Antiderivatives for exponential functions

[Pages:5]1. Antiderivatives for exponential functions

Recall that for f (x) = ecx, f (x) = c ecx (for any constant c). That is, ex is its own derivative. So it makes sense that it is its own antiderivative as well!

Theorem 1.1 (Antiderivatives of exponential functions). Let f (x) = ecx for some

constant c.

Then

F (x)

=

1 c

ecc

+

D,

for

any

constant

D,

is

an

antiderivative

of

f (x).

Proof.

Consider

F (x) =

1 c

ecx

+

D.

Then

by

the

chain

rule,

F

(x)

=

c

1 c

ecx

+

0

=

ecx.

So F (x) is an antiderivative of f (x).

Of course, the theorem does not work for c = 0, but then we would have that f (x) = e0 = 1, which is constant. By the power rule, an antiderivative would be F (x) = x + C for some constant C.

2.

Antiderivative

for

f (x) =

1 x

We have the power rule for antiderivatives, but it does not work for f (x) = x-1.

However, we know that the derivative of ln(x) is

1 x

.

So it makes sense that the

antiderivative

of

1 x

should

be

ln(x).

Unfortunately,

it

is

not.

But

it

is

close.

Theorem

2.1

(Antiderivative of

f (x)

=

1 x

).

Let

f (x)

=

1 x

.

Then

the

antiderivatives

of f (x) are of the form F (x) = ln( x ) + C.

Proof. Notice that

F (x) = ln( x ) =

ln(x) ln(-x)

for for

x>0 x

0,

we

have

[ln(x)]

=

1 x

.

For

x

<

0,

we

have

from

the

chain

rule

that

[ln(-x)]

=

-

1 -x

=

1 x

.

So

ln(

x

)+C

is

indeed

an

antiderivative

of

f (x)

=

1 x

.

Why do we use F (x) = ln( x ) rather than ln(x)? Notice that ln(x) is defined only for x > 0, while ln( x ) is defined for all real numbers other than 0. Further

1

2

note that f (x) =

1 x

is defined also

for all real numbers other than 0.

It is so the

domains

of

1 x

and

its

antiderivatives

match

up.

3. Another definition of ln(x) We defined ln(x) as the inverse function of ex. There is another way to define ln(x) using calculus. We define

x1

ln(x) =

1

dt. t

In the interest of time, we will not go through the proof that shows our definition

of ln(x) and this new one are exactly the same funtion. But we should be aware of its existence as it is actually a very historical definition.

4. Integration by substitution

Suppose we would like to find

x + 3dx. How could we go about doing this?

We

need

to

find

a

function

F (x)

whose

derivative

is

x

+

3.

None

of

our

methods

so far seem to work.

We will make a substitution of the integral into a more friendly

variables. form. Let

The motivation for doing u = x + 3. Then we would

this is have

to utudrxn.

Unfortunately, we cannot take this derivative. Notice that the dx signifies that

we are integrating with respect to x. We need to somehow replace dx by du in

some way. Notice that u is a function of x, so we may differentiate it with respect

to of

x. x,

du dx

=

1.

So du

so we can just

= dx. make

So the

the differential of substitution. So

wueisgeitdenticuadl uto=th23 ue3d2iff+eCre.ntSiaol

we have taken the derivative, but we would like the function to be in terms of x.

This is simple to fix, just reverse the substitution. Remember, u = x + 3. Thus,

C23 us3

2

+C

=

2 3

(x

+

3)3

2

+ C,

which

above are not necessarily the

is our same,

antiderivative. We should note that the but as they are arbitrary numbers, it is

fine to call them both C.

We should rule to get F

check (x) =

that this

3 2

2 3

(x

+

3is)1ac2 tu1a=llyaxn+a3n.tiSdoer, iivtatisiviendofeefd(xa)n.

We use the chain antiderivative.

What we did is a method often called u-subsitution. Intuitively, it "undoes" the chain rule for derivatives. We know that

d dx

F

(g(x))

=

F

(g(x))

g(x)

=

f

(g(x))

g(x),

3

where F (x) = f (x). This is just the chain rule. Now integrate both sides to get

d dx F (g(x)) =

f (g(x)) g(x).

By

the

fundamental

theorem

of

calculus,

d dx

F

(g(x))dx

=

F

(g(x))

+

C,

so

we

get

f (g(x)) g(x)dx = F (g(x)) + C.

What this means is if you have a function of the form f (g(x)) g(x) for some functions f (x) and g(x), then f (g(x)) g(x)dx = F (g(x)) + C, where F (x) is

the antiderivative of f (x). So to use this method, we must watch for functions that are of the form a composition of functions multiplied by the derivative of the

insider function of that composition. This sounds convoluted, but with practice,

it becomes much more natural.

Example 1. Find 2x sin(x2)dx.

Here we have that sin(x2) is a composition of functions. If f (x) = sin(x) and g(x) = x2, then our function is of the form f (g(x)) g(x). So we can use the substitution method.

We get sin(x2) 2xdx = - cos(x2) + C. We can easily check this by differentiating.

That was a very nice example, so let's try something a little harder.

Example 2. Find

5x6 dx.

12x7 + 19

We

need

a

g(x).

for

simplicity,

say

g(x)

= u = 12x7 + 19.

Then

du dx

= 84x6,

which

gives du = 84x6dx. We can easily substitute in u to get

5x6 12x7 + 19 dx =

5x6 dx.

u

We need to somehow get rid of that 5x6dx and get some sort of du. Well, we

know

du

= 84x6dx.

So

5 84

du

=

5x6dx.

So

we

have

the

new

integral

5x6 12x7 +

19

dx

=

5 84

1 du

u

=

5 84

ln(

u

)

+

C

=

5 84

ln(

12x7

+

19

)

+

C.

4

We can check this by differentiating with the chain rule.

5. Integration by parts

We have an integration method that "undoes" the chain rule for derivatives. We now present a method that "undoes" the product rule for derivatives. The method is know as integration by parts.

From the product rule, we know that

[f (x) g(x)] = f (x)g(x) + g(x)f (x).

Rearranging, we get

f (x)g(x) = [f (x) g(x)] - g(x)f (x).

Now we integrate both sides to get

f (x) g(x)dx = f (x) g(x) - g(x) f (x)dx.

Finally, call u = f (x) and v = g(x).

Then

du dx

= f (x) and

dv dx

= g(x).

Then we

get the very important formula

u dv = u v - v du.

Example 3. Find ln(x)dx.

This may not look like a formula of the form u dv, but it is. If we let u = ln(x)

and

v

= x,

then

du =

1 x

dx

and

dv

= 1dx.

So

ln(x)dx = u dv = u v - v du = ln(x) x - 1dx = ln(x) x - x + C.

To reiterate, ln(x)dx = x ln(x) - x + C.

Example 4. Find x cos(x)dx.

5

The challenge of integration by parts problems is to determine which function

should be u and which should be dv. Let's see what happens when we let u = cos(x)

and

dv

= xdx.

Then

we

get

v

=

1 2

x2

and

du = -sin(x)dx.

Then

x cos(x)dx

=

1 x2 2

cos(x)

-

-

1 2

x2

sin(x)dx.

While this formula is true, our choices of u and dv yielded a much more complicated formula. So let's switch them. u = x and dv = cos(x)dx. So du = 1dx and v = sin(x). So we get

x cos(x)dx = x sin(x) - sin(x)dx = x sin(x) + cos(x) + C.

As with all antiderivative problems, we can check our answer with differentiation.

The integration by parts formula is worth memorizing. Some people use the mneumonic "ultra-violet voodoo" to help remembering it.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download