Question

graph y=cos(pi*x/2) and y=1-x^2 and use integration to find the area in between the curve.

Okay, so when I graph these two I see that they like overlap during the [-1,1] x interval. But maybe there is still a small gap in between? But I'm not sure if the [-1,1] x interval is correct. Please help!

Thanks in advance

Answer 1

I see what you mean, the two curves appear to coincide form -1 to 1
http://www.wolframalpha.com/input/?i=y%3Dcos(pi*x%2F2),+y%3D1-x%5E2

Because of the symmetry, I looked closer from 0 to 1 and it shows that the parabolo lies above the cosine curve for that domain, and of course must do the same for x from -1 to 0
http://www.wolframalpha.com/input/?i=y%3Dcos(pi*x%2F2),+y%3D1-x%5E2+,+for+0+to+1

Also notice that they meet at x = 0 , at (0,1)

So we will just take the area from 0 to 1 and double it

area - 2∫(1 - x^2 - cos(πx/2) dx from 0 to 1

take over, the integration is straight-forward.

Answer 2

It's not surprising that the two curves are so close. As you learn about functions, you will come across the Taylor Series. Any function can be approximated over a limited domain by a polynomial. This is handy, because polynomials are well behaved.

Anyway, you will find that

cos(u) = 1 - u^2/2! + u^4/4! - u^6/6! ...

Note how both functions are even functions. And the difference between cos(u) and 1-u^2/2! is very tiny -- 4th and higher powers. On the domain (-1,1) those higher powers of fractions are quite small.

Plug in u = π/2 x and you have

cos(π/2 x) ≈ 1 - (π/2)^2/2! x^2

On the interval [-1,1] that is just 1- π^2/8 x^2. Since π^2/8 > 1, that parabola lies just below 1-x^2. as seen here in the restricted graph:

http://www.wolframalpha.com/input/?i=y%3Dcos%28pi*x%2F2%29,+y%3D1-x^2+for+-1%3C%3Dx%3C%3D1