   Chapter 6.1, Problem 69E ### 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 67-72, find the present value of the income c (in dollars) over t 1 years at the given annual inflation rate r. See Examples 6 and 7. c = 00 , 00 + 4000 t , r = 5 % , t 1 = 10  years

To determine

To calculate: The present value of the income c (in dollars) over t1 years at the provided inflation rate r, given below as,

c=100,000+4000t, r=5%, t1=10 years

Explanation

Given Information:

The provided values are c=100,000+4000t, r=5%, t1=10 years, where c is the income (in dollars), t1 is time in years and r is the inflation rate.

Formula used:

When c is the continuous income function in dollars per year at the inflation rate r, then the present value over the t1 years is given by,

Present value=0t1c(t)ertdt

Integration by parts.

When u and v is assumed to be the differentiable functions of x then,

u dv=uvv du

Calculation:

Consider the values,

c=100,000+4000t, r=5%, t1=10 years.

The income function would be given by, c(t)=100,000+4000t

So, apply the formula to calculate the present value.

Substitute c(t)=100,000+4000t,  r=0.05 and t1=10 in the expression 0t1c(t)ertdt to calculate the present value as,

Present value=010(100,000+4000t)e0.05tdt=010100,000e0.05tdt+4000010te0.05tdt=2,000,000[e0.05t]010+4000010te0.05tdt=2,000,000(1e0.5)+4000010te0.05tdt

In order to solve the integral te0.05tdt apply integration by parts formula,

Let u=t and dv=e0.05dt, then

dv=e0.05dt

Apply integral on both sides of the above equation as,

dv=e0.05dtv=20e0.05t

Thus, v=20e0.05t,

Then differentiate both sides of the equation u=t as,

du=d(t)du=dt

So, du=dt

Now apply integration by parts formula

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