Q. 77

Question

Prove the Fundamental Theorem of Calculus in your own words. Use the proof in this section as a guide.

Step-by-Step Solution

Verified

Answer

Ans: If $f$ is continuous on $[a, b]$ and F is any antiderivative of $f$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a) .$

1Step 1. Given Information:

The Fundamental Theorem of Calculus .

2Step 2. Proving Inverse Fundamental Theorem of Calculus :

Before we get to the proofs, let’s first state the Fundamental Theorem of Calculus and the Inverse Fundamental Theorem of Calculus. When we do prove them, we’ll prove ${F T C}^{- 1}$ before we prove FTC.

$If F 0 is continuous on [a, b], then \int_{a}^{b} F^{'} (x) dx = F (b) - F (a) In other words, if F is an antiderivative of f, then \int_{a}^{b} f (x) dx = F (b) - F (a) A common notation for F (b) - f (a) is F (x) \begin{matrix} b \\ a \end{matrix} There are stronger statements of these theorems that don ’ t have the continuity assumptions stated here, but these are the ones we ’ ll prove .$

3Step 3. Proving the Fundamental Theorem of Calculus :

$I f f i s a c o n t i n u o u s f u n c t i o n o n t h e c l o s e d i n t e r v a l [a, b], a n d F i s i t s a c c u m u l a t i o n f u n c t i o n d e f i n e d b y F (x) = \int_{a}^{x} f (t) d t f o r x i n [a, b], t h e n F i s d i f f e r e n t i a b l e o n [a, b] a n d i t s d e r i v a t i v e i s f, t h a t i s, F' (x) = f (x) f o r x \in [a, b] . \frac{d}{d x} \int_{a}^{x} f (t) d t = f (x) P r o o f o f t h e F T C^{- 1} . F i r s t o f a l l, \sin c e f i s c o n t i n u o u s, i t ’ s i n t e g r a b l e, t h a t i s t o s a y, F (x) = \int_{a}^{x} f (t) d t W e n e e d t o s h o w t h a t F' (x) = f (x) . B y t h e d e f i n i t i o n o f d e r i v a t i v e s, F' (x) = \lim_{h \to 0} \frac{F (x + h) - F (x)}{h} = \lim_{h \to 0} \frac{1}{h} (\int_{a}^{x + h} f (t) d t - \int_{a}^{x} f (t) d t) \underset{h \to 0}{= \lim} \frac{1}{h} \int_{x}^{x + h} f (t) d t$

4Step 4. Proving F   ' ( x )   =   f ( x )

$We ’ ll now go on to prove the FTC from the {FTC}^{- 1} G (x) = \int_{a}^{x} F' (t) d t Then by ftc - 1, G 0 (x) = F 0 (x) . Therefore, G and F differ by a constant C, that is, G (x) - F (x) = C for all x \in [a, b] but, G (a) = \int_{a}^{a} F' (t) dt = 0 and G (a) - F (a) = C, so C = - F (a) . Hence, G (x) - F (x) = - F (a) for all x \in [a, b] . In particular, G (b) - F (b) = - F (a), so G (b) = F (b) - F (a), that is, \int_{a}^{b} F' (t) dt = F (b) - F (a) .$

Q. 76

Q. 78

Other exercises in this chapter

Q. 75

Use the Fundamental Theorem of Calculus to give alternative proofs of the integration facts shown in Exercises 72–76. You may assume that all functions he

View solution

Q. 76

Use the Fundamental Theorem of Calculus to give alternative proofs of the integration facts shown in Exercises 72–76. You may assume that all functions he

View solution

Q. 78

Prove that the Net Change Theorem (Theorem 4.27) is equivalent to the Fundamental Theorem of Calculus.

View solution

Q. 79

Prove that if two functions F and G differ by a constant, then [F(x)]ab = [G(x)]ab.

View solution