Find the area under the curve of y = 10 ( x + 2 ) ( x + 3 ) y=\frac{10}{(x+2)(x+3)} between x = 0 x=0 and x = 4 x=4 .

So here, we are asked to find the area under the curve of y equals 10 over x plus two times x plus three between x equals zero and x equals four. So how exactly could we solve a problem like this? Well, the thing we need to recall is how exactly we find the area under the curve of some function. Now let's say we're dealing with some graph, and we are given some function on this graph. Now I'm not saying this function is the same as what we were dealing with up here, but I'm just gonna create some random curve right here. If we wanted to find some area of this curve from a to b, and we're trying to find the area underneath the curve of that function given the the values of the x axis that we're between, then what we want to do is take our function, and we want to integrate it from a to b. So integrating our function would allow us to find this exact area that we're looking for underneath the curve. So we can take that same process of finding the area underneath this function that we're dealing with right here, or the curve, y equals 10 over x plus two times x plus three. So what I can do is I can integrate this function. So integrate 10 over x plus two times x plus three. And we're integrating that with respect to x. Now what I wanna do is take this and bound it from zero to four. So this is what we have so far. Now at this point, what I need to be thinking about is strategies for integrating this function. Well, I could try a variable substitution down here, but that's not going to get rid of the x that we have here. We tried variable substitution there, but we still have an x right about there, and this 10 on top is not going to be something we want to set to the variable anyways, u. So I need to think about a different strategy. Well, variable substitution is not going to work. I think it will be better since we have a rational function if I did some form of partial fraction decomposition. Now looking at this, we have just a constant on top and then these x's on the bottom, so we're not going to need to do a polynomial division. I already know that there's gonna be a higher degree x on bottom than on top. But what I can do is recognize that these are two distinct linear factors. So if I go ahead and focus on just this fraction portion for the moment here, I have 10 over x plus two times x plus three. Now I could split this into some constant a over the distinct linear factor x plus two. It's going to be plus some other constant b over the distinct linear factor x plus three. Now at this point, what I need to do is multiply both sides of this equation by the common denominator right here. So we have x plus two times x plus three, which we'll go ahead and distribute into the brackets that we have. Now first off, I'm going to multiply this 10 over x plus two times x plus three by the common denominator right about there. Now multiplying this in here, notice that the x plus two times x plus three will cancel entirely, So all that's going to leave me with on the left side is 10, and we'll have 10 is equal to this portion where I'll multiply this a by the common denominator, this a term over x plus two, and notice that's going to get the x plus two to cancel entirely. So all that's going to leave me with is a times x plus three. Now next, I'll move to this term with b. What I can do is multiply this out by the common denominator, and notice the x plus three will cancel entirely. So that's going to leave me with b times x plus two. So this is what we've broken down to. Now at this point, now that I have this equation, I'm trying to find these constants, a and b. How can I go about doing that? Well, what I'm going to try here is making strategic substitutions for x that is going to allow me to cancel or simplify some of my factors. Now first, I'm going to let x equal three. Or, actually, I'm gonna let x equal negative three. Now there's a reason I chose negative three. Notice if we had a negative three right there, we would get negative three plus three, which I know is zero. That's gonna set this whole piece to zero. So if I replace every x that I see with negative three, that's going to give me 10. 10 is equal to a times negative three plus three, plus b times negative three plus two. And what that will allow me to do is completely get rid of this term since negative three plus three is zero. So all we're going to have is 10 is equal to b times negative three plus two, which is the same because it's just negative one b. So we can say that b is equal to negative 10. So this b up here is negative 10. Now next, we need to find a. So we've gone ahead and found this constant b here, but what can I do to find a? Well, to find a, I'm gonna make another strategic substitution for x. This time, I'm gonna let x equal negative two. Because I know if the x that I had right here was negative two, negative two plus two is zero. Getting zeros is really nice because I can cancel entire factors, so let's go ahead and do that. So letting x equal negative two will have that 10 is equal to a times negative two plus three, replacing the x right there, minus 10 times negative two plus two, replacing the x right there. Now notice that negative two plus two is zero, which will eliminate this whole factor right about here. So all we're going to have is that 10 is equal to a times negative two plus three, which means 10 is equal to one a, so we can say that a is equal to 10. So that means that our a over here is 10, and now we've gone ahead and broken up these fractions. So what we ultimately get right here is that the integral from zero to four of this rational function right here is equal to the integral from zero to four of what we just broke this fraction into, to the integral of 10 over x plus two minus 10 over x plus three. And now we can integrate each one of these fractions separately. So doing that, I'm first going to deal with this integral right here. Now integrating this, that 10 is a constant, so I can just go ahead and pull that 10 on the outside. And the integral of one over x plus two is the natural log of x plus two. Don't forget these absolute value bars. The next we'll have minus the integral of 10 over x plus three. Well, that 10 is just a constant. It'll come on outside the integral, and integrating one over x plus three will give us the natural log of x plus three. So now we have this whole thing which is going to be bounded from zero to four. So let's go ahead and apply these bounds. Well, if I'm applying the bounds here, what I first need to do is plug in this four. So plugging in four here, we're going to have 10. That's going to be times the natural log of four plus two, which is six. We'll have minus 10 times the natural log of in this case, we'll have four plus three. And keep in mind, I'm just plugging in four here. So I have four plus three, which is seven. This is what we get right here, and then what I'm going to do is subtract off plugging in my lower bound of zero for everything. So plugging in zero, we're going to have minus, and that's going to be the natural or 10, I should say, right there, times the natural log of zero plus two, which is natural log of two minus. And then we have 10 times the natural log of zero plus three, which is the natural log of three. So that's what happens when we plug in our upper and our lower bound, and then we subtract those results. Now if you go ahead and plug this into a calculator exactly as you see it, you should get this as an approximate decimal result. So 2.51 is the approximate answer, and that is how you can find the area underneath the curve of this function. So hope you found this video helpful. This is the solution to the problem and how you go about solving it. Let me know if you have any questions, and I'll see you in the next one.

12. Techniques of Integration

Partial Fractions

Calculus

12. Techniques of Integration

Partial Fractions

Struggling with Calculus?

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Multiple Choice

Find the area under the curve of $y = \frac{10}{(x + 2) (x + 3)}$ between $x equals 0$ and $x equals 4$ .

$2.51$

$19.46$

$negative 1.54$

$37.38$

Verified step by step guidance

Step 1: Recognize that the problem involves finding the area under the curve, which is equivalent to computing the definite integral of the given function y = $ \frac{10}{(x+2)(x+3)} $ over the interval [0, 4].

Step 2: Set up the definite integral: $ \int_{0}^{4} \frac{10}{(x+2)(x+3)} \, dx $. This represents the area under the curve from x = 0 to x = 4.

Step 3: Simplify the integrand if possible. In this case, use partial fraction decomposition to rewrite $ \frac{10}{(x+2)(x+3)} $ as $ \frac{A}{x+2} + \frac{B}{x+3} $, where A and B are constants to be determined.

Step 4: Solve for A and B by equating $ \frac{10}{(x+2)(x+3)} $ to $ \frac{A}{x+2} + \frac{B}{x+3} $ and clearing the denominators. This will give you a system of equations to solve for A and B.

Step 5: Substitute the partial fraction decomposition back into the integral, and compute the definite integral term by term. Use the antiderivative of $ \frac{1}{x+c} $, which is $ \ln|x+c| $, and evaluate the result at the bounds x = 0 and x = 4 to find the area.