What is the integral from 2 to 3 of (16)/(sqrt(3-x)) ?

The integral from 2 to 3 of (16)/(sqrt(3-x)) is 32

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integral from 2 to 3 of (16)/(sqrt(3-x))

\int_{2}^{3} \frac{16}{3 - x} d x

32

Solution steps

\int_{2}^{3} \frac{16}{3 - x} d x

Take the constant out:

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

= 16 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

Apply u-substitution

= 16 \cdot \int_{1}^{0} - \frac{1}{u} d u

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

= 16 (- \int_{0}^{1} - \frac{1}{u} d u)

Take the constant out:

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

= 16 (- (- \int_{0}^{1} \frac{1}{u} d u))

Apply the Power Rule

= 16 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

Simplify

16 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 16 [2 u]_{0}^{1}

= 16 [2 u]_{0}^{1}

Compute the boundaries:

2

= 16 \cdot 2

Simplify

= 32

What is the integral from 2 to 3 of (16)/(sqrt(3-x)) ?
The integral from 2 to 3 of (16)/(sqrt(3-x)) is 32