Indefinite integration

This is also known as ``antidifferentiation'', but that word is really convoluted.¹

The process of finding an anti-derivative, or an indefinite integral, of a function is usually a lot harder than finding a derivative. You can tell this isn't just because we don't want to teach the methods, since you can buy a calculator that can find derivatives of functions, but even a big computer can't find all antiderivatives of any function you might write down. But, we can start off easy:

Find the function F(x) whose derivative is x⁴ . That is, find the antiderivative of x⁴ .

Now, you know that the derivative of x⁵ is 5x⁴ , (x⁵)¢ = 5x⁴ , so we only have to change the input a little to take care of that 5;

æ ç è 1 5 x5 ö ÷ ø ¢ = x4.

The only problem is that this is not the only function with that derivative. The others are only a trivial change from this one. For any constant c ,

æ ç è 1 5 x5+c ö ÷ ø ¢ = x4.

By the way, I don't like calling these things ``antiderivatives'', so we're going to start calling the antiderivative of a function f the indefinite integral of f . We write the indefinite integral of f , F by:

F(x) = ó õ f(x)dx+C.

The +C is there to remind you that the indefinite integral is only known up to a constant. Maybe we should call it an indefinite integral of f , but we don't.

Theorem 1 The indefinite integral of x^a is

ó õ xadx = 1 a+1 xa+1+C,

if a ¹ -1 .

There is no need to work through a formal proof of this, since all you have to do to check that it is correct is differentiate the right-hand side.

Exercise 1 Why doesn't the formula in this theorem work for a = -1 ?

5.3.1  Examples/Table

You need to know these particular integrals. In general, I do not like to tell students to memorize things, but for this I make an exception. Again, to verify that these are correct, differentiate the right-hand sides of the equations.

Proposition 1  

1. 
   

   ó õ cf(x)dx = c ó õ f(x)dx

2. 
   

   ó õ f(x)±g(x)dx = ó õ f(x)dx± ó õ g(x)dx

Example 1 Find

ó õ x2+3x4dx.

Example 2

ó õ sinx- 1 x dx = -cosx-ln|x|+C

Note the ln|x| in that answer. The integral of 1/x is just ln(x)+C for x positive, but is ln(-x)+C = ln|x|+C for x negative, as you can check by differentiating, as usual.

5.3.2  Motion

If the position of a particle at time t is given by s(t) , then we saw that its velocity was v = s¢(t) . The second derivative of the position, the rate of change of the velocity, is the acceleration

a(t) = v¢(t) = s¢¢(t).

The understanding of motion problems really involves, usually, working this relationship backwards. Many times you will know the acceleration (this will come out of the forces involved), and you'll have to find the position as a function of time. If the acceleration is a(t) , then, using the notation I mentioned earlier for antiderivatives,

v(t) = ó õ a(t)dt+C,

where I am adding the +C for emphasis, since the notation òdt is supposed to mean any antiderivative of whatever is inside, so the +C is really there already. But, here you might be able to figure out, from the problem, the velocity at some time (usually t = 0 ), and use that to solve for the C . Then,

s(t) = ó õ v(t)dt+C

and again you figure out the constant.

Example 3 Assume that a thingie is moving along a line so that its acceleration is 4feet/sec² . If at t = 0 the thingie is moving at 5feet/sec and is 7feet from origin, what is its position at any time t ?

As long as we stay near the surface of the Earth, the acceleration of a body downward due to gravity is 32 feet/sec² , (or 9.8 meters/sec² ), which can be taken to be a constant for our purposes. If we presume that the air resistance can be forgotten (not really a good assumption, but it makes the problem easier), then a body in free fall will have that acceleration. Use that in the problems below.

Exercise 2 I shot an arrow, into the air. It fell to earth, I know not where. However, I shot the arrow from a height of 5 feet off the ground, at an angle of 45^° to the horizontal, pointing due North, with an initial velocity of 50 feet/sec.

a) Find a formula for the height h(t) of the arrow at time t (in seconds after I shot the arrow). h(t) =

Hint: Only that part of the initial velocity which is upward pointing counts. Use trigonometry to decide what the magnitude of the vertical part of the velocity is.

b) Find when the arrow will hit the ground. t =

c) Find a formula for the horizontal position s(t) of the arrow at time t , how far North of my position it is at time t . s(t) =

d) Find the position when the arrow hits the ground. s =

Assume that there is no air resistance, and that the ground is perfectly flat. This will answer the poet's question.

Exercise 3 A mischievous child (certainly not my child) has heard about dropping pennies off the Empire State Building. So, he goes the story one better. While standing on the observation deck 1,250 feet above the street, he throws a penny straight down, with an initial velocity of 75 feet per second. How fast is it going when the penny hits the ground? Assume that there is no air in New York.

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