Calculus Question: Definite Integrals?

Question

I have this theory that x1 and x2 are two different numbers. Both between pi/2 and pi. Such that

the interval over (x1,0) of cos(x)dx = 0.686
AND
the interval over (x2,0) of cos(x)dx = 0.686

I don't get this.

When I draw the graph of cos(x):
the only value for x, between pi and pi/2, that could possibly equal 0.686 is 0.814

The only other x value, between pi and pi/2, is 2.322. But this would give me a -0.686.

Then, I have to draw a graph, f, such that

The integral over (4,1) of f(x)dx = integral of (6,1) of f(x)dx.

Can someone help me understand this? What am I missing?

You know what, that was an idiot error on my part. I actually read the question wrong. The question is ASKING me if there are ARE two different numbers x1 and x2 such that cos(x)dx=0.686.

Answer 1

OK, the second one is easy:
All you need is for the integral from 4 to 6 of the function to be zero.

So, draw an x axis and a y axis.
Draw a line from (1,4) to (4,4) and then to (6, -4).
The integral from 1 to 4 of this function is 3*4 or 12
The integral from 1 to 6 is 12 + the integral from 4 to 6, which is zero as you'll see.

I'll agree that the first makes no sense. Between pi/2 and pi, the cosine function is negative. I think there is only ONE number that will satisfy (integral from 0 to x of cosxdx = .686), and that is arcsin(.686), and I get 2.385 for a number between pi/2 and pi whose sine is .686. Are you sure about the limits?