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# How do you find area under a curve watch

1. How do you find area under a curve. I know you have to integrate, but i forgot what you do

The equation I have is:

it says:

1. Find the area from x=0 to x=1
2. Find the area from y=0 to y=1

What do I need to do?
2. Integrate.
3. ok intergrate it, then do [1 substituded for x in intergrated equation]-[0]

we've just started covering this as were doing s1 and m1 first but i'm guessing for the y question you have to intergrate to get x=y something?

someone plz explain rep available
4. Calculate the integral with the limits of the integral being 0 to 1 in the first case.

In the second case, you also integrate using 0 to 1 as limits but you need to integrate the function that gives you x from y so if y=f(x), you're looking for the function g such that x=g(y) and you integrate that between 0 and 1
5. Integrate it, and the solve it for 1 and 0, and subtract.
6. (Original post by SamTheMan)
Calculate the integral with the limits of the integral being 0 to 1 in the first case.

In the second case, you also integrate using 0 to 1 as limits but you need to integrate the function that gives you x from y so if y=f(x), you're looking for the function g such that x=g(y) and you integrate that between 0 and 1

so the second one you'd intergate x=rooty right?
7. (Original post by Ekpyrotic)
Integrate.
I know i said that, but i forgot what you do

for 1. I get:

= from o to 1

=
8. What do you do for the y?
9. Idiotic post.
10. You're trying to find

for the next part.
11. (Original post by Ekpyrotic)
Integrate for x then flip.

do you intergrate for x then flip or flip and theb intergrate.

or are they the same thing?
12. (Original post by Bateman)
do you intergrate for x then flip or flip and theb intergrate.

or are they the same thing?
Correct idea is to ignore me, I was diff'ing.

13. = from o to 1

=

??
14. The above is n. right.
15. (Original post by Bateman)
so the second one you'd intergate x=rooty right?
I concur. I hadn't actually seen that y=x^2 so indeed x=sqrt(y) and that's what you integrate between 0 and 1. I believe that gives you 2/3. (there's a chance I'm wrong about the result)
16. (Original post by G O D I V A)

= from o to 1

=

??
Now sketch the curve and see why that's not the whole answer.
17. (Original post by Swayum)
Now sketch the curve and see why that's not the whole answer.

huh?
18. Did you sketch y = x^2? What area do you think your integral worked out? Hint: x = -rooty is perfectly valid also.
19. (Original post by Swayum)
Did you sketch y = x^2? What area do you think your integral worked out? Hint: x = -rooty is perfectly valid also.
I did sketch the graph, but i still dont get that
20. (Original post by G O D I V A)
I did sketch the graph, but i still dont get that
As you've stated it, the question appears to want the area 'inside' the parabola from y = 0 to y = 1 (imagine pouring water into it up to the level of y = 1). But all you've done is integrate root y between 0 and 1, which will give you the area between the curve x = root y (the right-hand half of the parabola) and the y-axis. Shade this bit in on your diagram. What else do you need to find in order to work out the required area?

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