Definite And Improper Integral Test Calculator, Series And Sum Calculator With Steps

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We introduce the procedure of “Slice, Approximate, Integrate” and use it study thearea of a region between two curves using the definite integral.

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We can also use the procedure of “Slice, Approximate, Integrate” to set up integralsto compute volumes.
We use the procedure of “Slice, Approximate, Integrate” to develop the washermethod to compute volumes of solids of revolution.
We use the procedure of “Slice, Approximate, Integrate” to develop the shell methodto compute volumes of solids of revolution.
We can use the procedure of “Slice, Approximate, Integrate” to find the length ofcurves.
We compute surface area of a frustrum then use the method of “Slice, Approximate,Integrate” to find areas of surface areas of revolution.
We apply the procedure of “Slice, Approximate, Integrate” to model physicalsituations.
We learn a new technique, called integration by parts, to help find antiderivatives ofcertain types of products by reexamining the product rule for differentiation.
We can use substitution and trigonometric identities to find antiderivatives of certaintypes of trigonometric functions.
We can graph the terms of a sequence and find functions of a real variable thatcoincide with sequences on their common domains.
Alternating series are series whose terms alternate in sign between positive andnegative. There is a powerful convergence test for alternating series.
For a convergent geometric series or telescoping series, we can find the exact errormade when approximating the infinite series using the sequence of partialsums.
There is a nice result for approximating the remainder for series that converge by theintegral test.
The basic question we wish to answer about a series is whether or not the seriesconverges. If a series has both positive and negative terms, we can refine this questionand ask whether or not the series converges when all terms are replaced by theirabsolute values. This is the distinction between absolute and conditional convergence,which we explore in this section.
Separable differential equations are those in which the dependent and independentvariables can be separated on opposite sides of the equation.
Projections tell us how much of one vector lies in the direction of another and areimportant in physical applications.

There is a nice result for approximating the remainder for series that converge by theintegral test.

When we have a convergent geometric series or a convergent telescoping series, wecan find an explicit formula for the terms in the sequence of remainders since we canfind an explicit formula for the terms in the sequence of partial sums. One of theother important convergence tests we have studied so far is the integraltest.
Integral Test Suppose that is a sequence, and suppose that is an eventuallycontinuous, positive, and decreasing function with for all , where is an integer. Then, either both converge or both diverge.

The key to proving that a series converges by the integral test is to notethat if all of the terms in are eventually positive, then will be eventuallyincreasing.

When , and we have the assumptions for the integral test, the picture to keep in mindis below.

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When the improper integral converges, it can be used to establish an upper bound for. This means that will be bounded and monotonic and thus have a limit, which wecan determine without finding an explicit formula for ! From the picture, it shouldalso be clear that the series and the improper integral do not have the same valuesince the series is represented by the sum of the areas of all of the rectanglesthearea under the curve whereas the improper integral is represented by the sum of the areas of all of therectanglesthe area under the curve .

As such, we made the following important observation.

When the assumptions for the integral test are met, we can use the integral test to determine if a series converges, but we cannot ever use it to find the value to which the series converges!

What, then should we do? Thankfully, the integral test comes with a niceremainder result, which we will state then explore in the context of a familiarexample.

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Remainders and the Integral Test

Remainder Estimates for the Integral Test If is a function that is positive,increasing, and continuous for , and for every , and we know that converges,then we have an upper and lower bound for the error given by the formulabelow.

We also have bounds for the value of the infinite series , which are constructed fromthe error bounds.

The series can thus be approximated by any value ini the above interval.

Note that we actually have two results here.

The first set of inequalities gives bounds for the error.

The inequality tells us a lower bound for the error; this means we know that the error made using is at least the value of . That is, if we use to approximate :

The inequality gives us an upper bound for the error; this means we know that the error we make in our approximation can be no more than the value of . That is, if we use to approximate :

The second result gives us a sharper estimate for the value of the seriesthan the usual estimate using just .

Since we have a positivity assumption on the terms of the series, note that this meansthat for every , will be an overestimateunderestimate. Since we have a result for the minimum error made, we can add this to the value of to obtain the smallest possible value of the series.

Before commenting further, let’s explore how this test works in the context of anexample we have seen in previous sections.

We have seen that converges, but we have not discussed a way to find its value (andwe will not learn how to do so in this course).

What if we want to approximate this sum within an error of . How manyterms should we sum? Is it enough to sum the first ten terms? The first hundred? We want to find a number where we can be sure that While we coulddetermine the smallest such value for when we had an explicit formula for , weare not so fortunate here. However, since we want to use a computer toobtain the actual approximation, we do not need the smallest number ; weonly need one for which we can be sure that will approximate to within.

As usual, the error in approximating by will be .

Consider the following graph.

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Note that the area of the first rectangle is (which is precisely ) , the area ofthe second rectangle is , and so on. Thus, the sum of the areas of all ofthese rectangles is exactly . Comparing this to the area under the curve for yields

In other words, the remainder of the series must be less than the givenintegral!

We can perform a similar construction to find a lower bound for the error.

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and conclude

Putting these together yields

which is precisely the first inequality guaranteed by the theorem in the context ofthis example.

Now, suppose that we want to approximate the value of the series to within. One way we can do this is to approximate by and find by requiringthat the maximum error made is no more than . Using the fact that , wefind

It seems that we need to be at least . Using a computer to sum the first terms wefind to five decimal places that

We have previously mentioned that the value of this series actually is known to be ,so the above sum certainly approximates the series to within , but there’s a secondpart of the theorem that provides a better approximation.

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Note that all of the terms in our series are positivenegative, so the the approximation we found above must be an overestimateunderestimate. Since we know that the minimum error in this approximation made is , and we canreadily compute that

we know that the smallest possible value for the series is .

Similarly, the largest error made is

so the largest possible value for the series is . Hence, we conclude that

We may approximate the value of the series by any point in this interval.

Wait a minute! We wanted to approximate the series in the last example towithin of its actual value, but the actual bounds for the error led us tofind

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We thus actually know the value to within (and can verify this since we are given theactual value of the series). What’s going on here?

Let’s return to the theorem and make an important observation explicit.

The inequality gives us the minimum possible error made. The inequality gives us the maximum possible error made.

Since we have a positivity assumption on the terms, note that this means that forevery , will be an overestimateunderestimate. This leads to an important observation; if we want to approximate a convergentseries to within of its actual value, we can require that the difference between theupper estimate for the series and the lower estimate be no more than . Note that forevery ,

If the difference between the left and right sides of the inequality is less than , thenthe infinite series must be within of its minimum possible value (the lefthand side)and its maximum possible value (the righthand side).

Now, note that

In going from the penultimate to the last step, we are justified in our calculationbecause the integrals both converge. This in practice will greatly reduce the numberof terms needed to approximate the value of a series.

Estimating with the Integral Test To approximate the value of a series thatmeets the criteria for the integral test remainder estimates, use the followingsteps. Choose (or be given) a desired precision , meaning, determine how closely you want to approximate the infinite series. Find the value for from setting . Call this value . Approximate by noting that

Choose any value in the above interval to approximate the value of the series.

It’s very easy to get lost in the sea of symbols above, so we strongly encourage thereader to think of these results conceptually.

To help gain some intuition and see how this process works, let’s see how thissharpens our estimate for in the last example.

In the last example, we found that we needed to use to approximate towithin of its true value; that is, we needed to add up the first terms of theseries.

We can now compute .

In many cases, technology will be useful in finding the value of , but we can do thisone by hand.

We can use the quadratic formula to find (we ignore the negative root since must bepositive). Using a calculator, we find , so we should set .

We can finish the estimate by noting that to four decimal places, we have thefollowing.

. . .

Thus, we have the following interval of values in which the exact value of the seriesmust lie.

We can approximate the value of the series by choosing any value in this interval. Wewill again choose the midpoint, and declare .

Note that we only needed terms in the infinite series by using this sharpertechnique, whereas we needed terms if we only used the upper bound forthe error to tell us how many terms we need to add to obtain the desiredaccuracy.

We hope that the reader actually performed the computations in the previousexample. While it seems that the second result is “nicer” in the sense that adding terms versus terms is desirable, a computer can perform both of these calculationsalmost instantaneously. As such, we should we care to use the sharper estimate if wehave technology at our disposal?

One type of series that plays a major role in our study is that of a convergent -series, for . Approximating values for these to even a moderate degree of accuracy can becomputationally taxing for a computer if we only use the upper bound for the error.For example, suppose we want to compute to within of its exact value. The curiousreader can verify the following.

If the upper bound for the error is used to establish a value for so is within of the exact value of the series, then . It takes Maple (a powerful CAS) a little over minutes to compute this. If the method in the previous example is used to establish a value for , then . It takes Maple seconds to compute this.

If the same attempt is made for the series ,

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If the upper bound for the error is used to establish a value for so is within of the exact value of the series, then . Maple crashes trying to compute this. If the method in the previous example is used to establish a value for , then . It takes Maple seconds to compute this.

The point? Technology has limitations! A little bit of mathematical theory can oftengo a long way, even when computers are used to perform the calculations.

We can check that the seriesabove meets the hypothesis of the integral test, so we may applyit to investigate whether the series converges. We must compute

To find the necessary antiderivative, we use integration by parts.

Thus, we find

Now, we can evaluate the improper integral needed for the integral test.

By growth rates, we have that , so we find that convergesdiverges to .

II. Using the integral test (and a calculator or computer!), use and the resultingerror bounds to approximate the value of .

Using a calculator or a computer, we find the following values to four decimalplaces. . Upper error bound: . Lower error bound: .

and so we can construct the interval of values in which the value of the series mustlie.

We can now approximate the series by choosing a value in this interval. We will againchoose the midpoint, and obtain that .

The curious reader may note that had only the upper bound for error been used, wewould find that , which is two orders of magnitude worse than the aboveresult!
We need to find so . Since we know that , we find the following.

This equation is impossible to solve by hand, but we can use technology to establishthat we have equality when . As usual, is a little too small, so we choose

To find the approximate value, we first find and the error bounds.

. Upper error bound: . Lower error bound: .

We can now construct the interval of values in which the value of the series mustlie.

We can now approximate the series by choosing a value in this interval. We will againchoose the midpoint, and obtain that .

Had we wanted to approximate the series by and find using the upper bound forerror, we would find that . While a computer can perform both calculations quicklyhere, it is once again useful to note how much more efficient the algorithm we usedis!

Summary

When we can establish that a series converges, but we do not have an explicitformula for , we cannot find the exact value of the series. However, we cansometimes approximate it by considering the sequence of remainders. Thereare two important types of questions we have asked about remainders thusfar.

How bad is the error made when we approximate a convergent infinite series by its first several terms? How many terms should we specify if we want to know the value of a convergent series to obtain a desired precision?

In some cases, we use a formula or bounds for the remainder to answer both of these,but when the series can be approximated using the integral test, we have a muchmore efficient way to estimate. We attack the previous questions in the followingmanner.

To approximate using and the error bounds, do the following. Compute . Compute the minimum error from . Compute the maximum error from .

The approximate value for the series is then found from putting these together.

To specify the value of a convergent series to within a desired precision of , do the following. Find a value for from setting . Compute . Approximate by noting that

Choose a value in this interval as the approximate value of the series. Frequently, the midpoint provides a good choice.

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