IS THIS RIGHT?
Find the area bounded by y = cosx, y = sinx and the y axis
solution below

Question

IS THIS RIGHT?
Find the area bounded by y = cosx, y = sinx and the y axis
solution below

anonymous · Answer

$$A = cosx dx- sinx dx  A= sinx - (-cosx) A = \sin(\pi/4) - \sin(0) - ( cospi/4-\cos0) A = 1/\sqrt{2} - 0[- 1/\sqrt{2} - 1]$$

anonymous · Answer

|dw:1425475250402:dw|

anonymous · Answer

@Compassionate

anonymous · Answer

@nikvist

anonymous · Answer

and how to graph that one @nikvist

nikvist · Answer

$$S=\int\limits_{0}^{\pi/4}\left(\cos{x}-\sin{x}\right)dx$$

anonymous · Answer

how o graph that y - axis, cosx and sinx

anonymous · Answer

and i have question why the upper limit become pi/4 and the lower limit become 0?

anonymous · Answer

@nikvist

nikvist · Answer

anonymous · Answer

why 0 become the lower limit?

anonymous · Answer

is that the graph of all?

anonymous · Answer

@nikvist

nikvist · Answer

x=0 (y-axis) is a begin of this area.

anonymous · Answer

why pi/4 become pi/4?

nikvist · Answer

Because x=pi/4 is end of area. Intersection of graph sin(x) and cos(x) is point (pi/4,sqrt(2)).

nikvist · Answer

(pi/4,sqrt(2)/2)

anonymous · Answer

how about the 0?

nikvist · Answer

anonymous · Answer

cant understand it easily

anonymous · Answer

im havig difficulties in figuring out how pi/4 and 0 put in upper and lower limit

nikvist · Answer

Simple, this area bounded:
left x=0, right x=pi/4, up y=cos(x), down y=sin(x). See graph.

anonymous · Answer

the graph is incomplete , theres no y axis graph

mathmate · Answer

|dw:1425471549015:dw|
Above graph shows what @nikvist had explained.