# AP Calculus AB : Asymptotic and Unbounded Behavior

## Example Questions

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### Example Question #121 : Asymptotic And Unbounded Behavior

Explanation:

We can use the substitution technique to evaluate this integral.

Let .

We will differentiate  with respect to .

, which means that .

We can solve for  in terms of , which gives us .

We will also need to change the bounds of the integral. When , , and when , .

We will now substitute  in for the , and we will substitute  for .

The answer is .

### Example Question #122 : Asymptotic And Unbounded Behavior

Evaluate:

Explanation:

Set .

Then and .

Also, since , the limits of integration change to  and .

Substitute:

### Example Question #123 : Asymptotic And Unbounded Behavior

Evaluate the following integral:

Explanation:

First you must know that:

and

Therefore we can rewrite our problem in this form:

where .

Thus the integral becomes,

### Example Question #124 : Asymptotic And Unbounded Behavior

Evaluate:

.

Explanation:

Setting the limits from zero to two we can find that,

### Example Question #125 : Asymptotic And Unbounded Behavior

Evaluate:

.

Explanation:

Seeing that the equation contains an absolute value you should know that the graph must always remain positive therefore resulting in a V-shaped graph.

Since the equation is , when  then the vertex of the graph is at .

The graph contains a triangle ranging from 0 to 1 and a triangle from 1 to 3. Remebering that taking the interal of a function is the same as finding the area under the curve we can use these triangles to solve our problem.

The area of the triangle from 0 to 1 is,

.

The area of the triangle from 1 to 3 is,

.

Thus the evaluated integral must be these areas added together,

.

### Example Question #126 : Asymptotic And Unbounded Behavior

Evaluate:

.

Explanation:

For this problem we need to use the U Substitution Method.

Using the U-du Rule you can set  and .

Because we only have a dx in our problem we need to solve for dx, thus

.

When  and when

Therefore your new equation will be:

.

Plugging in our interval we get,

### Example Question #127 : Asymptotic And Unbounded Behavior

Evaluate:

Explanation:

The first step is to find the antiderivative, recalling that:

.

For this integral:

where the intergral would be evaluated from  to  (the absolute value bar is not necessary, since both limits of integration are greater than zero):

### Example Question #128 : Asymptotic And Unbounded Behavior

Evaluate the following indefinite integral:

Explanation:

Use substitution, where  and .  Thus, the integral can be rewritten as:

.

Substitution of  back into this expression gives the final answer:

Note that since this is an indefinite integral, the addition of a constant term (C) is required.

### Example Question #129 : Asymptotic And Unbounded Behavior

Evaluate the limit: