AP Calculus AB : Integrals

Example Questions

Example Question #1 : Definite Integral As A Limit Of Riemann Sums

True or False: If  is a negative-valued function for all ,

True

False

True

Explanation:

This is true. Since  is negative-valued, its graph is below the -axis, and the Riemann sums used to evaluate the area between  and the -axis have a negative value for height.

Example Question #2 : Definite Integral As A Limit Of Riemann Sums

is a continuous function on the interval  and is differentiable on the open interval .  If , then which of the following statements MUST be true:

at some point , where .

at some point , where .

over the interval .

over the interval .

over the interval .

at some point , where .

Explanation:

According to Rolle's Theorem, if a function is continuous over a closed interval  and differentiable on the open interval , and if , then there has to be some value  such that , where .  To put this another way, if a function is continuous and differentiable over a certain interval, and if the value of the function is the same at both endpoints of that interval, then at some point in between those endpoints, the function is going to have a slope of zero (i.e. its first derivative will be zero).  This does not apply to the second derivative, nor does it require that the slope of the first derivative be zero over the entire interval.

Example Question #3 : Definite Integral As A Limit Of Riemann Sums

You may use one or both of the following summation formulas:

Express the following definite integral as a limit of Riemman Sums. Then evaluate the integral by evaluating the limit.

Explanation:

First, let's remind ourselves what the formula for Riemann Sums is. Written in Summation form, it is:

when , the number of subintervals, is small, the area calculated by the Riemann Sums is not very accurate. However, as we increase the number of subintervals, the approximation becomes closer and closer to the exact area under the function. "Closer and closer" is a concept from Limits. So if we find the limit of the Riemann sum formula, with n approaching infinity, the result is the exact area. This is the essence of the Definite integral definition. What it effectively tells us to do is stick a limit on the Riemann sums formula to get:

To use this formula, we need to do three things:

(1) we need to find

(2) we need to develop a formula for

(3) we need to plug that  into the function inside the given integral.

To find , we use the same formula from Riemann sums.

, where and are the upper and lower bounds from the definite integral. For our integral, and . Since we are not stating a specific number of subintervals, we leave as it is. Nothing gets plugged into it.

Plugging in and , we get

Now we have . Next we find a formula for , which by the way represents the right endpoint of the "i-th" subinterval. We will revisit that sentence shortly to clarify.

First, we know that the left endpoint of the 1st subinterval is , and that this subinterval has a width of . So to get the right endpoint of the 1st subinterval, we just add that width to

To get the 2nd right endpoint, we just add a 2nd .

to get the "i-th" right endpoint, we just add "i" times.

This is the "i-th" concept. Now if we wanted the 50th rigth endpoint, we would just plug 50 in for "i". Fortunately, we don't need to.

Now we have and we have a formula for . Next we will find by plugging our formula in to the function from inside our integral. In other words our function is .

Plugging in to this function we get

Recall that we determined . Pluggin that in, we get

To simplify, we will combine the 1 and -1 to cancel them,  and then apply the square power to what is left.

Now we have all the pieces of the Limit of Riemann sums formula. Let's plug them in.

To evaluate the limit, we must first find the sum. To do this, we must realize that we can pull any common factors out of the summation, as long as they are constant with respect to the summation index . In other words, we can pull out any variable that is not an . This means we can write the 's on the left of the sum, while still writing them inside the limit,

like this:

By passing the 's to the left of the sum, we have matched our sum to one of the provided sum formulas, specifically:

This means we can evaluate the sum by replacing it using the above formula.

Doing this, we get:

Now we can start evaluating the limit.

Lets start by reducing the single from top and with one from the on bottom.

Now we will multiply our the two factors on top, and also multiply the through the other fraction.

Combining like terms, we get

Since the denominator is a single term, we can split the big fraction into three individual fractions, and reduce each one.

Remember that the limit of a constant is equal to that constant. Also remember when evaluating limits at infinity, a constant divided by a variable equals zero.

Applying these, we get

This gives us the correct answer of

Example Question #1 : Calculus I — Derivatives

Consider the function

Find the minimum of the function on the interval .

Explanation:

To find potential minima of the function, take the first derivative of  using the power rule.

Set the derivative to 0:

We solve for  to obtain  and then plug in 0.5 into the original function to obtain the answer of

We can double check that  is indeed a minimum by using the second derivative test

which means the function is concave up, so that the point we found is a minimum.

Example Question #6 : Techniques Of Antidifferentiation

Find the x coordinate of the minimum of  on the interval

Explanation:

First, find the derivative, which is:

Set that equal to 0 to find your critical points:

Evaluate f at each critical point and endpoint over [-2, 2]:

The least of those numbers is the minimum. Because f(-2)=-32, that is the minimum.

Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

What is the local minimum of  when ?

There is no local minimum in that range.

The y-value is constant throughout that range.

Explanation:

To find the maximum, we need to look at the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat  as  since anything to the zero power is one.

Notice that  since anything times zero is zero.

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Because we want the MINIMUM, we want to see where the derivative changes from negative to positive.

Notice that  has a root when . In fact, it changes from negative to positive at that particular point. This is the local minimum in the interval .

Example Question #7 : Techniques Of Antidifferentiation

Integrate,

Explanation:

Integrate

1) Apply the sum rule for integration,

2) Integrate each individual term and include a constant of integration,

Further Discussion

Since indefinite integration is essentially a reverse process of differentiation, check your result by computing its' derivative.

This is the same function we integrated, which confirms our result. Also, because the derivative of a constant is always zero, we must include "C" in our result since any constant added to any function will produce the same derivative.

Explanation: