In this lesson we are going to take a look at Integral Test. The integral test is a way of determining when a series converges or diverges. The way it often works is we will be given a series in certain form. What we do is convert that series into a function of x. There are three things that must be checked before applying the Integral Test: whether the function is continuous, always positive and decreasing. If we checked all three of those conditions, the integral test tells us that we can look at the integral of f(x) from 1 to infinity, and determine whether the series converges or diverges.
Suppose f (x) is a function and you want to know if
Check three conditions:
Is f (x) continuous?
Is f (x) always positive?
Is f (x) decreasing? (i.e.,
is f (x) negative?)
(x)dx converges, then
(x)dx diverges, then
Definition: A p-series
is a series of the form
Using the Integral Test, we can show that the series converges if p
> 1 and diverges to infinity if p ≤ 1.
Hints and tips:
important to remember that the Integral Test only applies to series
with positive terms. However, after you learn about absolute
convergence later, you may be able to use it for series with some
negative terms by taking their absolute value and seeing if they are
Integral Test is one of the few two-way tests − you can use it to
conclude that a series converges or that it diverges.
make the common mistake of confusing p-series with geometric
series. p-series has the n in the base and a constant
exponent. Geometric series has a constant base and n in the
exponent. When a geoemetric series converges, we have formulas to
tell us what it converges to. When a p-series converges, we
usually dont know what it converges to.
worth memorizing the rule for p-series, since these are often
used later in conjunction with the Comparison Test.
that the Integral Test never tells you exactly what a series
Can the Integral Test be applied on the ∑n = 1∞ [1/(n − 1)] ?
If f(n) = [1/(n − 1)], then f(x) is not continuous on the intertval [1,∞] on x = 1
Thus the Integral Test cannot be applied
Can the Integral Test be applied on the ∑n = 1∞ [(n3)/(n + 1)] ?
If f(n) = [(n3)/(n + 1)], then f(1) < f(2) and thus f(x) is increasing.
Thus the Integral Test cannot be applied.
Solve ∫ [(x3)/([(x4)/4] + 1)]dx
Use substitution with u = [(x4)/4] + 1
u = [(x4)/4] + 1
du = x3dx
Integrate with substitution
∫ [(x3)/([(x4)/4] + 1)]dx = ∫ [du/u] = ln|[(x4)/4] + 1| + C
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.