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Definite integrals and areas under the curve

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the area under the cosine curve y=cosx from x=0 to x=b, where 0 is less than b is less than or equal to pi/2.

    2. Relevant equations
    [tex]\Sigma[/tex]cos kx = [sin(1/2)(nx) cos(1/2)(n+1)x]/sin(1/2)x

    3. The attempt at a solution
    I let n be a large integer and divided the interval [0,b] by n equal subintervals of length b/n.

    The heights of the subintervals are:

    cos(b/n), cos(2b/n), ..., cos(nb/n)

    So, the sum of the areas of these subintervals (rectangles) is:

    Sn=cos(b/n)(b/n) + cos(2b/n)(b/n) + ..., + cos(nb/n)(b/n)

    Using the formula Sn=[tex]\Sigma[/tex]cos kx = [sin(1/2)(nx) cos(1/2)(n+1)x]/sin(1/2)x, with x=b/n, I got:

    Area=the limit as n approaches infinity of Sn

    Substituting x=b/n, Sn=(b/n)[sin(b/2)*cos((n+1)b/2n)]/sin(b/2n)

    Now, this is where I get confused. I get odd results when I apply the limit n approaches infinity:

    In the numerator, the cos becomes cos(b/2(1+1/n)), which approaches cos(b/2) as n approaches infinity. So, the numerator is sin(b/2)cos(b/2). In the denominator, sin(b/2n) approaches sin0 as n approaches infinity and sin0=0. There's something wrong with my denominator, I think. Also, the whole quotient is multiplied by (b/n), which would make the quotient approach 0 as n approaches infinity, right?

    The book says the following:

    If we make theta = b/2n, then theta approaches 0 as n approaches infinity, and using the limit as theta approaches 0 of sin(theta)/theta we see that:

    (b/n)(1/sin(b/2n) = 2(b/2n)/sin(b/2n) = 2(theta/sin(theta)), which approaches 2 as n approaches infinity. The book says that these facts lead the the answer: area=to the limit as n approaches infinity of Sn = 2sin(b/2)cos(b/2)=sin(b).

    I don't understand the operations in the line above--can someone explain? I don't see how (b/n)(1/sin(b/2n) = 2(b/2n)/sin(b/2n); where did that 2 come from?
    Last edited: Mar 21, 2010
  2. jcsd
  3. Mar 21, 2010 #2


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    The limit the book refers to is

    [tex]\lim_{\theta\rightarrow 0}\frac{\sin \theta} {\theta}=1[/tex]

    Note that the denominator is what appears in the argument of the sine function. In your problem, you have something similar:

    [tex]\lim_{n \rightarrow \infty}\frac{\frac{b}{n}}{sin\frac{b}{2n}}[/tex]

    but the numerator doesn't match the argument of the sine exactly. To use the previous result, you have to make them match, so you get

    [tex]\lim_{n \rightarrow \infty}\frac{\frac{b}{n}}{\sin\frac{b}{2n}}=\lim_{n \rightarrow \infty}\frac{2(\frac{b}{2n})}{\sin\frac{b}{2n}}=2 \lim_{n \rightarrow \infty}\frac{\frac{b}{2n}}{\sin\frac{b}{2n}}=2[/tex]
  4. Mar 21, 2010 #3
    Thank you, vela; your comment was very helpful, but I have a related question.

    Before your comment, I got as far as this:

    limit as n approaches infinity of (b/n)[sin(b/2)cos(b/2)]/sin(b/2n)

    Now I see that I can multiply the equation by one (2/2) and reorganize it as follows:


    Setting b/2n=theta and applying the limit as theta approaches 0 of theta/sin(theta), I get the book's answer: 2sin(b/2)cos(b/2)=sin(b).

    The only thing I'm still a little unclear on is how I know to apply that second limit, given that the only limit suggested by the equation I set up is n approaches infinity. Applying the second limit seems a significant change to the equation, unlike multiplying it by (2/2).
    Last edited: Mar 21, 2010
  5. Mar 22, 2010 #4


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    You know because when you let n go to infinity, you end up with something that looks like 0/(sin 0). What you're really doing is finding the limit of f(g(n)) as n goes to infinity. There's a theorem that tells you this is equal to the limit of f(x) as x approaches G, where G is the limit of g(n) as n goes to infinity. In this problem, f(x)=x/(sin x) and g(n)=b/2n.
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