172 CHAPTER 7. THE INTEGRAL

as this integral∫ x

a f (t)dt which is computed as A(x)−A(a) where A′ = f . The argumentis similar if f is decreasing.

Of course you should wonder whether you get the same thing for∫ b

a f (x)dx if you usesome other F̂ (x) with F̂ ′ (x) = f (x).

Proposition 7.0.2 Procedure 7.0.1 is well defined in the sense that any F satisfyingF ′ = f , on (a,b) with F continuous on [a,b] yields the same answer for

∫ ba f (x)dx.

Proof: Suppose both F ′ = f and F̂ ′ = f . Then let G(x) = F (x)− F̂ (x) . For any x,y ∈[a,b] , G(x)−G(y) = G′ (z)(x− y) for some z between x and y, this by the mean valuetheorem. However, G′ (z) = 0 and so G(x) = G(y) for every x,y. In particular, F (b)−F̂ (b)−

(F (a)− F̂ (a)

)= 0 which implies F (b)−F (a) = F̂ (b)− F̂ (a). Also, F (x) =

F̂ (x)+C for some constant equal to the common value of G.

Example 7.0.3 Find the area under the graph of the function y = x2 where 0 ≤ x ≤ 2.

You find an antiderivative. One which works is x3

3 because its derivative is x2. Then the

desired area is 23

3 − 03

3 = 83 .

Example 7.0.4 Find the area under y = sinx for x ∈[

π

2 ,π].

An antiderivative is −cos(x), so the desired area is −cos(π)−(−cos

(π

2

))= 1.

Of course the big question is whether there exists an antiderivative for an arbitrarycontinuous function. It turns out that the answer is yes, but sometimes we can’t find it.However, there is one kind of function for which this is an easy problem. I know, forexample that an antiderivative for p(x) = 1+ x+ 2x2 is x+ x2

2 + 2 x3

3 . Just look at it. Itworks. So is there an easy way to find an antiderivative? Yes for polynomials.

Lemma 7.0.5 Let p(x) = a0+a1x+ · · ·+an−1xn−1+anxn. Then if P(x) = a0x+a1x2

2 +

· · ·+ an−1xn

n + anxn+1

n+1 it follows that P′ (x) = p(x). Thus if p(x) = ∑nk=0 akxk, then an

antiderivative is ∑nk=0 ak

xk+1

k+1 .

Proof: This follows from the rules of differentiation.

Definition 7.0.6 ∫f (x)dx denotes all functions F such that F ′ (x) = f (x). Thus,

from Proposition 7.0.2,∫

f (x)dx = F (x)+C where F ′ = f and C is an arbitrary constantof integration.

There are many examples of how this can be used. Here is another one. Imagine a linenext to a three dimensional solid as shown in the next picture. For each y between a and b,let A(y) denote the area of the cross section of the solid obtained by intersecting this solidwith a plane through y perpendicular to the indicated line. Then V (y+h)−V (y)

h ≈ hA(y)h = A(y)

the approximation getting better as h gets smaller. Thus in the limit, V ′ (y) = A(y)