Karl's Calculus Tutor - Solutions to Problems in 11.2

# Solutions to Problems in 11.2

### Exercise 1:

The problem was to find the indefinite integral of

 ``` ``` ``` ______ Ö3x + 2 dx ```

Step 1) Which of the rules that we developed in the last section can be applied to simplify this? Look at equation 11.2-11b. Can you see how this problem is just a special case of the general rule given in that equation? You just have to identify what the parts are. Equation 11.2-11b has the following parts that you have to identify with parts of the integral in this problem. They are: A, the constant by which the integrand is scaled; b, the constant by which x is scaled, and f, the constant by which bx is skewed. You also have to identify f, the function that is wrapped around all this.

Step 2) Identify A. Since there is no scalar shown in this problem that multiplies over the entire integrand, that means that  A = 1

Step 3) Identify b. The thing in this problem that scales x is 3. So  b = 3

Step 4) Identify f. The thing that is added to bx (which in this case is 3x) is 2. So  f = 2

Step 5) What is the function, f, that wraps this whole thing? Clearly it's the square root function. The argument to that f is  3x + 2.  The rule from equation 11.2-11b says that if you know the antiderivative of  f(x),  then you can find the indefinite integral of  f(3x + 2)  by treating the expression,  3x + 2,  as a single entity -- that is treat it as if it were itself a variable,  u = 3x + 2.  Then you find the antiderivative of:

```            _
f(u)  =  Öu
```

Step 6) What is an antiderivative of f? Look it up on the table if you don't already have that table memorized.

```            2
F(u)  =    u3/2
3
```

Step 7) Put it all together using the rule from equation 11.2-11b. That is put  3x + 2  in for u, and multiply the antiderivative by A/b (which is 1/3 in this case). And then don't forget to add the undetermined constant:

 ``` ``` ``` ______ 1 2 2 Ö3x + 2 dx = (3x + 2)3/2 + C = (3x + 2)3/2 + C 3 3 9 ```
With just a little practice you should be able to apply the rule in equation 11.2-11b on inspection without having to write down all the A, b, f, f, and u stuff. Go over how this rule was applied in this problem a few times until you can do it without having to go through all the intermediate steps.

### Exercise 2:

The problem was to find the indefinite integral of

 ``` ``` ``` (sin(px) + cos(3px)) dx ```

Step 1: Recognize that this is a sum. That means that the rule in equation 11.2-13b applies. This rule breaks the integral into the sum of two simpler integrals:

 ``` ``` ``` (sin(px) + cos(3px)) dx = ``` ``` ``` ``` sin(px) dx + ``` ``` ``` ``` cos(3px) dx ```

Step 2: Recognize that the rule in equation 11.2-11b applies to each of the summands. So identify what the parameters of the rule are in each summand. In both summands you have  A = 1.  Also can you see that in both summands you have  f = 0?  But in the first summand you have  b = p,  and in the second summand you have  b = 3p

Step 3: What are the f and g functions here that you can find in the table? Clearly if you let f be sin and g be cos, then what you have is

 ``` ``` ``` sin(px) + ``` ``` ``` ``` cos(3px) = ``` ``` ``` ``` f(px) + ``` ``` ``` ``` g(3px) ```
So what does the table tell you about what the antiderivatives of f and g are? It tells you that F is -cos and G is sin.

Step 4: Apply the rule from equation 11.2-11b. In equations, that means that

 ``` ``` ``` f(px) + ``` ``` ``` ``` 1 1 g(3px) = F(px) + G(3px) p 3p ```
(Notice again how we combined the two undetermined constants into just one undetermined constant, because an undetermined constant plus an undetermined constant is just yet another undetermined constant)

Step 5: Put in the functions for F and G. So you get as your answer

 ``` ``` ``` 1 1 (sin(px) + cos(3px)) dx = - cos(px) + sin(3px) + C p 3p ```

### Exercise 3

Step 1: Recognize that this is a sum. So again the rule in equation 11.2-13b applies. This rule breaks the integral into the sum of two simpler integrals:

 ``` ``` ``` 2 + e-nx dx = 1-x ``` ``` ``` ``` 2 dx + 1-x ``` ``` ``` ``` e-nx dx ```

Step 2: Recognize that the rule in equation 11.2-11b applies to each of the simpler integrals. So you need to determine A, b, and f for each of them. For the left integral, first turn it around into

 ``` ``` ``` 2 dx -x+1 ```
which makes it clear that in this one we have  A = 2,   b = -1,  and  f = 1.  For the right integral you have
 ``` ``` ``` e-nx dx ```
You should be able to see that in this one we have  A = 1,   b = -n,  and  f = 0

Step 3: What are the f and g functions here that you can find in the table? The  2/(-x+1)  is certainly a  1/x  style function, which is on the table. So we'll say that  f(x) = 1/x.  Likewise  e-nx  is an  ex  style function. So we'll let  g(x) = ex.  The table gives you antiderivatives for f and g as well. They are

```   F(x)  =  ln|x|

G(x)  =  ex
```

Step 4: Apply the rule in equation 11.2-11b to both summands. You have, according to the rule:

 ``` ``` ``` 2f(-x+1) dx + ``` ``` ``` ``` 1 g(-nx) dx = -2F(-x+1) - G(-nx) + C n ```

Step 5: Put in the functions for F and G.

 ``` ``` ``` 2dx + -x+1 ``` ``` ``` ``` 1 e-nx dx = -2ln|-x+1| - e-nx + C n ```
The expression to the right of the equal in the above is the indefinite integral of the original problem.

### Exercise 4

The problem was to find the indefinite integral,

 ``` ``` ``` x4 + 8x3 + 12x2 - 4x - 3 dx ```

Step 1: Recognize that this (and every polynomial) is nothing more than the sum of power terms. Since the integral of a sum is the sum of the integrals, you can break this down into five separate integrals according to the rule in equation 11.2-13b. So the following is exactly equivalent to the original problem:

 ``` ``` ``` x4 dx + ``` ``` ``` ``` 8x3 dx + ``` ``` ``` ``` 12x2 dx - ``` ``` ``` ``` 4x dx - ``` ``` ``` ``` 3dx ```

Step 2: Apply the rule in equation 11.2-3 to each integral that has a constant multiplier. This gives you

 ``` ``` ``` x4 dx + 8 ``` ``` ``` ``` x3 dx + 12 ``` ``` ``` ``` x2 dx - 4 ``` ``` ``` ``` x dx - 3 ``` ``` ``` ``` dx ```

Step 3: Recognize that each of the integrals is now a case of xr. And you can find that in the table. So apply that to each integral and you get

```   1        8        12        4
x5  +    x4  +     x3  -    x2  -  3x  +  C
5        4         3        2
```
I'll leave it to you to reduce the fractional coefficients to lowest terms on your own (if I were grading your paper, I wouldn't take off if you neglected to do that step -- after all, 12/3 is just as legitimate a way of expressing a coefficient as 4).

### Exercise 5

The problem was to find the indefinite integral,

 ``` ``` ``` n å Bk(kx + 1)k dx k=1 ```
The hints pretty much gave away how to do this whole thing. The summation sign indicates that this is the sum of n terms, each in the form of
```   Bk(kx + 1)k
```
where the Bk is constant in every case. The hints told you that you needed only to find an antiderivative to this, and then you could put that back into the summation and have your answer. The rule in equation
11.2-11b tells you how to do this. The hints gave you a reason why, for the purposes of integration, you could treat k as a constant. That means that to set up the rule you have  A = Bk,   b = k,  and  f = 1.  The function from the table that you need to use here is  f(x) = xr,  where in this case  r = k.  So for an antiderivative of f you have from the table:
```             1
F(x)  =      xk+1
k+1
```
Taking the indefinite integral of  Bkf(kx + 1)  using equation 11.2-11b gives
 ``` ``` ``` Bk Bk f(kx + 1) dx = F(kx + 1) + C k ```
When you put it altogether you get
```   n    Bk
å         (kx + 1)k+1  +  C
k=1 k(k+1)
```
If you didn't get this, go back over your work. If you still don't have a clue on this one but you got through the other exercises, then go back and review them, looking to see how what you applied in them is applied here.