Total Revenue from 1996-2002: Calculating Marginal Revenue Growth

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Homework Help Overview

The problem involves calculating total revenue based on a given marginal revenue function, which is expressed in terms of years since 1980. The specific time frame of interest is from the beginning of 1996 to the end of 2002.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the marginal revenue function to find total revenue, with some confusion regarding the limits of integration and the correct form of the integrand. Questions arise about the accuracy of the integration process and the interpretation of the variables involved.

Discussion Status

There is ongoing clarification regarding the limits of integration and the correct application of integration techniques. Some participants have pointed out errors in the integration attempts and are seeking further assistance to resolve these misunderstandings.

Contextual Notes

Participants note that the beginning of 1996 corresponds to a specific value of x, which is 16, and there is confusion about the correct setup for the integration process. The discussion reflects a need for careful attention to detail in mathematical expressions and operations.

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Homework Statement


According to data, the marginal revenue of a product(in billions of dollars per year) is approximated by 3.96+.01x+.0012x^2, where x=0 corresponds to 1980. What was the total revenue from the beginning of 1996 through the end of 2002?


Homework Equations





The Attempt at a Solution


∫23 on top, 17 on bottom (3.96+.01x+.012^2)=(3.96.02x^2+.004x^3)|23 on top, 17 on bottom
63.208-29.392=33.816
 
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jodd8782 said:

Homework Statement


According to data, the marginal revenue of a product(in billions of dollars per year) is approximated by 3.96+.01x+.0012x^2, where x=0 corresponds to 1980. What was the total revenue from the beginning of 1996 through the end of 2002?


Homework Equations





The Attempt at a Solution


∫23 on top, 17 on bottom (3.96+.01x+.012^2)=(3.96.02x^2+.004x^3)|23 on top, 17 on bottom
63.208-29.392=33.816
Do you have a question?
 
yes the answer i got 33.816 was wrong, can you help with the solution
 
The beginning of 1996 corresponds with x = 16, not 17. The other limit of integration, 23, looks OK.

Also, your integration is wrong.

$$\int cx^n dx = \frac{c}{n+1}x^{n+1}$$
I omitted the constant of integration since you're working with a definite integral.
What you're doing looks like you are differentiating, not integrating.
 
ok I am still confused
 
For the integration part, the integrand is 3.96+.01x+.012^2 (the last term should be .012x[/color]^2).

The antiderivative you showed is 3.96.02x^2+.004x^3.

1. In the antiderivative you have two terms munged together (3.96.02x^2).
2. The antiderivative of 3.96 is NOT 3.96.
3. The antiderivative of .01x is NOT .02x^2.
 

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