Hi, Hope you are all well! In the 2013 CT1 Chapter 5 Core Reading there is an example of how to calculate the present value of a five-year continuous cashflow; \[p(v) = 100 \times (0.8)^t\] and constant force of interest 8% pa. I understand why in this particular circumstance I need to integrate I just don't understand this specific example of integration. Could someone walk me through the following integration to calculate the present value please? i.e which rules of integration have been employed here please? \[ \int_0^5 (e^{-0.08t} \times 100 \times 0.8^t) dt = 100 \int_0^5 ((e^{-0.08}) \times 0.8)^t = \Bigg [ \frac{100(e^{-0.08} \times 0.8)^t}{log(e^{-0.08} \times 0.8)} \Bigg ]^5_0 \] Let me know if there is any further information I can be providing. Many thanks, Paddy
Hey, I'm sorry, a little ambiguity here. You want to understand just the integration technique applied in this particular question? Or the concept of using integration in continuous payments? anyway, in this ex, they have considered (e^(−0.08))×0.8 as one entity, let say "a". So integrate "a^t", u get a^t/log a.
Hi Woozie.Boozie, Thank you for your reply. - It's the substitution method I am not understanding here: If (e^(−0.08))×0.8 = a how does the integral of a^t become a^t/log t please? My understanding of Integration by substitution is: intgrl [ f(g(x))g'(x) ] dx = intgrl [ f(u) ] du , where u = g(x) .. Thank you again for all your help, Paddy
Integrating a^t gets you a^t/log(a) I'm sorry, that was a typo, I corrected it. Does it make sense now? It's like (a^t)(b^t) = (ab)^t Now considering ab as one entity, say c.. U can integrate c^t, which would turn out to be c^t/log c. Hope it's clear now
Hiya, Apologies I am still not following o) : How does intrgl C^t become c^t/log c please? Where does the log c come from please? Thank you for your patience. Many thanks, Paddy
Hi Woozie.Boozie, I understand now thank you so much for your help and patience! Warmest Regards, Paddy
