How do you find area under a curve

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

Reply 4

Rainfaery

12

Integrate it, and the solve it for 1 and 0, and subtract.

Reply 5

Bateman

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?

Reply 6

OP

Ekpyrotic

Integrate.

I know i said that, but i forgot what you do

for 1. I get:

\displaystyle\int^1_0 x^2 \, dx

=

\frac{x^3}{3}

from o to 1

=

\frac{1}{3}

Reply 7

OP

What do you do for the y?

Idiotic post.

You're trying to find

\int^1_0 \sqrt{y} \ \mathrm{d}y

for the next part.

Reply 10

Bateman

Ekpyrotic

Integrate for x then flip.

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

or are they the same thing?

Reply 11

Ekpyrotic

13

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.

Reply 12

OP

\displaystyle\int^1_0 y^\frac{1}{2}, dy

=

\frac{2}{3} \times y^\frac{3}{2}

from o to 1

=

\frac{2}{3}

??

Reply 13

Ekpyrotic

13

The above is n. right.

Reply 14

SamTheMan

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)

Reply 15

Swayum

18

G O D I V A

\displaystyle\int^1_0 y^\frac{1}{2}, dy

=

\frac{2}{3} \times y^\frac{3}{2}

from o to 1

=

\frac{2}{3}

??

Now sketch the curve and see why that's not the whole answer.

Reply 16

OP

Swayum

Now sketch the curve and see why that's not the whole answer.

huh?

Reply 17

Swayum

18

Did you sketch y = x^2? What area do you think your integral worked out? Hint: x = -rooty is perfectly valid also.

Reply 18

OP