What is the integral from-1 to 0 of 3/(3x-2) ?

The integral from-1 to 0 of 3/(3x-2) is ln(2)-ln(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 0 of 3/(3x-2)

\int_{- 1}^{0} \frac{3}{3 x - 2} d x

ln (2) - ln (5)

Decimal Notation

- 0.91629 \dots

Solution steps

\int_{- 1}^{0} \frac{3}{3 x - 2} d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} \frac{1}{3 x - 2} d x

Apply u-substitution

: \int_{- 5}^{- 2} \frac{1}{3 u} d u

= 3 \cdot \int_{- 5}^{- 2} \frac{1}{3 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{- 5}^{- 2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{- 5}^{- 2}

Simplify

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{- 5}^{- 2} : [ln ∣ u ∣]_{- 5}^{- 2}

= [ln ∣ u ∣]_{- 5}^{- 2}

Compute the boundaries:

ln (2) - ln (5)

= ln (2) - ln (5)

f (x) = - 3 x^{2} + 6 x + 4

\frac{d}{d x} ((1 + 2 x)^{l n (x)})

h \to 0 lim (\frac{0}{h})

\frac{s}{( s ^{2} + 4 ^{2} ) ^{2}}

\int sin (n x + x) d x

What is the integral from-1 to 0 of 3/(3x-2) ?
The integral from-1 to 0 of 3/(3x-2) is ln(2)-ln(5)