Need help with an Area/Integration Problem (1 Viewer)

[blackops23]

Hi, I don't really know how to do these area questions, could some please explain how to do:

Q1. Find the area under y=cosx - (1/2) BETWEEN x=-pi and x=pi

It is the (-1/2) part added to the graph that is causing me problems.

Could some please demonstrate to me how to do this question?

Thank you very much.

 

[Drongoski]

Hi, I don't really know how to do these area questions, could some please explain how to do:

Q1. Find the area under y=cosx - (1/2) BETWEEN x=-pi and x=pi

It is the (-1/2) part added to the graph that is causing me problems.

Could some please demonstrate to me how to do this question?

Thank you very much.

If you are familiar with graph of y = cos x, it is symmetrical about the y-axis cutting the x-axis at +- pi/2 and an infinite number of other points further out. So it is an EVEN function. y = cos x - 0.5 is just the same graph pulled down half a unit. Between + - pi/2, the graph now intersects the x-axis at +- pi/3. So now we have 1 part below the graph betwen +- pi/3 and 2 equal parts under the x-axis, each with area between curve and pi/3 and pi/2. (this is "negative" area.)

The question itself is ambiguous. Does it want the total area between the curve y = cos x - 0.5 and the x-axis within the said limits?? If so, by symmetry of even function:

Area =



From this I got (if no error)

 

[ohexploitable]

You just find the x-intercepts then find area as usual.

 

[blackops23]

Thanks for the quick helps guys, appreciate it heaps!

I got 2(root3) + pi/3. Can someone confirm whether this is correct?

 

[hscishard]

The bounds for the 2nd integral is pi/3 to pi