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Using $f(t)=(1/2)e^{-t/2}$ for $1<t<3$. Solve the integral for $f(t)$

$$P(1<T<3) = \int_1^{3} f(t) dt\tag{i}$$

$$P(1<T<3) = \int_1^{3} (1/2)e^{-t/2} dt\tag{ii}$$

$$P(1<T<3) = (1/2)\int_1^{3} e^{-t/2} dt\tag{iii}$$

$$P(1<T<3) = e^{-1/2}-e^{-3/2} \tag{iv}$$

My question is how do we show the middle steps between $(iii)$ and $(iv)$. I forgot how to solve this type of integrals. Can someone please show me the correct way with all the steps please?

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$$\int \mathrm{e}^{at}dt = \frac{1}{a}\mathrm{e}^{at} + C$$ then assuming you know how to use limits for definite integrals you can proceed. – Chinny84 Mar 17 '14 at 18:05

1 Answer 1

up vote 0 down vote accepted

$$\int_1^{3} (1/2)e^{-t/2} dt$$

Let $u = -\frac 12 t$. Then $\,du = -\frac 12\,dt$.

If we make our substitution as indicated above, we should change the bounds of integration:

When $t = 1, \; u = -\frac 12 t = -\frac 12$.

When $t = 3, \; u = -\frac 12(3) = -\frac 32.$

So $$\begin{align}\int_1^{3} (1/2)e^{-t/2} dt & = -\int_1^3 e^{\overbrace{{-t/2}}^{u}}\underbrace{\left(-\frac 12\, dt\right)}_{du}\\ \\ & = -\int_{-1/2}^{-3/2} e^u\,du \\ \\ & = -e^u\Big|_{-1/2}^{-3/2}\\ \\ & = -(e^{-1/2} - e^{-3/2}) = e^{-3/2} - e^{-1/2}\end{align}$$

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Thank you! I remember now. This is actually a simple way of thinking about it step by step. Thanks again. – 9959 Mar 17 '14 at 18:35
You're welcome! – amWhy Mar 17 '14 at 19:19

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