What is the integral from 2 to 3 of (18)/(6-x) ?

The integral from 2 to 3 of (18)/(6-x) is 18(2ln(2)-ln(3))

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integral from 2 to 3 of (18)/(6-x)

\int_{2}^{3} \frac{18}{6 - x} d x

18 (2 ln (2) - ln (3))

Decimal Notation

5.17827 \dots

Solution steps

\int_{2}^{3} \frac{18}{6 - x} d x

Take the constant out:

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

= 18 \cdot \int_{2}^{3} \frac{1}{6 - x} d x

Apply u-substitution

: \int_{4}^{3} - \frac{1}{u} d u

= 18 \cdot \int_{4}^{3} - \frac{1}{u} d u

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

= 18 (- \int_{3}^{4} - \frac{1}{u} d u)

Take the constant out:

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

= 18 (- (- \int_{3}^{4} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 18 (- (- [ln ∣ u ∣]_{3}^{4}))

Simplify

= 18 [ln ∣ u ∣]_{3}^{4}

Compute the boundaries:

2 ln (2) - ln (3)

= 18 (2 ln (2) - ln (3))

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What is the integral from 2 to 3 of (18)/(6-x) ?
The integral from 2 to 3 of (18)/(6-x) is 18(2ln(2)-ln(3))