# AP Calculus AB : Integrals

## Example Questions

### Example Question #11 : Applications Of Antidifferentiation

Solve the separable differential equation

given the initial condition

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

Next, we integrate both sides:

The integrals were solved using the following rules:

The two constants of integration were combined to make a single constant.

Now, exponentiate both sides to isolate y, and use the properties of exponents to rearrange the integration constant:

(The exponential of the constant is another constant.)

Finally, we solve for the integration constant using the initial condition:

Our final answer is

### Example Question #12 : Applications Of Antidifferentiation

Solve the separable differential equation:

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

Next, we integrate both sides:

The integrals were solved using the following rules:

The two constants of integration were combined to make a single one.

Now, we solve for y:

Because the problem statement said that y is negative - and y cannot be zero - our final answer is

### Example Question #21 : Applications Of Antidifferentiation

Solve the separable differential equation:

and at

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

Next, we integrate both sides:

The integrals were solved using the following rules:

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y:

Using the properties of exponents, we can rearrange the integration constant:

(The exponential of the constant is itself a constant.)

Using the given condition, we can solve for C:

Our final answer is

### Example Question #22 : Applications Of Antidifferentiation

The rate of a chemical reaction is given by the following differential equation:

,

where  is the concentration of compound  at a given time, .  Which one of the following equations describes  as a function of time?  Let  be the concentration of compound  when .

Explanation:

To begin with, the differential equation needs to be rearranged so that each variable is one side of the equation:

.

Then, integrate each side of the rate law, bearing in mind that  will range from  to , and time will range from  to :

After integrating each side, the equation becomes:

.

The left side has to be evaluated from  to , and the right side is evaluated from  to :

.  This becomes:

.

Finally, rearranging gives:

### Example Question #41 : Integrals

Given that  and , solve for . What is the value of ?

1

1

Explanation:

This is a separable differential equation. The simplest way to approach this is to turn  into , and then by abusing the notation, "multiplying by dx" on both sides.

We then group all the y terms with dy, and all the x terms with dx.

Integrating both sides, we find

Here, the first integral is found by using substitution of variables, setting . In addition, we have chosen to only put a +C on the second integral, as if we put it on both, we would just combine them in any case.

To solve for y, we multiply both sides by two and raise e to both sides to get rid of the natural logarithm.

(Note, C was multiplied by two, but it's still just an arbitrary constant. If you prefer, you may call the new C value .)

Now we drop our absolute value signs, and note that we can take out a factor of  and stick in front of the right hand side.

As  is just another arbitrary constant, we can relabel this as C, or  if you prefer. Solving for y gets us

Next, we plug in our initial condition to solve for C.

Leaving us with a final equation of

Plugging in x = 4, we have a final answer,

### Example Question #24 : Applications Of Antidifferentiation

Solve the separable differential equation:

given the condition at

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

Next, we integrate both sides:

The integrals were solved using the identical rules.

The two constants of integration are now combined to make a single one:

Now, exponentiate both sides of the equation to solve for y, and use the properties of exponents to rearrange C:

Finally, we solve for the integration constant using the given condition:

Our final answer is

### Example Question #1 : Interpretations And Properties Of Definite Integrals

If f(1) = 12, f' is continuous, and the integral from 1 to 4 of f'(x)dx = 16, what is the value of f(4)?

27

12

28

16

4

28

Explanation:

You are provided f(1) and are told to find the value of f(4). By the FTC, the following follows:

(integral from 1 to 4 of f'(x)dx) + f(1) = f(4)

16 + 12 = 28

### Example Question #1 : Interpretations And Properties Of Definite Integrals

Find the limit.

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n+ 6)

0

4

nonexistent

–6

1

4

Explanation:

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n+ 6)

Use L'Hopitals rule to find the limit.

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n+ 6)

lim as n approaches infiniti of ((12n2) – 6)/((3n2) – 4n + 6)

lim as n approaches infiniti of 24n/(6n – 4)

lim as n approaches infiniti of 24/6

The limit approaches 4.

### Example Question #1 : Interpretations And Properties Of Definite Integrals

If a particle's movement is represented by , then when is the velocity equal to zero?

Explanation:

The answer is  seconds.

now set  because that is what the question is asking for.

seconds

### Example Question #1 : Definite Integral Of The Rate Of Change Of A Quantity Over An Interval Interpreted As The Change Of The Quantity Over The Interval

A particle's movement is represented by

At what time is the velocity at it's greatest?

Explanation:

The answer is at 6 seconds.

We can see that this equation will look like a upside down parabola so we know there will be only one maximum.

Now we set  to find the local maximum.

seconds