Page 121 - Calculus Demystified

CHAPTER 4
108 As a result, the Riemann sum for the partition P is The Integral
k 2
R(f, P) = 4j · 2
j=1 k 2 k
k 2 k
= 8j = 8 j .
2
j=1 k 3 k 3 j=1
Now formula II above enables us to calculate the last sum explicitly. The result
is that
3 2
R(f, P) = 8 · 2k + 3k + k
k 3 6
= 8 + 4 + 4 2 .
3 k 3k
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In sum, 8Y
2 2 4 4 8
0 x dx = lim R(f, P) = lim + k + 3k 2 = .
k→∞ 3
3
k→∞

We conclude that the desired area is 8/3.
You Try It: Use the method presented in the last example to calculate the area
under the graph of y = 2x and above the x-axis, between x = 1 and x = 2. You
should obtain the answer 3, which of course can also be determined by elementary
considerations—without taking limits.

The most important idea in all of calculus is that it is possible to calculate an
integral without calculating Riemann sums and passing to the limit. This is the
Fundamental Theorem of Calculus, due to Leibniz and Newton. We now state the
theorem, illustrate it with examples, and then briefly discuss why it is true.
Theorem 4.1 (Fundamental Theorem of Calculus)
Let f be a continuous function on the interval [a, b].If F is any antiderivative of
f then
b f(x) dx = F(b) − F(a).
a
EXAMPLE 4.5

Calculate
2 x dx.
2
0

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