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When we find the indefinite integral of a function f(x) we say that we integrate the integrand f(x). Thus, the process of evaluating an integral is referred ...
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Integration is a formalization of the process of antidifferentiation. When we differentiate a function, we lose some information about the function - although we know how it is changing we don’t know where it started from. Thus, when we go through the reverse process of differentiation, we end up with a family of functions rather than just a single function. Because of this we need an initial value to find a unique solution to a differential equation. We define the indefinite integral in order to encompass the fact that we have a family of functions.
Definition 0.1.1 (Indefinite Integral). We call the set of all antiderivatives of f the indefinite integral of f , denoted by (^) ∫
f (x)dx.
The symbol
is an integral sign. The function f is the called the integrand and x is the variable of integration. We say that dx is a differential of x.
If we know any antiderivative F to a given function f , then we simply find the indefinite integral as (^) ∫
f (x)dx = F (x) + c,
because the expression on the right-hand side represents all possible antiderivatives of f (x), if we let c be an arbitrary constant. When we find the indefinite integral of a function f (x) we say that we integrate the integrand f (x). Thus, the process of evaluating an integral is referred to as integration. Unfortunately, the notation we have introduced for the indefinite integral probably looks rather arcane. As we move further into the study of integration, the reason for the notation will become more clear. For the moment let us note that this notation is based on the initial developments of calculus, where the notion of infinitesimal and differential were prevalent. The resemblance of
and an elongated S is not accidental. This notation reflects the relationship between integration and summation. Integration is essentially the summation of the area underneath the integrand f (x) over intervals of infinitesimal length. The differential dx represents an infinitesimal change in x, which represents the intervals of infinitesimal length over which the summation involved in integration occurs. When it will not lead to confusion we will refer to the indefinite integral as simply the integral. The term indefinite integral is used to distinguish the process of indefinite and definite integration (which we will consider soon). Although both of these concepts are related to the area underneath a function, the indefinite integral is a function, whereas the definite integral is a constant, which is given by the area underneath a function over a set interval (defined by the limits of integration, which are not present in an indefinite integral). Since the process of (indefinite) integration is an inverse to differentiation, we can derive many rules for integration using rules we already know for differentiation. For instance
d dx xn+1^ = (n + 1)xn
so we see that d dx
( (^) xn+ n + 1
= xn^ forn 6 = − 1.
This leads us to the product rule for integrals.
Theorem 0.1.1 (Power Rule for Integrals).
∫ xndx = xn+ n + 1
for every n 6 = − 1 (n ∈ R \ {− 1 }).
Example 1. Evaluate the indefinite integral of x^2. Solution In this case we use the product rule, to see that ∫ x^2 = x2+ 2 + 1
c = x^3 3
c.
Example 2. Evaluate the indefinite integral of
x. Solution Once we rewrite
x = x^1 /^2 we see that ∫ x^1 /^2 =
x^1 /2+ 1 /2 + 1
x^3 /^2 + c.
Example 3. Evaluate the indefinite integral of x−^3. Solution We find that (^) ∫ x−^3 = x−3+ −3 + 1
x−^2 + c
Example 4. Evaluate
dt. Solution Rewriting the integrand we find ∫ dt =
t^0 dt = t0+ 0 + 1
Just like differentiation, integration is a linear operation. What this means is that integration satisfies the following two properties.
Theorem 0.1.2 (Linearity of Integration). Let a ∈ R.
αf (x)dx = α
f (x)dx.
(f + g)(x)dx =
f (x)dx +
g(x)dx.
Example 5. Evaluate the indefinite integral of − 2 x−^3 + 4x^1 /^2. Solution We can use the constant product rule and sum rules in conjunction to find ∫ − 2 x−^3 + 4x^1 /^2 = − 2
x−^3 + 4
x^1 /^2 = −2(−
x−^2 ) + 4(
x^3 /^2 ) + c = x−^2 +
x^3 /^2 + c
In the above analysis we do not write the result as − 2 c + 4d, because we can easily enough choose c and d so that we have any value for the arbitrary constant. Thus, it is cleaner to just replace − 2 c + 4d with a single c, as both are capable of representing any arbitrary constant.