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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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 = 30 , 00 + 500 t , r = 7 % , t 1 = 6  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=30,000+500t, r=7%, t1=6 years

Explanation

Given Information:

The provided values are c=30,000+500t, r=7%, t1=6 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=30,000+500t, r=7%, t1=6 years.

The income function would be given by, c(t)=30,000+500t

So, apply the formula to calculate the present value.

Substitute c(t)=30,000+500t,  r=0.07 and t1=6 in the expression 0t1c(t)ertdt to calculate the present value as,

Present value=06(30,000+500t)e0.07tdt=0630,000e0.07tdt+50006te0.07tdt=3,000,0007[e0.07t]06+50006te0.07tdt=3,000,0007(e0.421)+50006te0.07tdt

In order to solve the integral te0.07tdt apply integration by parts formula,

Let u=t and dv=e0.07dt, then

dv=e0.07dt

Apply integral on both sides of the above equation as,

dv=e0.07dtv=1007e0.07t

Thus, v=1007e0.07t,

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

du=d(t)du=dt

So, du=dt

Now apply integration by parts formula.

Substitute u=t, dv=e0.07dt, v=1007e0

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