   Chapter 6, Problem 15RE ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Present Value In Exercises 13–16, find the present value of the income c (in dollars) over t1 years at the given annual inflation rate r. c = 24 , 000 t ,     r = 5 % ,     t 1 = 10   years

To determine

To calculate: The present value of the income (c, in dollar) of 24,000t (c, in dollar) over a time period (t1) of 10 years at an annual inflation rate (r) of 5%.

Explanation

Given Information:

The value of income, c, is 24,000t dollar.

The time period, t1, is 10 years.

The rate of the annual inflation, r, is 5%.

Formula used:

The expression for the present value is written as,

Present value=0t1c(t)ertdt

Here, r is the rate of inflation and c(t) represents the income which is a function of time t.

The method of integration by parts:

If v and u are two differentiable function of x. Then,

udv=uvvdu

Steps to solve the integral problems:

Step1: At first find the most complicated portion of the integrand and try to letter it as dv so that it can fit a fundamental integration rule. Then, the remaining factor or factors of the integrand will be u.

Step2: First find the factor whose derivative is simple and consider it as u and then the remaining factor or factors of the integrand will be dv and dv should always include the term dx of the original integrand.

Calculation:

Recall the expression of the present value.

Present value=0t1c(t)ertdt

Substitute 10 for t1, 24,00t for c(t) and 0.05 for r.

Present value=010(24,00t)e0.05tdt=01024,00te0.05tdt=24,00010te0.05tdt

In the above integrand, the simplest portion of the integrand is t. So, consider, u=t and the remaining factors as dv=e0.05tdt.

Therefore,

du=dt

And,

dv=e0.05tdt

Integrate the above expression for v.

dv=e0

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