Total Revenue from 1996-2002: Calculating Marginal Revenue Growth

[jodd8782]

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

 

[Mark44]

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?

 

[jodd8782]

yes the answer i got 33.816 was wrong, can you help with the solution

 

[Mark44]

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.

 

[jodd8782]

ok I am still confused

 

[Mark44]

About what?

 

[Mark44]

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.