Express the integrand as a sum of partial fractions and evaluate the integrals. ∫-1^0 (x^3 d x)/

Express the integrand as a sum of partial fractions and...

Key Concepts

Evaluation of Definite Integrals

This concept involves computing the antiderivative of a function and then using the boundaries defined by the integral limits to determine the exact area under the curve. Attention to the limits after integration is essential, as it affects how the final answer is derived from the antiderivative.

Polynomial Long Division

When the degree of the numerator is equal to or greater than the degree of the denominator, polynomial long division is used to simplify the rational function before applying partial fractions. It transforms the integrand into a polynomial plus a proper fraction, which can then be handled using standard integration techniques.

Partial Fractions Decomposition

This concept involves breaking down a complex rational expression into a sum of simpler fractions that are easier to integrate. In the context of integration, partial fractions allow the integrand to be expressed as a combination of basic terms whose antiderivatives are known, thus simplifying the evaluation of the integral.

Factorization of Quadratic Expressions

Recognizing and factoring quadratic expressions, such as identifying a perfect square, is crucial in simplifying the integrand. Factorization helps in rewriting the denominator in a form that is more amenable to partial fraction decomposition or other integration methods.

Transcript

00:01 In this problem, we have the integral from negative 1 to 0 of x cubed over x squared minus 2x plus 1 dx.

00:13 And we want to express this as a sum of partial fractions and then evaluate it.

00:21 So first off, notice that the numerator has a greater degree than the denominator.

00:28 So first we needed a long divide this out.

00:34 So we're dividing x cubed plus 0x squared plus x, so 0x plus 0.

00:43 So i just put these in for placeholders.

00:46 And we're dividing x squared minus 2x plus 1 into that.

00:51 Okay, so first we need to apply by an x, so that'll get us x cubed minus 2x squared plus x.

00:59 So when we subtract that, so it'll be zero, this will get 2x squared minus x.

01:05 Okay, so now we need a positive 2.

01:08 So that'll be 2x squared minus 4x plus 2.

01:13 So when we subtract that out, so that'll be 0, that'll be 3x minus 2.

01:19 Okay, so we get this plus 3x minus 2 over x squared minus 2x plus 1.

01:29 Okay, so let's rewrite the interval so we know kind of what we just did.

01:32 So we have negative 1 to 0 of, and so we're going to have x plus 2 plus 3x minus 2 over x squared minus 2x plus 1 dx.

01:50 All right, so far so good.

01:52 So now looking at this, so we have more fractions.

01:55 So we want to kind of break this up.

01:57 So what is our denominator? so our denominator, we can reverse foil that to be, so this is negative 1 squared, and this is 2 times negative 1.

02:08 So we'll have x minus 1, 1 squared.

02:12 Okay, so, and then if we look at the numerator, we have something very close to x minus 1.

02:18 So we could do a minus 1 plus 1 here.

02:23 And so that'll make this negative 1 to 0, x plus 2.

02:28 We have three times x minus one because you got the negative two and the negative one that makes the negative three that is over x minus one quantity squared and then we have this plus one over the denominator so we'll have plus one over x minus one quantity squared dx okay so uh let's go ahead and we need a plus here gosh let's tidy this up, and then we're going to start doing some substitutions and whatnot.

03:02 Okay, so we have negative 1 to 0, x plus 2, plus 3 times 1 over x minus 1, plus 1 over x minus 1 quantity squared dx.

03:18 Okay, so now let's go ahead and break off this into two integral.

03:22 So we're going to have the integral of things without fractions.

03:26 And the integral of things with fractions with a denominator of x minus one in it.

03:30 So you do negative 1 to 0, x plus 2 dx, and negative 1 to 0 of 3, usually 3 on top, x 3 over x minus 1 plus 1 over x minus 1 quantity squared d x.

03:49 Okay, so we're going to do a u substitution.

03:51 U is going to be x minus 1, du is going to equal to d x.

03:56 Okay, so if x is negative 1, so that means, so if x is from negative 1 to 0, then u is going to be negative 2 to negative 1.

04:14 Looks good, looks good...

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