to 12 of 9/(sqrt(12-x)) ?
The integral from 0 to 12 of 9/(sqrt(12-x)) is 36sqrt(3)