If your solution to a stated
problem does not match the given solution, you should self-critique per
instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation
of the problem along with a statement of what you do or do not understand about
it. This response should be given, based on the work you did in completing the
assignment, before you look at the given solution.
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Question:
`q001. Integrate e^(3 t) with respect to t.
Your solution:
Confidence Assessment:
Given Solution:
If we understand the chain rule, it isn't
hard to figure out this antiderivative without using any special integration
techniques:
We know that, by the chain rule, the
derivative of e^(k t)
is (k t) ' * e^(k t) = k e^(k t).
So for example the derivative of e^(9
t) = 9 e^(9 t).
We therefore know that the derivative
of e^(3 t) is 3 e^(3 t).
So it's easy to figure out what we
need to take the derivative of to get 3 e^(3 t):
Thus 1/3 e^(3 t) is an antiderivative
of e^(3 t) and our general integral is
int(e^(3t) dt) = 1/3 e^(3 t) + c,
where c is an arbitrary integration
constant.
This type of reasoning works in a number
of simple cases. However not all integrals are as transparent as this one,
and we will need to develop some techniques to help us with more difficult
integrals.
We therefore use this example to
illustrate the process of integration by substitution, which basically
'reverses' the chain rule:
Our integral is of the form int(e^(3
t) dt).
We can change variables in this
expression from t = u as follows:
e^(3 t) is of the form e^u, with u
= 3 t.
In order to change the variable
from t to u, we must have an expression for dt.
The derivative of u with respect
to t is just (3 t) ' = 3.
We know that the derivative of u
with respect to t can be written du/dt = 3.
du/dt isn't really a fraction, but
for our present purposes we can treat it as if it is.
-
If we do so, we can solve du/dt
= 3 for dt.
-
We multiply both sides by dt
to get du = 3 dt, then we solve for dt, obtaining dt = du / 3.
Now we can replace e^(3 t) by e^u,
and dt by du/3 obtaining
int(e^u * du/3).
e^u * du / 3 is just 1/3 * e^u du,
and the 1/3 is a constant, so our expression is now
1/3 int(e^u du)
int(e^u du) = e^u + c, where c is
a constant, so our expression becomes
1/3 e^u + c.
Now u is still equal to 3 t, so
our expression can be written
1/3 e^(3 t) + c.
Self-critique (if necessary):
Self-critique Rating:
Question:
`q002.
Consider the integral int( e^(t^2) dt). It looks like we could do this one
in the same way as the first, this time letting u = t^2 and transforming the
integrand to the form e^u. However this doesn't work.
To see why, go
ahead and try it:
Let u = t^2.
Your solution:
Confidence Assessment:
Given Solution:
First let u = t^2.
First we find du/dt:
du/dt = 2 t
Next we solve for dt:
We first rearrange du/dt = 2 t to the
form du = 2 t dt, then solve for dt. We obtain
Now we try to
rewrite int(e^(t^2) dt) in terms of u. Using u = t^2 and dt = du / (2 t dt)
we get the expression
int(e^u *
du / (2 t) ).
This is
correct, but it doesn't help much. Our integral contains the variable t,
but the integral needs to be performed with respect to u.
There's still
hope: Since u = t^2, it turns out that u = sqrt(t) (actually u could
equally well be - sqrt(t), but let's not worry about that right now; it won't
matter in this case). Can't we just replace t by sqrt(u)?
It turns
out that we can, though as it turns out in this example that won't help.
We get the expression
int(e^u
du / (2 sqrt(u) ), which is the same as 1/2 int(e^u / sqrt(u) du)
That's at
least as bad as what we started with. We just can't integrate that.
We
could try. We could integrate e^u, and we could integrate sqrt(u),
but as you should know by this point that won't help us integrate e^u /
sqrt(u).
If you
need to convince yourself of that, go ahead. Then take the
derivative of whatever you get. Your derivative will be pretty
messy, and it won't be equal to e^u / sqrt(u).
I can be proven, in fact, that it's
impossible to express int(e^(t^2) dt) in terms of elementary functions.
COMMON FALLACY:
e^(t^2) is an antiderivative of e^(t^2).
WHY THIS ISN'T SO:
The derivative of e^(t^2) is (t^2) ' *
e^(t^2) = 2 t e^(t^2), not just e^(t^2).
It really is impossible to write down a
closed-form expression for the antiderivative of e^(t^2).
Self-critique (if necessary):
Self-critique Rating:
Question: `q003. Consider the
integral int( t e^(t^2) dt). It looks like this one would be as hopeless
at the preceding int(e^(t^2) dt). However this one works out.
Let u = t^2,
so that e^(t^2) can be written as e^u.
Before we go
on, look at what's left in the original expression after we've replaced e^(t^2))
by e^u:
So it's t * dt
that has to be replaced by an expression involving only the variable u.
Keep that in
mind as you do the following:
-
Find du/dt.
-
Solve for
du.
-
Can you
now solve for t * dt? You can. Do so.
-
Now
rewrite the expression int( t e^(t^2) dt) in terms of the variable u, with
no reference to the variable t.
Your solution:
Confidence Assessment:
Given Solution:
First let u = t^2.
First we find du/dt:
du/dt = 2 t
Next we solve for du:
We first rearrange du/dt = 2 t to the
form du = 2 t * dt
Now solve for t * dt:
Now we try to
rewrite int(t e^(t^2) dt) in terms of u. Using u = t^2 and t * dt = du /
(2 dt) we get the expression
int(e^u *
du / 2 ).
This is just
equal to 1/2 int( e^u * du), so our integral is 1/2 e^u + c.
Substituting
t^2 for u, we obtain our final result:
int( t
e^(t^2) dt) = 1/2 e^(t^2) + c.
To check, we
take the derivative of this expression:
(1/2
e^(t^2) + c) ' = 1/2 (e^(t^2) ' + c ' ).
c is a
constant so c ' = 0.
e^(t^2) '
= (t^2) ' e^(t^2) = 2 t e^(t^2), so
1/2
(e^(t^2) ' + c ' ) = 1/2 ( 2 t e^(t^2) + 0) = t e^(t^2).
The derivative
is our original integrand. So our substitution worked.
Self-critique (if necessary):
Self-critique Rating:
Question: 'q004.
Use a similar technique to find int( t^2 e^(t^3) dt).
Your solution:
Confidence Assessment:
Given Solution:
Substitute u = t^3, which transforms out
e^(t^3) into the form e^u, still leaving t^2 * dt in the expression.
We get du/dt = 3 t^2 so that du = 3
t^2 * dt.
We still have that t^2 * `dt in our
original expression. However our expression for du contains 3 t^2 * dt,
so we can easily solve for what we need:
du = 3 t^2 * dt, so to get an
expression for the t^2 dt we need we just divide both sides by 3:
t^2 dt = du / 3.
Our integral becomes
int( e^u * du/3) = 1/3 int(e^u du)
= 1/3 e^u + c.
Substituting for u we find that
int(t^2 e^(t^3) dt) = 1/3 e^(t^3)
+ c.
You should take the derivative of this
expression to verify that the result is the original integrand t^2 e^(t^3).
Self-critique (if necessary):
Self-critique Rating:
Question: `q005. Use
substitution to find the following:
Your solution:
Confidence Assessment:
Given Solution:
Solutions are given without a lot of
detail:
-
int(3 t^4 e^(t^5) dt) = 3/5 e^(t^5) +
c, obtained using u = t^5 so that du = 5 t^4 dt. Thus t^4 dt is equal
to du/5.
-
int( e^(sqrt(x)) / sqrt(x) dx) = 2 e^(sqrt(x))
+ c, obtained using u = sqrt(x) so that du = 1 / (2 sqrt(x) ) dx. Thus
dx / sqrt(x) is equal to 2 du.
-
int ( x e^(x^2 + 4) dx) = 1/2 e^(x^2 +
4) + c, obtained using u = x^2 + 4 so that du = 2 x dx. Thus x dx is
equal to du/2.
-
int ( 1/t^2 * e^(-1/t) dt) = e^(-1/t)
+ c, obtained using u = -1/t so that du = 1 / t^2 dt. Thus 1 / t^2 dt
is equal to du.
Self-critique (if necessary):
Self-critique Rating:
Question: `q006.
Find an antiderivative of e^(-.05 t)
Your solution:
Confidence Assessment:
Given Solution:
If we let u =
-.05 t we get du /dt = -.05 so that du = -.05 dt and dt = du / (-.05) = -20 du.
Thus
int
(e^(-.05 t) dt = int(e^u * (-20 du) ) = -20 int(e^u du) = -20 * e^u + c =
-20 e^(-.05 t) + c
To check:
The derivative of
-20 * e^(-.05 t)
+ c is -.20 * ( -.05
e^(-.05 t)) = e^(-.05 t).
Thus -20
e^(-.05 t) + c is an antiderivative of e^(-.05 t).
STUDENT
QUESTION
I would take integral of e^-0.05t dt let u = -.0.05
t
and du/dt = -0.05 which would be e^u * du/dx = e^-0.05t(-0.05) =
-0.05e^(-0.05t).
Not understanding where or how
they got -20.
INSTRUCTOR
RESPONSE
The derivative of -.05 * e^(-.05 t) is -.05 * ( -.05 e^(-.05 t)) = .0025 e^(-.05
t). This isn't equal to the original integrand, so your integration has an
error.
Important note: Always take the derivative of your antiderivative to check your
integration. It's always worth the time and effort, an greatly accelerates the
process of learning integration technique by connecting the process to the
result.
To integrate e^-0.05t dt you are right to let u = -.0.05 t so that du/dt =
-0.05.
This gives you du = -.05 dt.
The integral int(e^(-.05 t) dt) is in terms of dt, so you need an expression for
dt. Since du = -.05 dt, we have dt = du / (-.05), so that our integral becomes
int(e^u * du / (-.05) ).
Since 1/(-.05) = -20, this can be rewritten (using the constant rule) as
-20 int(e^u du), which is just
-20 * e^u + c. Substituting for u our integral becomes
-20 e^(-.05 t) + c.
Self-critique (if necessary):
Self-critique Rating:
Question: `q007. If we let u
= 3 x^2 + 1 in the integral int(x / (3
x^2 + 1)^2 dx), then the 3 x^2 + 1
in the denominator can be replaced by u.
-
If we make this substitution, what
expression remains to be expressed in terms of u?
-
What is our expression for du/dx?
-
What do we get when we solve for du?
-
Can we solve for the remaining
expression in terms of u and du?
-
What then is our integral in terms of
u?
Complete the integration.
Your solution:
Confidence Assessment:
Given Solution:
If u = 3 x^2 + 1 we get du/dx = 6 x.
Thus int( x / (3 x^2 + 1)^2 dx ) becomes int( x / u^2 * dx), leaving x * dx
still in terms of x.
Solving du/dx = 6 x for du we get
du = 6 x
* dx.
This is easily solved for our 'remaining
expression' x * dx:
x * dx = du / 6.
Since u = 3 x^2 + 1 and x dx = du/6, our
integral becomes
int(1 / u^2 * du/6), or just
1/6 * (1/u^2 du).
This is easily integrated:
1 / u^2 is the p = -2 power function u^-2,
so its antiderivative is the p = -1 power function -u^-1. We get
1/6 * (-1 / u) + c, or -1 / (6 u) + c.
Substituting 3 x^2 + 1 for u our result
becomes
int( x / (3 x^2 + 1) dx) = -1 / (6 ( 3
x^2 + 1) ) + c.
You should take the derivative to confirm
that this integral is correct.
Self-critique (if necessary):
Self-critique Rating:
Question: `q008.
Integrate the following:
Your solution:
Confidence Assessment:
Given Solution:
To integrate x / sqrt( 3
x^2 + 1), let u = 3 x^2 + 1.
The expression x * dx is equal to du / 6 and the integral becomes 1/6 int(du /
sqrt(u)) = 1/6 int(u^-(1/2) du) = 1/6 ( 2 u^(1/2) ) + c = 1/6 ( 2(3x^2 +
1)^(1/2) ) + c = sqrt( 3 x^2 + 1) / 3 + c.
When you integrate x (x^2 + 3)^5
you will get 1/6 int( u^5 du) = u^6 / 36 + c = (3 x^2 + 1)^6 + c.
When you integrate x / (x^2 + 3)^4
you will get 1/6 ( u^-4 du) = ... = -(3 x^2 + 1)^-3 / 18 + c = -1 / (18 ( 3 x^2
+ 1) ^3) + c.
Self-critique (if necessary):
Self-critique Rating: