What is the integral from 3/2 to 5/2 of e^{-st} ?

The integral from 3/2 to 5/2 of e^{-st} is -(e^{-(5s)/2}-e^{-(3s)/2})/(s)

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integral from 3/2 to 5/2 of e^{-st}

\int_{\frac{3}{2}}^{\frac{5}{2}} e^{- s t} d t

- \frac{e ^{- \frac{5 s}{2}} - e ^{- \frac{3 s}{2}}}{s}

Solution steps

\int_{\frac{3}{2}}^{\frac{5}{2}} e^{- s t} d t

Apply u-substitution

: \int_{- \frac{3 s}{2}}^{- \frac{5 s}{2}} - \frac{e ^{u}}{s} d u

= \int_{- \frac{3 s}{2}}^{- \frac{5 s}{2}} - \frac{e ^{u}}{s} d u

Take the constant out:

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

= - \frac{1}{s} \cdot \int_{- \frac{3 s}{2}}^{- \frac{5 s}{2}} e^{u} d u

Use the common integral:

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

= - \frac{1}{s} [e^{u}]_{- \frac{3 s}{2}}^{- \frac{5 s}{2}}

Simplify

- \frac{1}{s} [e^{u}]_{- \frac{3 s}{2}}^{- \frac{5 s}{2}} : - \frac{[ e ^{u} ] _{- \frac{3 s}{2}}^{- \frac{5 s}{2}}}{s}

= - \frac{[ e ^{u} ] _{- \frac{3 s}{2}}^{- \frac{5 s}{2}}}{s}

Compute the boundaries:

e^{- \frac{5 s}{2}} - e^{- \frac{3 s}{2}}

= - \frac{e ^{- \frac{5 s}{2}} - e ^{- \frac{3 s}{2}}}{s}

y^{^{'}} = 4 - y

\int (\frac{2}{x ^{3}} + \frac{3}{x ^{2}} + 5) d x

\frac{e ^{- s}}{s ( s + 2 )}

x \to 0 lim (x^{2} cos (\frac{9}{x}))

\int 14 - 20 - 3 16 - 2 x d x

What is the integral from 3/2 to 5/2 of e^{-st} ?
The integral from 3/2 to 5/2 of e^{-st} is -(e^{-(5s)/2}-e^{-(3s)/2})/(s)