What is the integral from 0 to t-1 of (-e^{-2x}) ?

The integral from 0 to t-1 of (-e^{-2x}) is 1/2 (e^{-2(-1+t)}-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t-1 of (-e^{-2x})

\int_{0}^{t - 1} (- e^{- 2 x}) d x

\frac{1}{2} (e^{- 2 (- 1 + t)} - 1)

Solution steps

\int_{0}^{t - 1} - e^{- 2 x} d x

Take the constant out:

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

= - \int_{0}^{- 1 + t} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 2 (- 1 + t)} - \frac{1}{2} e^{u} d u

= - \int_{0}^{- 2 (- 1 + t)} - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{0}^{- 2 (- 1 + t)} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- \frac{1}{2} [e^{u}]_{0}^{- 2 (- 1 + t)})

Simplify

= \frac{1}{2} [e^{u}]_{0}^{- 2 (- 1 + t)}

Compute the boundaries:

e^{- 2 (- 1 + t)} - 1

= \frac{1}{2} (e^{- 2 (- 1 + t)} - 1)

\int_{0}^{1} 2 y^{2} - 2 y^{3} d y

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

\int_{0}^{b} cos (x) d x

\int_{0}^{\frac{1}{2} l n (3)} e^{2 x} d x

\int_{- 4}^{3} ∣ x + 3 ∣ d x

What is the integral from 0 to t-1 of (-e^{-2x}) ?
The integral from 0 to t-1 of (-e^{-2x}) is 1/2 (e^{-2(-1+t)}-1)