Evaluate the definite integral. ∫ 1 2 ( x − 3 ) ( x 2 − 6 x ) 2 d x \int_1^2\left(x-3\right)\left(x^2-6x\right)^2dx ∫ 1 2 ( x − 3 ) ( x 2 − 6 x ) 2 d x

this problem we're asked to evaluate the definite integral from one to two of x minus three times x squared minus six x squared d x. Now because here this is an integral that we have to use substitution to evaluate and it's also a definite integral, there are two different ways that we could go about solving this problem. Now depending on your textbook or your professor, you may have learned just one method but not the other, or you may just have a preference for one over the other. So here, I'm going to go through both ways that you could have gone about solving this problem, so feel free to skip around accordingly. We're gonna start with this first method, what we're referring to as method one, where we're really going to treat this as an indefinite integral to start with. And then once we plug back in our original function, we're going to evaluate at those original bounds here from one to two. So getting started with our first method here, we wanna choose u and find du. Now looking at this particular problem, it can be a little bit confusing to determine what u should be here because we have two separate functions, both enclosed in parentheses, so you may perceive both of them to be these sort of inside functions. Now most often when this happens, we're going to want to choose the function that has a little bit more going on, meaning it might not just be in parentheses, but it's in parentheses and raised to a power or raised to a higher power than some other function. Now here we have this x minus three by itself. It's not raised to a power, but it's being multiplied by this x squared minus six x, which is raised to the power of two. So that's usually a good clue that this is what we should choose as u. So getting started with our first method here, method one, if I choose that x squared minus six x as u, so if u is equal to x squared minus six x, what does that make du? Well it makes du equal to that derivative, so two x minus six and then times dx. Now you may be wondering here how is this going to work out in a substitution looking at what we have left over in our integral. But this two x minus six, if I go ahead and factor a two out of there, this becomes two times x minus three dx. So that x minus three is clearly in my integral here, so we know that this is going to work out. So having chosen u and found du, we move on to that second step, rewriting our integral only in terms of that u and du. Now because we know that that u is that x squared minus six x, we know du is two times x minus three d x. So that means that we need to go ahead and multiply it by a constant and its reciprocal in order to get this to work because right now all we have is x minus three d x. We need to multiply by two. Now, if I multiply that by two, I also need to multiply by one half so that I'm not really changing the value of my integral. So now that I have that two times X minus three DX, I do have DU in my integral. So rewriting my integral here, this is going to be one half times the integral of u squared du. Now, if you need some help envisioning this here, I can actually rewrite this integral. So if I have that one half on the outside and I do some rearranging on the inside, this is x squared minus six x squared. And then over here I have two times x minus three d x. So I have u, that's that u square that we rewrote here, and then I have du. So this is what my rewritten integral is. Remember in method one, we're ignoring those bounds for now. So we can go ahead and move on to step three here, integrating with respect to u. So what is this gonna give me here? Well, I end up with one third u cubed, but remember I have that one half on the outside. So this is one half times one third u cubed having applied the power rule. I'm gonna go ahead and stick that constant of integration on there just for now. We know that we're not gonna need it later because this is a definite integral. Now if I do some simplification here, going ahead and multiplying those fractions, one half and one third, that gives me one sixth u cubed plus that constant of integration c. Then I wanna go ahead and replace u with my original function of x here. That is x squared minus six x. So that's what we're plugging in here for u. So this is one sixth times x squared minus six x cubed. Now because we've already replaced u with g of x, we can go ahead and bound our integral with those original bounds from one to two. So this is what we now want to apply our bounds to using the fundamental theorem of calculus. I'm gonna go ahead and move it down here just so that we can continue on with enough space. So if I go ahead and apply those bounds, I'm plugging in that upper bound first. So this becomes a one six times two squared minus six times two cubed, and then a minus going ahead and plugging in that one as well. This is going to be one squared minus six times one cubed. Then that's everything that I have there. We just needed to do some algebra. You can plug all of this into your calculator at once or just looking at what's going on here. Two squared is four minus six times two which is 12 that is negative eight. Negative eight cubed is going to be negative 512. And then minus one squared is one, six times one is six, one minus six is negative five, negative five cubed is going to give me negative 125. So negative five twelve minus negative 125, and then all of that is getting multiplied by that constant one six. Let's not forget about our constant there. Now if I go ahead and do some simplification, this is one six times negative 387. So one six times negative three eighty seven As a fraction, if I go ahead and reduce this fraction to its most simple form, this is gonna be negative one twenty nine over two if I go ahead and divide both of those values by three. If you don't wanna report this as a fraction, we can put this in its decimal form and this ends up being equal to negative 64.5 for the answer to this definite integral. Now that was our first method. Let's go ahead and use our second method in order to solve this problem as well. So remember, this is the definite integral. So here we're getting started with method two. Method two is where we're actually going to change our bounds along with the rest of our integral when we're making that substitution. So this is the definite integral from one to two of x minus three times x squared minus six x squared d x. So we're starting off in mostly the same way choosing u finding du so it's gonna be the exact same that we saw previously. So if I go ahead and choose u to be that x squared minus six x that then makes du equal to two x minus six d x Or if I go ahead and factor a two out of there, that's two times x minus three d x. So that makes it a little bit more clear to see what's going on in our integral here. So going ahead and moving on to step two here, we're rewriting our entire integral, including our bounds, in terms of u and du. Now we can see here that this two is missing from our integral currently, so we need to go ahead and multiply by that constant and its reciprocal in order to make this substitution work. So multiplying by that two on the inside, I'm gonna go ahead and stick the one half that I don't need inside my integral on the outside. So one half times two, I'm just multiplying by constant and its reciprocal in order to get this du to appear in my integral. Then we also need to go ahead and find those new bounds. So the way that we do that is we plug them into our choice for u. So in this case, that was x squared minus six x. So here, I need to find g of one and g of two. Those are my two original bounds. I'm finding my new bounds by plugging them into x squared minus six x. So plugging in one, that's one squared minus six x or six times one, plugging in plugging in one for both instances of x here. So one squared is one minus six is negative five so that is negative five for my new lower bound. And then g of two plugging two in that's two squared minus six times two which is really four minus 12 that gives me negative eight. So here I have that lower bound of negative five, upper bound of negative eight. So going ahead and rewriting my integral here, I'm rewriting this with all of my substitutions. So this is equal to one half times the integral from negative five to negative eight of u squared du. So we can see that our balance here go from negative five to negative eight. That doesn't make a ton of sense because we're starting at a higher number and going back to a lower number. But remember using our integration rules, we can go ahead and flip our bounds here if we stick a negative on our integral. So this is equal to negative one half times the integral from negative eight to negative five of u squared du. So continuing on with our steps here, moving on to step three, integrating with respect to u. So this is going to be equal to negative one half times a one third u cubed using the power rule here and then bounded from negative eight to negative five. So going ahead and applying my bounds here, I'm also going to get a bit more space by just shrinking my original work here. So this is then equal to negative one six. We're going to go ahead and combine that. So negative one six, and then we're just plugging in these bounds to that u cube because we went ahead and multiply those together. So this becomes negative five cubed minus negative eight cubed. So this is negative one six, negative five cubed is going to be negative 125, so this is negative 125 minus negative five twelve because that's negative eight cubed so that's really plus five twelve. Now when we do this here negative 125 plus five twelve is going to be positive three eighty seven but we still have this negative on the outside so this is really negative three eighty seven over six just doing our next algebraic step. Now if I go ahead and simplify this fraction this is going to end up giving me negative 129 over two. If I go ahead and divide both my numerator and denominator by three But if we want this in its decimal form this ends up being equal to negative 64.5 which is of course the same exact answer that we got when using method one. No matter which method you should you you use you should get the same answer. Feel Feel free to let us know if you have any questions here, and I'll see you in the next one.

8. Definite Integrals

Substitution for Definite Integrals

Business Calculus

8. Definite Integrals

Substitution for Definite Integrals

Struggling with Business Calculus?

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Multiple Choice

Evaluate the definite integral.
$\int_1^2\left(x-3\right)\left(x^2-6x\right)^2dx$

-64.5

533

-2.33

-1.17

Verified step by step guidance

Rewrite the integral in a more manageable form. The given integral is ∫₁² (x - 3)(x² - 6x)² dx. Start by letting u = x² - 6x, which simplifies the expression. Compute du/dx = 2x - 6, so du = (2x - 6) dx.

Substitute u and du into the integral. Replace (x² - 6x) with u and (2x - 6) dx with du. Adjust the limits of integration accordingly: when x = 1, u = 1² - 6(1) = -5, and when x = 2, u = 2² - 6(2) = -8.

The integral now becomes ∫ from u = -5 to u = -8 of (x - 3) * u² * (1 / (2x - 6)) du. Simplify the expression further by substituting x - 3 in terms of u using the original substitution u = x² - 6x.

Simplify the resulting integral and integrate with respect to u. Use the power rule for integration: ∫ uⁿ du = (uⁿ⁺¹) / (n + 1), where n ≠ -1. Be careful to handle constants and coefficients properly.

After integrating, substitute back the original limits of integration (u = -5 and u = -8) into the antiderivative. Evaluate the definite integral by subtracting the value of the antiderivative at the lower limit from its value at the upper limit.