Find the area bounded by the curve #2

  • Context: MHB 
  • Thread starter Thread starter shamieh
  • Start date Start date
  • Tags Tags
    Area Bounded Curve
Click For Summary

Discussion Overview

The discussion centers on finding the area bounded by the curve defined by the equation $$y = 6x - x^2$$ and the axes. Participants explore different approaches to calculating this area, including the use of integration and the even function rule.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the even function rule to calculate the area, suggesting the integral $$2 \int^2_0 (6x - x^2) dx$$ and arrives at a result of $$\frac{56}{3}$$.
  • Another participant questions the limits of integration and suggests confirming the roots of the function, indicating that the area should be calculated between the roots of the curve.
  • A later reply corrects the limits of integration to $$0$$ and $$6$$, stating that the area should be calculated as $$2 \int^6_0 (6x - x^2) dx$$, leading to a result of $$72$$.
  • One participant points out that the function $$f(x) = 6x - x^2$$ is not an even function, challenging the initial use of the even function rule.
  • Another participant reiterates the correction regarding the limits and the nature of the function, suggesting a transformation to make the function even, leading to a new area calculation of $$36$$ using the modified function $$y(x+3) = 9 - x^2$$.

Areas of Agreement / Disagreement

Participants express disagreement regarding the use of the even function rule and the correct limits of integration. Multiple competing views on how to approach the problem remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the function's properties, the validity of the even function rule in this context, and the accuracy of the integration limits.

shamieh
Messages
538
Reaction score
0
Find the area bounded by the curve $$x = 6x - x^2$$ and the y axis.

So can I use the even function rule to get:

$$2 \int^2_0 6x - x^2 dx$$

I just need someone to check my work.

$$2 [ 3x^2 - \frac{1}{3}x^3 ] | 2, 0$$
$$
2 [ 12 - \frac{8}{3}]$$
$$2 [ \frac{36}{3} - \frac{8}{3} ]$$
$$2 * 28/3$$

= $$\frac{56}{3}$$

is this correct?
 
Physics news on Phys.org
I assume the curve is $y=6x-x^2$ and you are to find the area bounded by it and the $x$-axis. Correct?

Where did you get the limits of integration? What are the roots of the function?

If I wanted to use the even function rule here, I would transpose the graph of the curve 3 units to the left.
 
I knew I was doing something wrong...

So this is what I should have

6x - x^2 = 0

x (6 - x ) = 0

x = 0 x = 6
$$
2 \int^6_0 6x - x^2 dx$$

which equals 72
 
Did you read what Mark said? \displaystyle \begin{align*} f(x) = 6x - x^2 \end{align*} is NOT an even function, as \displaystyle \begin{align*} f(-x) = -6x - x^2 \neq f(x) \end{align*}.
 
shamieh said:
I knew I was doing something wrong...

So this is what I should have

6x - x^2 = 0

x (6 - x ) = 0

x = 0 x = 6
$$
2 \int^6_0 6x - x^2 dx$$

which equals 72

Your limits are correct, however as Prove It showed, the function is not even, so the factor of 2 you added has doubled the correct result. As I stated, if you want to use the even function rule, transpose the graph 3 units to the left:

$$y(x+3)=6(x+3)-(x+3)^2=6x+18-x^2-6x-9=9-x^2$$

Now you have an even function and you may write:

$$A=2\int_0^3 9-x^2\,dx=2\left[9x-\frac{1}{3}x^3 \right]_0^3=2\cdot18=36$$
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K