Question

find area under curve

Answer 1

\[y=\cos(\pi x/2)\]

Answer 2

\[y=1-x^2\]

Answer 3

So you have to calculate 2 areas?

Answer 4

those are curves first i need to find the intersection and then integrate it

Answer 5

I still do not understand completely: do you have to calculate the area between the 2 curves?

Answer 6

yup

Answer 7

Normally, finding the intersection of a 2nd degree polynomial function and a cos is nearly impossible, but today is your lucky day!
Can you see what the period of cos(pix/2) is?

Answer 8

i got 0 ,1,-1

Answer 9

OK, you got these right. Now shall we just integrate from 0 to 1, because from -1 to 0 gives the same result?
Now you only have to know which of the two is biggest on the interval (0, 1).

Answer 10

x^2 is bigger than cos(pix/2) on (0, 1), so the integral is: \[\int\limits_{0}^{1}(x^2-\cos(\frac{ \pi }{ 2 }x))dx\]
Do you know how to do this?

Answer 11

should that be 1-x^2

Answer 12

Sorry, you are right:\[\int\limits_{0}^{1}(1-x^2-\cos(\pi x/2))dx\]

Answer 13

but i dont know how to integrate cospix/2

Answer 14

is that sinpix/2

Answer 15

i got -1/3

Answer 16

Never mind (for now) the pi x/2 part. It's just a cos-function.
Primitive is sin(pix/2) --almost, because what happens if you calculate its derivative, just to check the answer?

Answer 17

If you check:\[(\sin(\frac{ \pi }{ 2 }x))'=\cos(\frac{ \pi }{ 2 }x) \cdot \frac{ \pi }{ 2 }\] :(

Because of the Chain Rule, you get an extra factor pi/2, but you don't want that!
Now if you just change sin(pix/2) to:\[\frac{ 2 }{ \pi }\sin(\frac{ \pi }{ 2 }x)\]you'll see that the extra pi/2 you got cancels out with the 2/pi!
Just see it as a correction factor that you need.
I do it precisely the same way: guess the primitive function, differentiate it to see what goes wrong (if so) and then apply a correction constant. Works fine.

Answer 18

\[\frac{ 2\pi-6 }{ 3\pi }\]

Answer 19

All in all I get:\[\left[ x-\frac{ 1 }{ 3 }x^3-\frac{ 2 }{ \pi }\sin(\frac{ \pi }{ 2 }x) \right]_{0}^{1}=\]\[1-\frac{ 1 }{ 3 }-\frac{ 2 }{ \pi }\sin \frac{ \pi }{ 2 }-(0-0-0)=\frac{ 2 }{ 3 }-\frac{ 2 }{ \pi }=\frac{ 2\pi-6 }{ 3\pi }\]

OK, I see you've already figured it out. Well done!

Answer 20

instead of 0-1 can i directly do from -1 to 1

Answer 21

You can, but because of the symmetry is is much easier to multiply by 2.
Going from -1 to 1 leads often to calculation errors: you have to calculate F(1)-F(-1).
See: all these minuses make you go nuts ;).
2*(F(1)-F(0)) is much easier, because F(0) =0, nothing to calculate there....

Answer 22

ok thx an dhappy new year

Answer 23

you too, thanks!