AP Calculus AB : Riemann sums (left, right, and midpoint evaluation points)

Example Questions

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Explanation:

Explanation:

Explanation:

Explanation:

Example Question #207 : Integrals

Let .

A relative maximum of the graph of   can be located at:

The graph of  has no relative maximum.

Explanation:

At a relative minimum  of the graph , it will hold that  and

First, find . Using the sum rule,

Differentiate the individual terms using the Constant Multiple and Power Rules:

Set this equal to 0:

Either:

, in which case, ; this equation has no real solutions.

has two real solutions,  and

Now take the second derivative, again using the sum rule:

Differentiate the individual terms using the Constant Multiple and Power Rules:

Substitute  for :

Therefore,  has a relative minimum at .

Now. substitute  for :

Therefore,  has a relative maximum at .

Example Question #22 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Estimate the integral of  from 0 to 3 using left-Riemann sum and 6 rectangles. Use

Explanation:

Because our  is constant, the left Riemann sum will be

Example Question #41 : Numerical Approximations To Definite Integrals

Use Left Riemann sums with 4 subintervals to approximate the area between the x-axis, , , and .

Explanation:

To use left Riemann sums, we need to use the following formula:

.

where  is the number of subintervals, (4 in our problem),

is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that will be 1, 2, 3, and then 4)

is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

, is the width of each subinterval, which we will determine shortly.

and means add all versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing , 4 times; 1 time each for , , , and . This gives us

Think of as the base of each box, and  as the height of the 1st box.

This is basically , 4 times, and then added together.

Now we need to determine what and are.

To find we find the total length between the beginning and ending x values, which are given in the problem as and . We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short,

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from to , we'll just use the smaller one, , for our first .

Now we know that .

To find the next endpoint, , just increase the first x by the length of the subinterval, which is . In other words

And repeat to find

Now that we have all the pieces, we can plug them in.

plug each value into and then simplify.

This is the final answer, which is an approximation of the area under the function.

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