Area

[chjopl]

Find area between y=Sin x and y=Cos x on intervals [Pi/4, 15Pi/4].

[faust9]

Area can be found by \int_{x(0)}^{x(1)} (F(x)-G(x))dx

Simply let one of the functions be F(x) and the other be G(x). The absolute value of your integral over your limits will be the area (we don't owe the universe area if it comes out negative).

Also, use symetry to your advantage here. from pi/4 to 5pi/4 is one area block. You have 3.75 area blocks thus you can say 3xOne area block plus the area of the last block (13pi\4 to 15pi\4)

good luck.

[garytse86]

how do you that the areas won't cancel out? If one is above and the other below the x-axis?

[chjopl]

I can do the intregal part i am just having trouble finding the points of intersection. I know there are 4 different intregals that need to be used.

[faust9]

sin and cos functions intersect every (x)pi/4

[faust9]

how do you that the areas won't cancel out? If one is above and the other below the x-axis?

The answer here depends on how the question is actually phrased. I gave a general response to finding an area using integrals; however, the usage depends on the question. If the absolute area is desired then you must break the function into discrete areas and sum the individual areas. If the question wants the differential area then simple application of the area formula suffices.

[garytse86]

when u talk about differential area do you mean just normal intergration with limits regardless of "cancelling the area"?

[CartoonKid]

Integration is just a convenient way to find the area; however, we cannot always use pure integration in finding out the area. The best way to visualise is to draw the graph. You will see that there will be some common areas between this 2 equation. The trick is that we must avoid adding them twice. By finding out the points of intersection, we can eventually find the area.

[CartoonKid]

To find the point of intersection
sin x = cos x
sin x = sqrt(1-(sin x)^2)
then continue the working and you should be able to find all points of intersection.