What is the integral from-3 to 0 of 5e^{-jωt} ?

The integral from-3 to 0 of 5e^{-jωt} is -(5(1-e^{3ωj}))/(ωj)

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integral from-3 to 0 of 5e^{-jωt}

\int_{- 3}^{0} 5 e^{- jω t} d t

- \frac{5 ( 1 - e ^{3 ωj} )}{ωj}

Solution steps

\int_{- 3}^{0} 5 e^{- jω t} d t

Take the constant out:

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

= 5 \cdot \int_{- 3}^{0} e^{- ωj t} d t

Apply u-substitution

: \int_{3 ωj}^{0} - \frac{e ^{u}}{ωj} d u

= 5 \cdot \int_{3 ωj}^{0} - \frac{e ^{u}}{ωj} d u

Take the constant out:

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

= 5 (- \frac{1}{ωj} \cdot \int_{3 ωj}^{0} e^{u} d u)

Use the common integral:

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

= 5 (- \frac{1}{ωj} [e^{u}]_{3 ωj}^{0})

Simplify

5 (- \frac{1}{ωj} [e^{u}]_{3 ωj}^{0}) : - \frac{5 [ e ^{u} ] _{3 ωj}^{0}}{ωj}

= - \frac{5 [ e ^{u} ] _{3 ωj}^{0}}{ωj}

Compute the boundaries:

1 - e^{3 ωj}

= - \frac{5 ( 1 - e ^{3 ωj} )}{ωj}

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What is the integral from-3 to 0 of 5e^{-jωt} ?
The integral from-3 to 0 of 5e^{-jωt} is -(5(1-e^{3ωj}))/(ωj)