The Indefinite Integral
Integration
We begin with a question.
Question: List two functions F(x) such that F'(x) = x
Answer: 1/2 x^{2} and 1/2 x^{2} + 3
We can see that if
F'(x) = x
then
F(x) = 1/2 x^{2} + C
for some constant C.
We call F(x) the antiderivative or integral of f(x) and write
In general, if
F'(x) = f(x)
then we write
From the derivative formula
d
x^{n} = nx^{n1}
dx
We get the integral formula
The Power Rule

Just like with derivatives, to find an antiderivative of a sum or difference, we can take the antiderivative of each term. Also like derivatives, the antiderivative of the product or quotient is not easily found.
Example
Which of these has an easy to find antiderivative
A. 8x^{3}  6x
B.
x
x^{5} + 2
Solution
We can find the antiderivative of part A. easily, by finding the antiderivative of 8x^{3} and 6x separately. The antiderivative is
8(1/4 x^{4})  6(1/2 x^{2}) + C = 2x^{4}  3x^{2} + C
B. This one, on the other hand, is a quotient. We do not have a way of finding its antiderivative.
Exercise
Find the following integrals:
(x
+ x^{2} )dx
1/
x^{2} dx
(12 x )^{2} dx
1
/ dx
(1  2x)^{20 }dx
Particular Solutions
We have seen that an integral produces a whole family of solutions parameterized by C. In most applications, we are given an initial or other condition and hence find the value of C. The antiderivative with known C is called a particular solution.
Example
Find a solution to
F'(x) = 4x  3
given that
F(1) = 2
Solution:
We first find an antiderivative:
F(x) = 2x^{2}  3x + C
Now plug in 1 for x and 2 for F to get:
2 = 2(1)^{2}  3(1) + C = 1 + C
So that C = 3. The particular solution is
F(x) = 2x^{2}  3x + 3.
Example
Find the solution to the differential equation
dy/dx = 3x^{2}
 4x + 2
Solution
We find the antiderivative of
3x^{2}
 4x + 2
We can find this antiderivative by finding the antiderivative of x^{2},
x, and 2 separately.
3(1/3 x^{3})
 4(1/2 x^{2}) + 2x
= x^{3}
 2x^{2} + 2x
Notice that since the derivative of a constant is zero, adding a constant of an
antiderivative results in another antiderivative for the same function. We
can write the final answer as
x^{3} 
2x^{2} + 2x + C
where C represents any constant.
Applications
Since the acceleration of gravity is a constant a = 32, we can derive the physics equations.
Example
Suppose that we kick a football with an initial upward velocity of 100 feet per second how long will it take to hit the ground?
Solution
We have
v(t) = 32 dt = 32t + C
v(0) = 100 = C
s(t) = (32t + 100)dt = 16t^{2} + 100t + C
s(0) = 0 = C
hence
s(t) = 16t^{2} + 100t = t(16t + 100)
So that
s(t) = 0 when 16t + 100 = 0
or
t = 100/16 = 6.25
It will take 6.25 seconds to hit the ground.
Example
Suppose the marginal revenue for a ski resort is
M = 50  0.01 x
And suppose that at $50 per ticket, the ski resort will have 2,000 skiers.
Find the demand equation.
Solution
Since the marginal revenue is the derivative of the revenue, the revenue is the antiderivative of the marginal revenue.
R = (50  0.01x)dx = 50x  0.005 x^{2} + C
The revenue is equal to the price times the quantity. That is
50x  0.005 x^{2} + C = px
Now find C by noting that when p = 50, x = 2,000.
50(2,000)  0.0005 (2,000)^{2} + C = (50)(2,000)
80,000 + C = 1,000,000
C = 920,000
Substituting the C into our equation and dividing by x gives the demand equation
p = 50  0.005 x + 920,000/x
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