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# Finding area beneath/below x-axis; integration watch

1. Hi guys;

Finding myself somewhat baffled by the attached questions.

For example, the first one. The only two values I see, are 3, and 0; then equation.

I integrated the equation; on the basis of the 3 and 0 values.

From there; I identified the answer as 6-9/2=3/2

The answers listed are given as 2 and 2.5 respectively.

I know that when it comes to the total shaded area, you find the area of the area below the x axis, then the area above the x axis, disregard the negatives; and add the two values together.

However, given that there seems to be only 3 and 0 given as values in that diagram, I added the 3/2 and 3/2 together to give me 9/4, which obviously is totally wrong.

I realise I have probably muddied the waters, and explained things clumsily, but just trying to give a full breakdown of the extent of my knowledge, and the steps I've taken thus far to try and find the solution.
Attached Images

2. (Original post by apronedsamurai)
...
For the first one, solve it geometrically.

For the second:
You need to find the point where it crosses the x axis. Call this point .

3. Im very sorry.

I don't quite follow what you mean, "solve it geometrically." Can you expand on that please?
4. (Original post by apronedsamurai)
Im very sorry.

I don't quite follow what you mean, "solve it geometrically." Can you expand on that please?
If you work out the points where it crosses the axes and the value of y at x = 3, you have two triangles which you can work out the area of.
5. Looking at the questions in the attached sheet, are you sure you're integrating correctly? I take it you tried integrating y = 2-x to get the equation in your post?
6. That is correct.

To be clear, the equation listed in my original post, refers to the first question, of the attached document.
7. Be careful with your integration method. should integrate to give .

Add one to the index of x and divide by the new index value for each term.
8. (Original post by apronedsamurai)
That is correct.

To be clear, the equation listed in my original post, refers to the first question, of the attached document.
It appears you made a mistake on that one.

Even so; using the integration equation, with a value of 3; gives us 3/2
10. (Original post by apronedsamurai)

Even so; using the integration equation, with a value of 3; gives us 3/2
This is where integrating the parts of the graph above and below the x-axis separately comes in. The section from 0 to 2 will integrate to give 2. The section from 2 to 3 will integrate to give -0.5. So if you simply integrate between 0 and 3 you'll get 1.5 which is wrong - area is always positive. So integrate them separately, add them ignoring the negative sign, and you get 2+0.5 =2.5
11. (Original post by Actaeon)
The section from 0 to 2 will integrate to give 2. The section from 2 to 3 will integrate to give -0.5.
Im sorry for being such a retard; but, where did you get the 2 from?

0 to 2, 2 to 3; where did the 2 come from?
12. (Original post by apronedsamurai)
Im sorry for being such a retard; but, where did you get the 2 from?

0 to 2, 2 to 3; where did the 2 come from?
That's where it crosses the x axis.
13. How did you determine it crosses the x axis; and at that point? :|
14. (Original post by apronedsamurai)
Im sorry for being such a retard; but, where did you get the 2 from?

0 to 2, 2 to 3; where did the 2 come from?
Not at all! 2 is the x-coordinate where the graph cuts the x axis which us why the question asks you to find it first. Take a look at the graph - for x>2 it's below the x axis, for x<2 it's above the axis.
15. (Original post by apronedsamurai)
How did you determine it crosses the x axis; and at that point? :|
From the graph it is clear that it crosses the x axis once.

16. Equation is y=x(x-2)
Graph cuts at x axis, therefore at that point y coordinate should be 0.
therefore , the coordinates of the point where curve cuts (a) = 2 [x(x-2)=y=0 => x-2 = 0 => x=2)

(Original post by apronedsamurai)
Im sorry for being such a retard; but, where did you get the 2 from?

0 to 2, 2 to 3; where did the 2 come from?
17. (Original post by Raysorb)
Equation is y=x(x-2)
Graph cuts at x axis, therefore at that point y coordinate should be 0.
therefore , the coordinates of the point where curve cuts (a) = 2 [x(x-2)=y=0 => x-2 = 0 => x=2)
Equation is actually y = 2 - x.
18. (Original post by morgan8002)
Equation is actually y = 2 - x.
According to his attachment, y = x(x-2)
19. (Original post by Raysorb)
According to his attachment, y = x(x-2)

There are three questions in total.

I was asking for help regarding the first question, so that I might be better enabled to work on the other two myself.
20. (Original post by Raysorb)
According to his attachment, y = x(x-2)
The first question in the attachment, the one that the previous 14 posts have been referring to has the equation y = 2 - x.

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