What is the integral from-1 to 0 of te^{-jwt} ?

The integral from-1 to 0 of te^{-jwt} is (-1-wje^{wj}+e^{wj})/(w^2j^2)

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 te^{-jwt}

\int_{- 1}^{0} t e^{- j wt} d t

\frac{- 1 - w j e ^{w j} + e ^{w j}}{w ^{2} j ^{2}}

Solution steps

\int_{- 1}^{0} t e^{- j wt} d t

Apply u-substitution

: \int_{e^{w j}}^{1} \frac{ln ( u )}{w ^{2} j ^{2}} d u

= \int_{e^{w j}}^{1} \frac{ln ( u )}{w ^{2} j ^{2}} d u

Take the constant out:

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

= \frac{1}{w ^{2} j ^{2}} \cdot \int_{e^{w j}}^{1} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{e^{w j}}^{1}

= \frac{1}{w ^{2} j ^{2}} [u ln (u) - \int 1 d u]_{e^{w j}}^{1}

\int 1 d u = u

= \frac{1}{w ^{2} j ^{2}} [u ln (u) - u]_{e^{w j}}^{1}

Compute the boundaries:

- 1 - w j e^{w j} + e^{w j}

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

Simplify

= \frac{- 1 - w j e ^{w j} + e ^{w j}}{w ^{2} j ^{2}}

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

\int_{- 4}^{0} 2 x^{3} + 12 x^{2} + 16 x d x

\int_{0}^{36} \frac{1}{x} d x

\int_{5}^{x} t^{4} d t

\int_{- 1}^{3} (- 2 x^{2} + 4 x + 6) d x

What is the integral from-1 to 0 of te^{-jwt} ?
The integral from-1 to 0 of te^{-jwt} is (-1-wje^{wj}+e^{wj})/(w^2j^2)