9 FURTHER APPLICATIONS OF INTEGRATION

SPECIAL PROJECT FOR SECTION 9.1
How to Define π
(Idea due to H. Wu, The Joy of Lecturing, pages 5–6)
In this age of calculators, people sometimes get confused about how the number π is defined. The constant
π is not the number 22 7
, nor 3.1415926, nor even
3.14159265358979323846264338327950288419716939937510582097
All of those are just approximations.
One can find many expressions that equal π. Forexample,
( )
2 × 2 × 4 × 4 × 6 × 6 × 8 ×···
π = 2
1 × 3 × 3 × 5 × 5 × 7 × 7 ×···
π = 4
(1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 +···)
11
π =
√
6
1 2 + 6 2 2 + 6 3 2 + 6 4 2 +···
π = 3
(1 + 12
4 × 6 + 1 2 × 3 2
4 × 6 × 8 × 10 + 1 2 × 3 2 × 5 2
+···)
4 × 6 × 8 × 10 × 12 × 14
Usually, however, people define π as the ratio of the circumference of a circle to its diameter. Then we say
that the circumference of a circle is 2πr, and its area is πr 2 . But it is easy to forget how amazing an idea this
originally was; that the circumference, the area, and the radius of a general circle are so tightly related. In
this exercise, we go back to basics. We define π as the length of a semicircle of radius 1 (that is, one-half the
circumference) and then prove that it is equal to the area of a circle of radius 1.
1. Use the arc length formula to show that the length of the semicircle of radius 1, x 2 + y 2 = 1, y ≥ 0, can
be written as
∫ 1
−1
dx
√ .
1 − x 2
2. Show that the statement “The length of a semicircle of radius 1 equals the area of a circle of radius 1” can
be written as
∫ 1
−1
∫
dx
1 √
√ = 2 1 − x 2 dx
1 − x 2
−1
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