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Numerade Educator



Problem 17 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_1^2 \frac{4y^2 - 7y - 12}{y (y + 2)(y - 3)}\ dy $


$\frac{27}{5} \ln 2-\frac{9}{6} \ln 3$


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Video Transcript

Let's evaluate the definite integral From 1 to 2, we have a polynomial divided by a polynomial, and the denominator is already factored. So using what the textbook were called, case one here in the section they refer to this as case one. We have three distinct and linear factor. So the partial fraction decomposition should look something like over y be over y plus two see over y minus three. And then we could go ahead and multiply both sides by this denominator over here. So after multiplying four y squared minus seven Y minus 12 on the left, on the right, we have a no y plus two y minus three. And then for B, we have y and y minus three. And finally for C. Why? But why in a Y plus two. So let's just go ahead and rewrite this left hand or the right hand side. Just multiply things out. Y squared minus Y minus six. This is for a and then for me. Why squared minus three y and for C y squared plus two y. Now, one more thing to do here on this right hand side is to just rewrite it by factoring out first, we'll factor Ry Square. We'll have a plus B plus C one here, one here, one there. Yeah, Then we'll have. For after we found her to lie, we'll have a minus a a minus three B plus to see And then finally, the constant term. On the right hand side, it's just minus six. A. So on the left hand side, the term the constant in front of the White Square is a four on the right hand side, A plus B plus C that gives us an equation. Similarly, for the white term, we have a negative seven on the left, and this is what we have on the right. It gives us our second equation minus a minus three B plus to see equals minus seven. And then on the left hand side, the constant term was minus 12. Whereas on the left hand side it was minus six A. So we can go ahead. And those are three equations. We have a three by three system and A B and C. So, taking this last equation here we see that A s two. So there's already one of our values. And now let's just go ahead and plug this a value into the previous two equations. So we'll replace that A with the two and this over here that will become minus two. And then we can rewrite these equations. So this first one here that just becomes B plus C equals two. And for the other equation, add that negative two to the other side. We have minus three B plus two C equals negative. Seven plus two is minus five. So let's go on to the next page, running out of room here. But we will have this two by two system and B and C to solve, and then we'll have our coefficients. So we had two equations from the previous page. We had B Plus C is, too, and a minus three B plus to see is minus five. So let's just go ahead and solve this for B or C. So taking this first equation C equals two minus B and then plug this into the second equation. We'll have negative three b plus two, and then R C is two minus B, and the right hand side is negative. Five. So let's just go ahead and solve this for B, so we'll have minus three B and then minus two more B. That's minus five B and then we have a plus four. So maybe add the five to the left and then push the five beats on the other side. And then we C B is just 9/5 and then using this equation up here, let's solve proceed. So plugging B equals 9/5. In we have C equals two, which is 10/5 minus B, which is 9/5. That leaves us with 1/5. So now we've found our coefficient A, B and C. On the previous page, we found that a was to sell me. Just rewrite that here from page one. And then Now let's go on to the next page. Mhm. So let's rewrite the original integral 1 to 2. Yeah, why? Why? Plus two y minus three. This is what we intended to evaluate. We just did the partial fraction decomposition. We found a B and C. So now let's go ahead and plug those in. So a over Y becomes too over y be over Y plus two becomes 9/5 over y plus two. And then we found that C was 1/5 over Y minus three d Y Yeah, and this is a much easier integral for us. If this plus two in minus three or throwing you off, you go ahead. And do you use up here? U equals Y plus two or for the second one, U equals my mind. History. Yeah, let's go ahead and evaluate those integral the first one to natural log Absolute value y The second one That's a five in the denominator. 9/5 and that natural log y plus two plus 1/5. Natural log. Absolute value. Why my mystery and then our endpoints 1 to 2. Let's go ahead and plug in those endpoints. So plug in the two first we have natural log of four plus 1/5, and then we have two minus three, which is minus one. But then absolute value gives us a one so natural log of one. Now we plug in one for why we have 1/5 natural log of negative, too. But then we take absolute value. That's a two there. And now we cancel and simplify as much as we can. We know that natural log of one is zero. So these go away zero and zero. That's the whole the whole term there. And now we combine as much as we can, so we have to natural log two minus 1/5. Natural log, too. We can write. That is nine number five l N two. And then we have 9/5. Ln four minus 9/5 l N three. Mm. And here we can. That can be our final answer. It's a number, but we can go ahead and combine. So here, let's write this for is two squared. And then using the properties of the algorithm, we can pull out this two in front of the nine and then multiply. So we have 9/5 Ln two and then two times nine. So we have 18 there, over five, minus 9/5. That should have been Ellen. Mm. Ln of two minus 9/5 l n three. And then just combine those first two fractions and we have 27/5 natural log of two minus nine, Number five Ln of three. And there's a final answer