Antiderivatives

The Indefinite Integral

Integration

We begin with a question.

Question:  List two functions F(x) such that F'(x) = x

Answer:      1/2 x²         and          1/2 x² + 3

We can see that if

        F'(x) = x 

then 

        F(x) = 1/2 x² + C 

for some constant C.

We call F(x) the antiderivative or integral of f(x) and write

       \( \int f(x)dx = 1/2 x² + C \)

In general, if

         F'(x)  =  f(x) 

then we write

         \( \int f(x)dx = f(x) + C \)

From the derivative formula

              d
                      xⁿ  =  nx^n-1 
             dx

We get the integral formula

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.  8x³ - 6x

B.          x
                                
          x⁵ + 2

Solution

We can find the antiderivative of part A. easily, by finding the antiderivative of 8x³ and 6x separately.  The antiderivative is

        8(1/4 x⁴) - 6(1/2 x²) + C  =  2x⁴ - 3x² + 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:

1. (x + x² )dx          
    

2. 1/ x² dx               
    

3. (12 x )² dx          
    

4.  \( \frac{1}{\sqrt{x}dx\) 
    

5. (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) = 2x² - 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) = 2x² - 3x + 3.

Example

Find the solution to the differential equation 

        dy/dx = 3x² - 4x + 2


Solution

We find the antiderivative of 

        3x² - 4x + 2

We can find this antiderivative by finding the antiderivative of x², x, and 2 separately.

        \( 3(\frac{1}{3}x^3 - 4(\frac{1}{2}x^2) + 2x \)

        \( = x^3 - 2x^2 + 2x \) 

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

        x³ - 2x² + 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  =  -16t² + 100t  + C

        s(0)  =  0  =  C

hence

        s(t)  =  -16t² + 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 x² + C

The revenue is equal to the price times the quantity.  That is

        50x - 0.005 x² + 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        

Back to the Integration and Applications Page