What is the integral from-e^5 to-e^3 of 1/t ?

The integral from-e^5 to-e^3 of 1/t is -2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-e^5 to-e^3 of 1/t

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

- 2

Solution steps

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

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= [ln ∣ t ∣]_{- e^{5}}^{- e^{3}}

Compute the boundaries:

- 2

= - 2

\int_{a}^{b} \frac{1}{( x - a ) ^{p}} d x

\int_{0}^{t^{4}} u^{5} d u

\int_{5}^{12} \frac{1}{x - 5} d x

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

\int_{2}^{9} x - 1 d x

What is the integral from-e^5 to-e^3 of 1/t ?
The integral from-e^5 to-e^3 of 1/t is -2