# Area of 2 curves -Integration

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• Oct 8th 2013, 08:33 AM
Darrylcwc
Area of 2 curves -Integration
I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.

Find the area of the region trapped between the 2 given curves and the indicated lines.

f(x) = 3x
g(x) = x^2
x = -1, x = -2

int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)

Take z(-2) - z(-1) = results

What I also did was to integrate each function separately and sub in the the range of x and find the difference.

Apparently, the results I obtain does not tally with the answer sheet that says 31/6

I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.

My answer works out to be 3/2.
• Oct 8th 2013, 10:08 AM
SlipEternal
Re: Area of 2 curves -Integration
First of all, when is $\displaystyle 3x \ge x^2$ and when is $\displaystyle x^2 \ge 3x$? To figure this out, find out when they are equal.

$\displaystyle x^2 = 3x$ when $\displaystyle x=0, x=3$.

So, at $\displaystyle x=-1, 1 = (-1)^2 > 3(-1) = -3$. Hence $\displaystyle x^2 \ge 3x$ on the interval $\displaystyle (-\infty,0]$.

How about on $\displaystyle [0,3]$? Well, $\displaystyle 1 = (1)^2 < 3(1) = 3$, so $\displaystyle 3x \ge x^2$ on the interval $\displaystyle [0,3]$.

So, the correct set-up for this would be:
$\displaystyle \int_{-1}^0 (x^2-3x)dx + \int_0^2 (3x-x^2)dx$.
• Oct 8th 2013, 01:36 PM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.

Find the area of the region trapped between the 2 given curves and the indicated lines.

f(x) = 3x
g(x) = x^2
x = -1, x = -2

int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)

Take z(-2) - z(-1) = results

What I also did was to integrate each function separately and sub in the the range of x and find the difference.

Apparently, the results I obtain does not tally with the answer sheet that says 31/6

I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.

My answer works out to be 3/2.

Graph the functions to visualzse what you are doing

Attachment 29415

Do you see what you can do now: integrate under the parabola and subtruct the area of the triangle
• Oct 8th 2013, 05:09 PM
Darrylcwc
Re: Area of 2 curves -Integration
Where did the negative one came from?
Equation is equal only at 0 ll -3.
• Oct 8th 2013, 05:25 PM
SlipEternal
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
Where did the negative one came from?
Equation is equal only at 0 ll -3.

You are told to bound the functions between the lines x=-1 and x=2. Finding where the equations are equal only tells you the points where it "may" switch which one is "above" the other.
• Oct 8th 2013, 05:26 PM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
Where did the negative one came from?
Equation is equal only at 0 ll -3.

The integration in the orginal post is area from -2 to -1.
if your question is about the line y = 3x, at x=-2 y=-6, at x=-1 y=3*(-1) = -3
• Oct 8th 2013, 05:38 PM
SlipEternal
Re: Area of 2 curves -Integration
Quote:

Originally Posted by votan
The integration in the orginal post is area from -2 to -1.
if your question is about the line y = 3x, at x=-2 y=-6, at x=-1 y=3*(-1) = -3

Oh, oops! I thought it was -1 and 2. Good catch! So then the integral should be

$\displaystyle \int_{-2}^{-1}(x^2-3x)dx$
• Oct 8th 2013, 07:19 PM
Darrylcwc
Re: Area of 2 curves -Integration
It just doesn't make sense.
Since x = 0 and x =3 is where the entire curve intersect, then why can't I use x = [0,3] as the range?
graph y=x^2 && y=3x - Wolfram|Alpha=
It just doesn't make sense mathematically to cover the domain outside of 0 and 3 because the area is not enclosed.

I'm solving this, otherwise I'm not sleeping.
• Oct 8th 2013, 07:35 PM
SlipEternal
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
It just doesn't make sense.
Since x = 0 and x =3 is where the entire curve intersect, then why can't I use x = [0,3] as the range?

I'm solving this, otherwise I'm not sleeping.

Check out the graph of the functions (link). You are asked to find the region bounded by four functions. $\displaystyle y=x^2, y=3x, x=-1, x=-2$. Where does the line $\displaystyle x=-1$ intersect the graph $\displaystyle y=x^2$? It intersects it at $\displaystyle x=-1$. Similarly for $\displaystyle y=3x$. And $\displaystyle x=-2$ intersects both functions at $\displaystyle x=-2$. Figuring out where the two graphs intersect can help you find "a" bounded region. It is not the only one. There are several bounded regions. Since you need to use all four functions as bounds, the only bounded region with $\displaystyle x=-2$ and $\displaystyle x=-1$ as bounds has area as I gave in the previous post.
• Oct 8th 2013, 07:38 PM
Darrylcwc
Re: Area of 2 curves -Integration
Hang on. Isn't x the domain? Why is it a function?
• Oct 8th 2013, 07:40 PM
SlipEternal
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
Hang on. Isn't x the domain? Why is it a function?

Because you said it was:

Quote:

Originally Posted by Darrylcwc
Find the area of the region trapped between the 2 given curves and the indicated lines.

f(x) = 3x
g(x) = x^2
x = -1, x = -2

• Oct 8th 2013, 08:00 PM
Darrylcwc
Re: Area of 2 curves -Integration
In this case, shouldn't the set up thus be

[-1,-2]int (x^2-3x) + [0,3]int (x^2 - 3x) ?
• Oct 8th 2013, 08:09 PM
votan
Re: Area of 2 curves -Integration
Quote:

Originally Posted by Darrylcwc
In this case, shouldn't the set up thus be

[-1,-2]int (x^2-3x) + [0,3]int (x^2 - 3x) ?

notice athe -2 < -1, you could integrate from -1 to -2 if you place a negative sign infront of the integral.

I couldn't post a new reply, I append this post by the image.

Attachment 29422
• Oct 8th 2013, 08:11 PM
Darrylcwc
Re: Area of 2 curves -Integration
I understand that. This set up I gave should be correct. But it still doesn't tally.

Is something wrong? It's can't be the case. This is supposed to be a straight forward and simple abstract question.
• Oct 8th 2013, 08:28 PM
Darrylcwc
Re: Area of 2 curves -Integration
Quote:

Originally Posted by SlipEternal
First of all, when is $\displaystyle 3x \ge x^2$ and when is $\displaystyle x^2 \ge 3x$? To figure this out, find out when they are equal.

$\displaystyle x^2 = 3x$ when $\displaystyle x=0, x=3$.

So, at $\displaystyle x=-1, 1 = (-1)^2 > 3(-1) = -3$. Hence $\displaystyle x^2 \ge 3x$ on the interval $\displaystyle (-\infty,0]$.

How about on $\displaystyle [0,3]$? Well, $\displaystyle 1 = (1)^2 < 3(1) = 3$, so $\displaystyle 3x \ge x^2$ on the interval $\displaystyle [0,3]$.

So, the correct set-up for this would be:
$\displaystyle \int_{-1}^0 (x^2-3x)dx + \int_0^2 (3x-x^2)dx$.

I contrast my set up with yours and I don't understand why the second set up ranges from 0,2. The point of intersection is at 0,3.
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