In this activity, we will review the concept of Improper Integrals.
First, let's review the meaning of the integral.[br][br]In the integral [math]\int^x_af\left(t\right)dt,[/math] what does the [math]a[/math] represent ?
In the integral [math]\int^x_af\left(t\right)dt,[/math] what does the [math]x[/math] represent ?
Explain why the integral [math]\int_a^{\infty}f\left(t\right)dt[/math] is [i]improper.[/i]
What's another example of an improper integral?
Improper Integrals can also be formed if the integrand is discontinuous on the region of integration.[br]For example, [math]\int_0^x\frac{1}{t-2}dt[/math] is improper for [math]x\ge2[/math] because the function [math]f\left(t\right)=\frac{1}{t-2}[/math] is undefined (and therefore discontinuous) at [math]t=2[/math].
The Fundamental Theorem of Calculus does not hold (in general) for Improper Integrals. That is, we cannot simply evaluate [math]\int_a^bf\left(t\right)dt=F\left(b\right)-F\left(a\right),[/math] where [math]F\left(x\right)[/math] is an antiderivative of [math]f\left(x\right)[/math].[br][br]If we can find an antiderivative, we can evaluate indefinite integrals that involve discontinuities by [b]separating into two (or more) integrals and evaluating left-hand or right-hand limits.[br][/b][br]We can evaluate indefinite integrals that involve unbounded regions of integration [b]by finding limits at infinity[/b].
Let's try an example. [br][br][br] Evaluate [math]\int_0^5\frac{1}{t-2}dt.[/math][br][br]We split the integral into the sum of two integrals:[br][br][math]\int_0^2\frac{1}{t-2}dt+\int_2^5\frac{1}{t-2}dt[/math][br][br][br]Now each integral only has a discontinuity on the boundary (t=2).[br]If we can find an antiderivative of [math]\frac{1}{t-2}[/math] , we can apply the Fundamental Theorem of Calculus, taking the left or right hand limit when applying to the endpoint t=2.[br][br]What is an antiderivative of [math]\frac{1}{t-2}[/math]?[br][br][br]
[math]\ln\left|t-2\right|+C[/math]
To evaluate [math]\int_0^2\frac{1}{t-2}dt[/math], you find the limit of the antiderivative as [math]t[/math] approaches 2 from the left, the value of the antiderivative at t=0, and subtract them. What is [math]\int_0^2\frac{1}{t-2}dt[/math]?
To you evaluate [math]\int_2^5\frac{1}{t-2}dt[/math], you find value of the antiderivative at t=5, the limit of the antiderivative as [math]t[/math]approaches 2 from the right, and subtract them. What is [math]\int_2^5\frac{1}{t-2}dt[/math]?
What can you conclude about [math]\int_0^5\frac{1}{t-2}dt[/math] ? Explain.
Most of the improper integrals we will encounter in this class will be a different type.[br]Explain why the integral [math]\int_1^{\infty}\frac{1}{t^2}dt[/math]is improper.
What is an antiderivative of [math]f\left(t\right)=\frac{1}{t^2}[/math] ?
[math]F\left(t\right)=-\frac{1}{t}+C[/math]
To evaluate the integral, you take the limit of the antiderivative as [math]t[/math] approaches [math]\infty[/math], and then subtract the value of the antiderivative at [math]t=1.[/math][br][br]What is the value of [math]\int_1^{\infty}\frac{1}{t^2}dt[/math] ?
Manipulate the slider to vary the value of [i]t [/i](upper limit of integration) in the integrals below. Then answer the questions.
Describe how [math]\int_1^t\frac{1}{\sqrt{x}}dx[/math] changes as [math]t[/math] increases.
Contrast with how [math]\int_1^t\frac{1}{x^2}dx[/math] changes as [math]t[/math] increases.
Suppose the function [math]h\left(x\right)\ge0[/math] and [math]h\left(x\right)\le x^{-2}[/math] for all [math]x\ge1[/math]. What can you conclude about [math]\int_1^{\infty}h\left(x\right)dx[/math]?
Suppose the function [math]j\left(x\right)\ge0[/math] and [math]j\left(x\right)\ge\frac{1}{\sqrt{x}}[/math] for all [math]x\ge1[/math]. What can you conclude about [math]\int_1^{\infty}j\left(x\right)dx[/math]?
Justify your reasoning for the last two questions.