INTRODUCTION TO CALCULUS
MATH 1A
Unit 18: Fundamental theorem
Lecture
18.1. The fundamental theorem of calculus for differentiable functions allows us
in general to compute integrals nicely. You have already made use of this theorem in the
homework for today. Earlier in the course, we saw that Sf (x) = h(f (0) + · · · + f (kh))
and Df (x) = (f (x + h) − f (x))/h we have SDf = f (x) − f (0) and DSf (x) = f (x)
if x = nh. This now becomes the fundamental theorem. It assumes f 0 to be
continuous.
Rx Rx
0
f 0 (t) dt = f (x) − f (0) and d
dx 0
f (t) dt = f (x)
Proof. Using notation of Euler, we write A ∼ B. We say ”A and B are close” and mean
that A − B → 0 for h → 0. 1 From DSf (x) = f (x) for x = khRwe have DSf (x) R x ∼ f (x)
x
for kh < x < (k + 1)h because f is continuous. We also know 0 Df (t) dt ∼ 0 f 0 (t) dt
because Df (t) ∼ f 0 (t) uniformly for all 0 ≤ t ≤ x by the definition of the derivative
and the assumption that f 0 is continuous and using Bolzano on the bounded interval.
We also know
R x SDf (x) = f (x)−f (0) for R xx = kh. By definition of the Riemann integral,
Sf (x) ∼ 0 f (t) dt and so SDf (x) ∼ 0 Df (t) dt.
Z x Z x
f (x) − f (0) ∼ SDf (x) ∼ Df (t) dt ∼ f 0 (t) dt
0 0
as well as Z x Z x
d
f (x) ∼ DSf (x) ∼ D f (t) dt ∼ f (t) dt .
0 dx 0
R5 8 8
Example: 0 x7 dx = x8 |50 = 58 . You can always leave such expressions as your final
result. It is even more elegant than the actual number 390625/8.
R π/2 π/2
Example: 0 cos(x) dx = sin(x)|0 = 1.
Rπ
Example: Find 0 sin(x) dx. Solution: The answer is 2.
R2
Example: For 0 cos(t + 1) dt = sin(x + 1)|20 = sin(2) − sin(1), the additional term
+1 does not make matter as when using the chain rule, it goes away..
R π/4
Example: π/6 cot(x) dx. This is an example where the anti-derivative is difficult
to spot. It becomes only easy when knowing where to look: the function log(sin(x)) has
1Bolzano or Weierstrass would write A ∼ B as ∀ > 0, ∃δ > 0, |h| < δ ⇒ |A − B| < . Parse this!
MATH 1A
Unit 18: Fundamental theorem
Lecture
18.1. The fundamental theorem of calculus for differentiable functions allows us
in general to compute integrals nicely. You have already made use of this theorem in the
homework for today. Earlier in the course, we saw that Sf (x) = h(f (0) + · · · + f (kh))
and Df (x) = (f (x + h) − f (x))/h we have SDf = f (x) − f (0) and DSf (x) = f (x)
if x = nh. This now becomes the fundamental theorem. It assumes f 0 to be
continuous.
Rx Rx
0
f 0 (t) dt = f (x) − f (0) and d
dx 0
f (t) dt = f (x)
Proof. Using notation of Euler, we write A ∼ B. We say ”A and B are close” and mean
that A − B → 0 for h → 0. 1 From DSf (x) = f (x) for x = khRwe have DSf (x) R x ∼ f (x)
x
for kh < x < (k + 1)h because f is continuous. We also know 0 Df (t) dt ∼ 0 f 0 (t) dt
because Df (t) ∼ f 0 (t) uniformly for all 0 ≤ t ≤ x by the definition of the derivative
and the assumption that f 0 is continuous and using Bolzano on the bounded interval.
We also know
R x SDf (x) = f (x)−f (0) for R xx = kh. By definition of the Riemann integral,
Sf (x) ∼ 0 f (t) dt and so SDf (x) ∼ 0 Df (t) dt.
Z x Z x
f (x) − f (0) ∼ SDf (x) ∼ Df (t) dt ∼ f 0 (t) dt
0 0
as well as Z x Z x
d
f (x) ∼ DSf (x) ∼ D f (t) dt ∼ f (t) dt .
0 dx 0
R5 8 8
Example: 0 x7 dx = x8 |50 = 58 . You can always leave such expressions as your final
result. It is even more elegant than the actual number 390625/8.
R π/2 π/2
Example: 0 cos(x) dx = sin(x)|0 = 1.
Rπ
Example: Find 0 sin(x) dx. Solution: The answer is 2.
R2
Example: For 0 cos(t + 1) dt = sin(x + 1)|20 = sin(2) − sin(1), the additional term
+1 does not make matter as when using the chain rule, it goes away..
R π/4
Example: π/6 cot(x) dx. This is an example where the anti-derivative is difficult
to spot. It becomes only easy when knowing where to look: the function log(sin(x)) has
1Bolzano or Weierstrass would write A ∼ B as ∀ > 0, ∃δ > 0, |h| < δ ⇒ |A − B| < . Parse this!
