What is the integral from 3 to 4 of 1/(2-5t) ?

The integral from 3 to 4 of 1/(2-5t) is 1/5 (ln(13)-ln(18))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to 4 of 1/(2-5t)

\int_{3}^{4} \frac{1}{2 - 5 t} d t

\frac{1}{5} (ln (13) - ln (18))

Decimal Notation

- 0.06508 \dots

Solution steps

\int_{3}^{4} \frac{1}{2 - 5 t} d t

Apply u-substitution

: \int_{- 13}^{- 18} - \frac{1}{5 u} d u

= \int_{- 13}^{- 18} - \frac{1}{5 u} d u

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

= - \int_{- 18}^{- 13} - \frac{1}{5 u} d u

Take the constant out:

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

= - (- \frac{1}{5} \cdot \int_{- 18}^{- 13} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{5} [ln ∣ u ∣]_{- 18}^{- 13})

Simplify

= \frac{1}{5} [ln ∣ u ∣]_{- 18}^{- 13}

Compute the boundaries:

ln (13) - ln (18)

= \frac{1}{5} (ln (13) - ln (18))

\int_{0}^{2} (3 x + 1) e^{2 x} d x

\int_{1}^{4} (- 5 - x^{2}) d x

\int_{0}^{x} 1 d y

\int_{- 1}^{2} (\frac{4}{3} x^{3} + 4 x^{2} + x) d x

\int_{0}^{5} x (\frac{2 x}{25}) d x

What is the integral from 3 to 4 of 1/(2-5t) ?
The integral from 3 to 4 of 1/(2-5t) is 1/5 (ln(13)-ln(18))