Question

Find the area of the region in the first quadrant bounded below by the x-axis and above by the curves of sinx and cosx.

Answer 1

@dan815

Answer 2

@thomaster @iambatman

Answer 3

@Zarkon @.Sam.

Answer 4

have you sketched this situation?
Where do the sine and cosine curves intersect in the 1st quadrant?

Two integrals are needed to compute the area in question.

Answer 5

they intersect first at pi/4, so i think my interval would be 0 to pi/4

Answer 6

\[\int\limits_{0}^{\frac{ \pi }{ 4}}cosx-sinx dx\]

Answer 7

i think

Answer 8

sin pi/4 = cos pi/4, so I agree with you that far.

As for your integral, \[\int\limits\limits_{0}^{\frac{ \pi }{ 4}}cosx-sinx dx\]... Where's the differential (dx)?
Again, please sketch the functions y=cos x and y=sin x. Next, shade the areas between the graphs of cos x and sin x and the x-axis. How would you find the area under the cosine curve but above the sine curve, on the interval [0,pi/4]?

Answer 9

so then the integral would equate to\[\sin x+\cos x \]

Answer 10

on interval 0 to pi/4

Answer 11

this gives me sin(pi/4)+cos(pi/4)-1

Answer 12

Would you please sketch y=sin x on the interval [0,pi/4]. How would you find the area under this curve over the given interval?