Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #71 : How To Find Midpoint Riemann Sums

A Riemann Sum approximation of an integral  follows the form

.

Where n is number of points/subintervals used to approximate the integral.

Knowing this, imagine a modified style of Riemann Sum, such 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:

 ?

Possible Answers:

Correct answer:

Explanation:

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=20 will give a closer approximation than n=10:

Reimannsumcomparison

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.

Consider the function  over the interval . The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

So the derivative is

It might be intuitive to see that the steepness is positive and that it gets progressively greater over the specified interval, although to be precise, we may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

Since this derivative is positive over the specified interval, , and since the slope was shown to be positive as well, the slope is ever-increasing and growing steeper.

The intervals should in turn grow increasingly thinner.

Example Question #72 : How To Find Midpoint Riemann Sums

A Riemann Sum approximation of an integral  follows the form

.

Where n is number of points/subintervals used to approximate the integral.

Knowing this, imagine a modified style of Riemann Sum, such 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:

 ?

Possible Answers:

Correct answer:

Explanation:

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=100 will give a closer approximation than n=50:

Reimannsumcomparison

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.

Consider the function  over the interval . The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

So the derivative is

It can be seen that the slope is negative over the specified interval of . We may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

Since this derivative is positive over the specified interval, and since the slope is negative, It means that the slope is flattening out

The intervals should in turn grow increasingly wider.

Example Question #73 : How To Find Midpoint Riemann Sums

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

Possible Answers:

Correct answer:

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 asked to approximate 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 #74 : How To Find Midpoint Riemann Sums

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

Possible Answers:

Correct answer:

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 asked to approximate 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 #75 : How To Find Midpoint Riemann Sums

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

Possible Answers:

Correct answer:

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 to approximate 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 #76 : How To Find Midpoint Riemann Sums

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

Possible Answers:

Correct answer:

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 approximating 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 #71 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral   takes the form

.

Where  is the number of points/subintervals, and each subinterval is of uniform width .

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:

 ?

Possible Answers:

Correct answer:

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:

Reimannsumcomparison

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.

Considering the function , finding the sign of the slope will help determine if it becomes steeper or shallower. The slope is given by the derivative of the function:

The slope is negative on the interval 1 to 10. Now, to determine whether or not it's growing steeper or leveling off, take the derivative of this slope function. If the sign is the same (negative) on the interval, the slope is becoming steeper. If it is positive, the slope is flattening.

This function is positive on the interval 1 to 10, so the function is flattening out. The intervals should start thin and grow wider.

Example Question #78 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral   takes the form

.

Where  is the number of points/subintervals, and each subinterval is of uniform width .

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:

 ?

Possible Answers:

Correct answer:

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=50 will give a closer approximation than n=25:

Reimannsumcomparison

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.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

We are considering the integral

.

So take the derivative of the function

.

The natural log of any value between 0 and 1 is negative, so the derivative will be positive on the interval 0.01 to 1.

Now, take the derivative once more:

Again, the natural log of any value between 0 and 1 is negative, so the second derivative is negative.

Since the 1st derivative is positive and the 2nd derivative is negative, the slope is getting flatter. 

The intervals should begin narrow, and then get progressively wider.

Example Question #79 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral

 using three midpoints.

Possible Answers:

Correct answer:

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  .

We are asked to approximate 

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

Example Question #80 : How To Find Midpoint Riemann Sums

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

Possible Answers:

Correct answer:

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 approximating is 

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

 

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