What is e^{-a(t+1)}\H(t+1)e^{-st}dt ?

The solution to e^{-a(t+1)}\H(t+1)e^{-st}dt is e^{-a(t+1)-ts}\H(t+1)dt

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simplify e^{-a(t+1)}\H(t+1)e^{-st}dt

simplify

e^{- a (t + 1)} H (t + 1) e^{- s t} d t

e^{- a (t + 1) - t s} H (t + 1) d t

Solution steps

e^{- a (t + 1)} H (t + 1) e^{- s t} d t

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{- a (t + 1)} e^{- s t} = e^{- a (t + 1) - t s}

= H (t + 1) e^{- a (t + 1) - s t} d t

52.891 \cdot 1 0^{- 4}

78 (28.69) e^{- 0.000025 (28.69)}

1.6 \cdot 1 0^{- 12}

(g h^{3}) (g^{2} h^{5})

3 Y^{2} (4 X Y)

What is e^{-a(t+1)}\H(t+1)e^{-st}dt ?
The solution to e^{-a(t+1)}\H(t+1)e^{-st}dt is e^{-a(t+1)-ts}\H(t+1)dt