Question

how do i find the area in the first quadrant enclosed by the graphs of y=sinx, y =cosx, and the y-axis for 0 (less than or = to x sign here) x (less than or equal to sign here) pi/2. help please!

Answer 1

You can integrate between 0 and pi/2. For example the area of sinx in the region (0, pi/2) is:
\[\int\limits_0^{\pi/2} \sin x = [-\cos x]_0^{\pi/2} = -(0 - 1) = 1\]

Answer 2

just in case this is the problem here...

Answer 3

Is that the area enclosed between the curve and the x axis or the curve and the y axis?

Answer 4

its the area enclosed by the graphs of y=sinx, y=cosx and the y-axis for 0 is less than or equal to x and that same x is less than or equal to pi / 2....its in the picture...

Answer 5

The end of the lines aren't on the picture sorry!

Answer 6

most is its just mising the os x for y=cos x

Answer 7

Right it's between the curve and the y-axis then!
In that case you could consider taking away the area between the curve and the x axis from the square of area pi/2. Or theres a formula for it., which I'm struggling to remember at the minute!

Answer 8

?

Answer 9

For sinx we worked out the area between the curve and the x axis which was 1. Then between the y axis and the curve the area is \[\frac{\pi}{2} - 1.\]

Answer 10

and this is totally correct?

Answer 11

Yes. I've done it again and got the same :)

Answer 12

You can use the formula \[A = \int\limits_{y=a}^{y=b} f(y)dy \] which in this case gives \[\int\limits_0^1 \arcsin y ~dy \] but that's not a nice integration since it requires integration by parts.

Answer 13

okay thanks..