# Calculus 1 : Functions

## Example Questions

### Example Question #61 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral  using five midpoints.

Explanation:

A Reimann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

We are given the function

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

The Reimann sum approximation is thus:

Or, in long form:

Note that

### Example Question #1091 : Calculus

Using the method of midpoint Reimann sums, approximate the integral  using three midpoints

Explanation:

A Reimann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1092 : Calculus

Using the method of midpoint Reimann sums, approximate the integral  using three midpoints.

Explanation:

A Reimann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

Our integral is

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1093 : Calculus

Using the method of midpoint Reimann sums, approximate the integral  using three midpoints.

Explanation:

A Reimann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

The function given is

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1094 : Calculus

Using the method of midpoint Reimann sums, approximate the integral  using three midpoints.

Explanation:

A Reimann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

Our function is

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1095 : Calculus

Using the method of midpoint Reimann sums, approximate the integral  using three midpoints.

Explanation:

A Reimann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

Our integral is

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1096 : Calculus

Using the method of midpoint Riemann sums, approximate the integral  using three midpoints.

Explanation:

A Riemann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

The integral we are given is

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1097 : Calculus

The method of Riemann sums is a way of approximating integrals, often for functions which do not have a defined integral.

Picture an integral

A number of rectangular areas are defined, this number being defined by the amount of points or subintervals selected. Generally, these rectangles have uniform widths (that of the subinterval), given by the formula  where n is the number of points/subintervals.

Each rectangle differs in height, however, with individual heights defined by the function value at a given point, .

The area of a particular rectangle is then

Adding each of these rectangles together then gives the integral approximation.

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as  ,

the integral approximation becomes

Which of the following parameters would give the closest integral approximation of the function:

?

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

Since the function  becomes progressively steeper, a better approximation for a fixed number of intervals would have the intervals grow thinner as  increases.

### Example Question #1098 : Calculus

Using the method of midpoint Riemann sums, approximate the integral  using three midpoints.

Explanation:

A Riemann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

The integral we're given is

So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are

### Example Question #1099 : Calculus

Using the method of midpoint Riemann sums, approximate the area of the region between the functions  and  over the interval  using three midpoints.

Explanation:

To find the area between two functions, the method traditionally used is to find the difference of the integral values between the larger and smaller function over the specified interval. However, neither of these functions can be integrated, so a method of approximation is useful.

A Riemann sum integral approximation over an interval  with  subintervals follows the form:

It is essentially a sum of  rectangles each with a base of length  and variable heights , which depend on the function value at a given point  .

We're told that we're integrating from 1 to 2.5, so the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are .

As  is greater than  over the given interval, our approximation looks as follows: