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JOhnJDC

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## Homework Statement

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.

## Homework Equations

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

## 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:

S

_{n}=cos(b/n)(b/n) + cos(2b/n)(b/n) + ..., + cos(nb/n)(b/n)

Using the formula S

_{n}=[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 S

_{n}

Substituting x=b/n, S

_{n}=(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 S

_{n}= 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?

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