Evaluate the following integrals. ∫1^2 (2)/(t^3(t+1)) d t |...

Name: Problem 44, Chapter 8: Calculus: Early Transcendentals
Uploaded: 2020-05-27T19:10:40-08:00
Duration: 9 min 49 s
Channel: Tom Greenwood
Description: Evaluate the following integrals. ∫1^2 (2)/(t^3(t+1)) d t

Key Concepts

Rational Functions

Rational functions are expressions that are the ratio of two polynomials. Integrals involving rational functions often require algebraic manipulation, such as partial fractions, to express the integrand in a form that is amenable to standard integration techniques. Understanding the behavior and properties of rational functions is essential for correctly setting up and evaluating such integrals.

Integration Techniques

Integration techniques refer to the various methods used to evaluate integrals, including substitution, integration by parts, and partial fraction decomposition. Choosing an appropriate technique depends on the form of the integrand, aiming to simplify the expression into terms that are easier to integrate.

Definite Integral

A definite integral represents the accumulation of a quantity, such as area under a curve, over a specific interval. It involves computing an antiderivative of the integrand and then applying the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from that at the upper bound.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational functions into a sum of simpler fractions whose antiderivatives are straightforward to compute. This technique is particularly useful when the denominator is a product of polynomials, allowing the integrator to handle each term separately.

Transcript

00:03 All right, we are going to break this down using our partial fraction decomposition, and that t to the third power tells us that we need three fractions with t as the denominator.

00:17 We need each power of t as a fraction, because we never know which of those were involved here with this.

00:27 And that shouldn't be a plus.

00:28 That should be an equals.

00:32 So we would have something over t plus something over.

00:38 T squared plus something over t cubed so those three fractions are all potential candidates for being in our decomposition because of the t to the third and because they're linear we just need something to represent the constant in the numerator so i'm going to use a b and c and then we need a fraction with our t plus one factor as the denominator and again that's going to be a linear just need a variable representing that numerator there.

01:14 So now we can multiply everything by our common denominator to get rid of the fractions.

01:20 And so we can figure out what these numerators are.

01:23 So everything divides out on the left -hand side.

01:28 And as we do the right -hand side, we're going to have a little something divide out each time.

01:32 And we're going to have our numerator times what was left over.

01:36 So one of the t's divides out with the first fraction.

01:39 So we've got t squared times t plus 1 times our a, which gives us a, t to the third, plus a, t squared.

01:52 And two of the t's will divide out for our next fraction, leaving us with b times t plus t plus 1.

02:01 So that'll give us plus b, t squared, plus b t.

02:07 And then all three ts reduce with this next fraction, giving us c times the t plus 1.

02:17 So that's plus ct plus c.

02:21 And the t plus 1 divides out in the last fraction, leaving us with d times the t to the third.

02:28 So there's that term.

02:31 So now i can set up my system of equations to find these numerators, just by setting the coefficients of each term.

02:40 Equal to each other.

02:42 So there's no t cubed on the left -hand side, but i do have some t -cubes on the right, so a plus d must equal zero.

02:55 There are no t -squared on the left -hand side, but i do have some t -squareds on the right, so zero must equal a plus b.

03:09 There are no t to the first power on the left -hand side, but i have some t to the first on the right, so zero must equal b plus c.

03:22 And then finally, i have a constant of two on the left, and there's just one constant term there on the right, that's c.

03:30 So c is two.

03:33 And then we can work our way back up.

03:35 That means b is negative two, and that means a is positive two, and that means d is negative two.

03:46 So that means we can take this integral and we can rewrite it using this decomposition as the integral from 1 to 2 of 2 over t minus 2 over t squared plus 2 over t to the third and then minus 2 over t plus 1 all times our dt.

04:23 And now i can integrate each of these individually, separating the addition and subtraction...

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