What is the integral from 2 to-1 of 1/(4-x) ?

The integral from 2 to-1 of 1/(4-x) is -ln(5)+ln(2)

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integral from 2 to-1 of 1/(4-x)

\int_{2}^{- 1} \frac{1}{4 - x} d x

- ln (5) + ln (2)

Decimal

- 0.91629 \dots

Solution steps

\int_{2}^{- 1} \frac{1}{4 - x} d x

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

= - \int_{- 1}^{2} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{5}^{2} - \frac{1}{u} d u

= - \int_{5}^{2} - \frac{1}{u} d u

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

= - (- \int_{2}^{5} - \frac{1}{u} d u)

Take the constant out:

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

= - (- (- \int_{2}^{5} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- (- [ln ∣ u ∣]_{2}^{5}))

Simplify

= - [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

= - (ln (5) - ln (2))

Simplify

= - ln (5) + ln (2)

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\int_{3}^{5} \frac{1}{2 x - 6} d x

What is the integral from 2 to-1 of 1/(4-x) ?
The integral from 2 to-1 of 1/(4-x) is -ln(5)+ln(2)