What is the integral from 0 to 2 of 1/(4-x) ?

The integral from 0 to 2 of 1/(4-x) is ln(2)

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integral from 0 to 2 of 1/(4-x)

\int_{0}^{2} \frac{1}{4 - x} d x

ln (2)

Decimal

0.69314 \dots

Solution steps

\int_{0}^{2} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{4}^{2} - \frac{1}{u} d u

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{2}^{4})

Simplify

= [ln ∣ u ∣]_{2}^{4}

Compute the boundaries:

ln (2)

= ln (2)

\int_{0}^{2} (x - 4)^{2} d x

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

\int_{0}^{1} (\frac{6}{x ^{2} + 1}) d x

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

\int_{0}^{3} e^{- t^{2}} d t

What is the integral from 0 to 2 of 1/(4-x) ?
The integral from 0 to 2 of 1/(4-x) is ln(2)