finding the area between two curves

A

Anthonyk2013

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using integration find the area between curves y=x2+1 and y=7-x

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x=-3 x=2

wondering if im on the right track
 
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Anthonyk2013 said:
using integration find the area between curves y=x2+1 and y=7-1

x2 +1=7-x
x2+1-7-x=0
x2-6-x=0
(x-3)(x+2)
x=3 x=-2

wondering if im on the right track
Click to expand...

Did you plug your answers back into the original equations to check them?

After you do that (and correct your sign error), then you can proceed.
 
wjm11 said:
Did you plug your answers back into the original equations to check them?

After you do that (and correct your sign error), then you can proceed.
Click to expand...


When I plug back in -3 I get 6 and when I plug in 2 I get 0, so not sure where im going wrong
 
Anthonyk2013 said:
using integration find the area between curves y=x2+1 and y=7-x
Click to expand...

If you have not seen a plot of both functions, then no you are not on the right track.

From the plot you can see that they cross at \(\displaystyle x=-3~~x=2\)

You can also see that \(\displaystyle x^2+1\) is lower curve and \(\displaystyle 7-x\) is the upper.

So \(\displaystyle \int_{ - 3}^2 {[(7 - x) - ({x^2} + 1)] dx} \) is your area.
 
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pka said:
If you have not seen a plot of both functions, then no you are not on the right track.

From the plot you can see that they cross at \(\displaystyle x=-3~~x=2\)

You can also see that \(\displaystyle x^2+1\) is lower curve and \(\displaystyle 7-x\) is the upper.

So \(\displaystyle \int_{ - 3}^2 {[(7 - x) - ({x^2} + 1)] dx} \) is your area.
Click to expand...

This tells me what is positive and what is negative right and this is important why?
 
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Anthonyk2013 said:
This tells me what is positive and what is negative right and this is important why?
Click to expand...
That tells me that you know nothing about this question!
Given that why are you even trying to do this question?

Can you simply find the number \(\displaystyle \int_{ - 3}^2 {[(7 - x) - ({x^2} + 1)] dx} \). Can you do it?

If not then drop out of this course, go back a level or two.
 
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Anthonyk2013 said:
using integration find the area between curves y=x2+1 and y=7-x

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x=-3 x=2

wondering if im on the right track
When I plug back in -3 I get 6 and when I plug in 2 I get 0, so not sure where im going wrong
Click to expand...
I think that you went wrong by not writing "=0" in (x-3)(x+2). If you had written (x-3)(x+2)=0 you might have realized that (x-3)(x+2) equals 0 instead of 6, which you wrote above. I know 100% that you know that 3-3=0. Yet you will allow yourself to be tricked into thinking that a number different from 3, like -3, minus 3 =0. Don't fall for this.
Most students know that x-x=0. Yet if they are faced with x being replaced with numbers they are completely confused or correctly make changes in signs and operation which is not necessary, etc. Since x-x=0 that means that (-7)-(-7)=0 and (-17)-(-17)=0. You MUST understand that when you are subtracting you are basically find how far apart two numbers are on a number line OR the difference between two numbers. Since -7 and -7 are the same number, so there is no difference and we have (-7)-(-7) = 0 and not 14 or -14. More importantly if x-y=0 THEN x=y (just add y to both sides to see this). So if x-3=0 then x=3. If 7y-5=0 then 7y=5, if 3x-9=0 then 3x=9, ....This should be immediate. Way easier than saying your own name.
Jomo
 
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pka said:
That tells me that you know nothing about this question!
Given that why are you even trying to do this question?

Can you simply find the number \(\displaystyle \int_{ - 3}^2 {[(7 - x) - ({x^2} + 1)] dx} \). Can you do it?

If not then drop out of this course, go back a level or two.
Click to expand...

Disgraceful comment feel free to mind your own business when I post and thanks to the gentlemen who try and help

The course im doing is designed to make you go home and study, put you under pressure, research and learn at home,the syllabus is large and we spent one and a half hours doing this these type of integration, I have a full time job, 3 kids and im doing my best to get through this course.

DROPPING BACK OR QUITING IS NOT AN OPTION. IF I ANNOY YOU DONT HELP ME, SIMPLE
 
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Jomo said:
I think that you went wrong by not writing "=0" in (x-3)(x+2). If you had written (x-3)(x+2)=0 you might have realized that (x-3)(x+2) equals 0 instead of 6, which you wrote above. I know 100% that you know that 3-3=0. Yet you will allow yourself to be tricked into thinking that a number different from 3, like -3, minus 3 =0. Don't fall for this.
Most students know that x-x=0. Yet if they are faced with x being replaced with numbers they are completely confused or correctly make changes in signs and operation which is not necessary, etc. Since x-x=0 that means that (-7)-(-7)=0 and (-17)-(-17)=0. You MUST understand that when you are subtracting you are basically find how far apart two numbers are on a number line OR the difference between two numbers. Since -7 and -7 are the same number, so there is no difference and we have (-7)-(-7) = 0 and not 14 or -14. More importantly if x-y=0 THEN x=y (just add y to both sides to see this). So if x-3=0 then x=3. If 7y-5=0 then 7y=5, if 3x-9=0 then 3x=9, ....This should be immediate. Way easier than saying your own name.
Jomo
Click to expand...

Ya i understand all of the above. I'm having technical trouble typing here.
The biggest problem i have is with typing on this forum, for some reason when posting here its like my laptop get a virus and its very hard type the correct info. I wont use this forum again until i sort this problem. thanks for your help jomo.
 
Anthonyk2013 said:
Disgraceful comment
Click to expand...
Actually, it is often caring advice. Mathematics is incremental; new material builds on older material. If one does not understand that older material, one is generally doomed to repeated failure at newer material. People who have too-often seen that tragedy tend to make the suggestion. While it is to be regretting that this caused you upset, I would respectfully suggest that it may be worthy of consideration. Dropping back one or two courses, and then succeeding, surely is a better option than flunking four or five times, and then dropping out entirely.

Apologies in advance. ;)
 
pka said:
That tells me that you know nothing about this question!
Given that why are you even trying to do this question?

Can you simply find the number \(\displaystyle \int_{ - 3}^2 {[(7 - x) - ({x^2} + 1)] dx} \). Can you do it?

If not then drop out of this course, go back a level or two.
Click to expand...
Hi,
Normally I agree that many Calculus students should go back a course or two. That is normally because they are WEAK with algebra. This student is having trouble with (sort of) a Calculus part of Calculus. Hence this student has not shown us (YET) that he/she is weak with algebra. The student is not clear yet that when we integrate we are adding up rectangles. Again, I do not think that knowing algebra better would help the student seeing that we are adding up rectangles.

So why do we do the top - bottom in the integral? Well think about this rectangle-the height goes from 7 to 12 and the width/base is 2. To find the height, which is always positive, we compute 12-7 (top -bottom) and NOT 7-12 (bottom-top) and then multiple by 2 (the base). The area is (12-7)*2=5*2=10. So in the integral we integrate top-bottom times the base (dx).
If we use the base as delta y then the height (or length) will be right - left.
Jomo
 
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