Finding the Area of a Curve Limited by x-Axis

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Homework Help Overview

The discussion revolves around finding the area of the region limited by the curve defined by the equation y=(x+2)(3x-3) and the x-axis. Participants are exploring how to determine the area under the curve, particularly focusing on the integration process and the identification of bounds.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to integrate to find the area but express uncertainty about the limits of integration (a and b). Some suggest solving for where the curve intersects the x-axis to determine these limits. Others mention the importance of sketching the curve to visualize the area.

Discussion Status

There is an ongoing exploration of the integration process, with some participants confirming the negative area calculated and discussing the implications of the curve being below the x-axis. Guidance has been provided regarding the evaluation of the function within the specified interval and the need to consider absolute values for area calculations.

Contextual Notes

Participants note the importance of understanding the behavior of the curve between its intersection points with the x-axis and question how to determine where the function is negative. There is also mention of the necessity to graph the function for better insight.

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Homework Statement


y=(x+2)*(3x-3)

Homework Equations


Find the area of the region limited by the curve y=(x+2)*(3x-3) and the axis x.

The Attempt at a Solution


we should do:
(x+2)*(3x-3)=3x2+3x-6
now we should differentiate ?
 
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chawki said:

Homework Statement


y=(x+2)*(3x-3)

Homework Equations


Find the area of the region limited by the curve y=(x+2)*(3x-3) and the axis x.

The Attempt at a Solution


we should do:
(x+2)*(3x-3)=3x2+3x-6
now we should differentiate ?
No. To find the area beneath a curve, integrate. Your textbook should have numerous examples.
 
ok i know we should integrate...but i was looking for kind of (integration from a to b) and we don't know that a and b. we just know axis x is y=0
and that will make ((x+2)*(3x-3))-0
if we integrate without a and b, the integration gives us:
x2+(3/2)*x2-6x
 
Have you drawn a sketch of the region? The region is between the graph of y = (x + 2)(3x - 3) and the x-axis. If you draw a reasonably accurate sketch, you will see what you need to use for a and b.
 
i draw the curve but didnt help or maybe i didn't know what to look after.
maybe we should solve the equation of y=0 ?
 
chawki said:
i draw the curve but didnt help or maybe i didn't know what to look after.
maybe we should solve the equation of y=0 ?
That's an excellent place to start. That will give you the x values at the points where the quadratic crosses the x-axis.
 
ok i found -2 and 1
and after integration we get -27/2
 
chawki said:
ok i found -2 and 1
and after integration we get -27/2

If the curve lies above the x-axis between -2 and 1 the area should be positive. Check your work for sign errors and make sure your bounds are in the correct places.
 
pergradus said:
If the curve lies above the x-axis between -2 and 1 the area should be positive. Check your work for sign errors and make sure your bounds are in the correct places.

maybe i was wrong in the integration ?
isn't x3+(3/2)*x2-6x ?
 
  • #10
chawki said:
maybe i was wrong in the integration ?
isn't x3+(3/2)*x2-6x ?

No, you were right. The integral of (x+2)*(3*x-3) between -2 and 1 is -27/2. That's fine. The curve is below the x-axis though. What is meant by 'area' is usually the absolute value of the integral, since you usually think of area as positive.
 
  • #11
The curve lies below the x-axis on [-2,1].

Edit: too late...
 
  • #12
That's the negative of the integrand. Since the region lies below the x-axis, the typical area element is [tex]\Delta A = [0 - (3x^2 + 3x - 6)]\Delta x[/tex]

This makes the integral
[tex]\int_{-2}^1 (-3x^2 -3x + 6)dx[/tex]

When you carry out the integration, you get +27/2.
 
  • #13
jhae2.718 said:
The curve lies below the x-axis on [-2,1].

Edit: too late...

Nothing wrong with being too late by 1 minute. What you say is still true. It's not a contest :).
 
  • #14
jhae2.718 said:
The curve lies below the x-axis on [-2,1].

Edit: too late...

how you know it is below and not above ?
 
  • #15
chawki said:
how you know it is below and not above ?

Graph it. You should always do that. The curve intersects the x-axis at 1 and -2. You do need to know what it does in between.
 
  • #17
Mark44 said:
That's the negative of the integrand. Since the region lies below the x-axis, the typical area element is [tex]\Delta A = [0 - (3x^2 + 3x - 6)]\Delta x[/tex]

This makes the integral
[tex]\int_{-2}^1 (-3x^2 -3x + 6)dx[/tex]

When you carry out the integration, you get +27/2.

how we find that 3x2+3x-6 < 0 ?
 
  • #18
When finding the area under a curve, if on some interval (a,b) f(x)<0 for all x in (a,b), the area of the curve is defined by [tex]A=-\int_a^b f(x) dx[/tex].

So, after finding the zeros, pick a test case in the interval you are interested in and evaluate f(x).

For this problem:
[tex]f(x)=(x+2)(3x-3) = 3x^2+3x-6[/tex].
[tex]f(x) = 0 \Rightarrow x \in \{-2,1\}[/tex]
Thus, we are interested in [-2,1]. Zero is contained in this interval, so let's evaluate f(0)=-6. This indicates that f(x) < 0 on [-2,1]. Then, evaluate the integrand and take the opposite as stated above.
 
Last edited:
  • #19
ok thank you everyone, i get it now :)
 
  • #20
chawki said:
how we find that 3x2+3x-6 < 0 ?
Look at your graph. For -2 < x < 1, 3x2+3x-6 < 0.
 
  • #21
Mark44 said:
Look at your graph. For -2 < x < 1, 3x2+3x-6 < 0.

Yep. thank you Mark :blushing:
thank you all again :)
 

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