Evaluate the definite integral. ∫ 1 2 x ln x d x \int_1^2x\ln xdx

we're asked to evaluate the definite integral. Now notice here we have this integral that's bounded from one to two, and we have the integral of the x times the natural log of x with respect to x. And notice that we have the product of two functions. And what we have here is a situation where we actually cannot do a variable substitution. Now I notice this because I can see that we both have x raised to only one power showing up here. But in addition to that, we also have this natural log, and that's going to just cause a lot of problems. So if we were to set this x equal to u, then its derivative would just be one dx, and that's not going to allow us to cancel anything with this x right here. And same thing if we set this x equal to u, we're just gonna get one dx, which is not gonna do anything to this x. So we're going to need to use integration by parts here. You can go ahead and try variable substitution, but you would see that it actually wouldn't work with this problem. So we're going to need to use the integration by parts strategy that we know about. So for integration by parts we first need to figure out what our u is going to be. U needs to be something that we can easily take the derivative of that's gonna make the function more simple when we do, and then I also need to find dv which is something that can be easily integrated. Now I'll start by finding my u here, and the u here that I'm going to choose is the natural log of x, so that's going to be my u, because I know that the derivative of the natural log of x is just going to be one over x with respect to x. One over x does seem a lot more simple than the natural log, so that's gonna be my u. Now it may have been tempting to say that this x would be u instead, since we know that we can take the derivative of x and that will just give us one with respect to x. Now while that could have been the case, notice that would have required us to eventually integrate this natural log. And integrating the natural log is not an easy thing to do. Now in this situation here, also, we know that x here can be easily integrated. So instead, I'm gonna make my d v equal to x. Now I'll just also mention here, typically, you will want to set your natural log equal to u if you have a logarithm in your problem. So in this case, we have our u and our d u, and we also have our d v set to x here. And now what I'm going to do is integrate d v with respect to x. So integrating dv will give me v, and the integral of x using the power rule is going to be x squared over two. So this right here is going to be the variables we need, and now I can use my formula here for integration by parts. So let's plug everything in. Now first off, we're going to have u times v, and that's going to be the natural log of x times x squared over two. So I'm gonna write this like this, we're gonna have x squared natural log of x over two, just combining these all into one. So that's what happens when we multiply there and then that's going to be minus the integral of v du, so that's going to be x squared over two times du which we have as one over x dx, and that's going to be our setup. Now keep something in mind here. This is a definite integral that we're dealing with. This goes from one to two. So I need to take my balance here and put it on this portion from one to two, I also need to put my bounds on this portion from one to two on the integral. So now we have everything set up. Now what I'm gonna do first here before I apply these bounds, and I'm gonna see if I can evaluate this integral first. But what I can see is that one of these x's is gonna cancel with that one. I also have a onetwo that I can pull out to the front of the integral. So rewriting this law of x squared natural log of x over two bounded from one to two, then we're going to have minus, that's gonna be onetwo, the integral from one to two, and then that's just going to leave us with x inside the integral. Now I know how to integrate just x, that's gonna be x squared over two. So to rewrite out this whole thing, it's gonna be x squared natural log of x over two, then we're gonna have minus, that's gonna be one half times the integral of x, which the integral of x using the power rule will just be again x squared over two. And then this whole thing is going to be bounded from one to two. So it's gonna go from one up to two. So that's what what we get when we do this. Now rather than that's why I didn't apply the bounds here immediately is because I know that once I integrated this, I could just apply the bounds to everything at the end, which which I think is gonna be a little more simple. Now what I'm gonna do is combine this one times x squared over two times two into x squared over four, because I can just multiply those denominators. That's gonna give me x squared over four, and now I can go ahead and apply my balance. Now first off, I need to plug in my high bound of two. So doing that is just taking every place I see x and replacing it with two. So we're gonna have this x or excuse me, two squared times the natural log of two all over two, and we're gonna have minus and it's gonna be two squared over four, plugging in our two right there, then we're going to subtract off one plugged in for x. So doing this we're going to have one squared times the natural log of one over two, and we're gonna have minus, and that's gonna be one squared over four. So now we have everything plugged in. Now at this point, it just turns into a process of simplifying all of this. So what I can do is I can cancel one of the twos with the two right there, and then I can also recognize that two squared over four is the same thing as four over four, which we'll just cancel. So all this is gonna leave me with is that we're gonna have, in this case, gonna be two times the natural log of two right up here, and we're going to have minus one because the two squared and the four cancel just give you one. Then we're going to have minus this whole quantity over here, which you're gonna have one squared times natural log of one over two, which is the same thing as just natural log of one over two, and then that's gonna be minus one over four. Now from here, I'll go ahead and distribute the negative sign into these brackets. So that's gonna give me two natural log of two minus one, and then we're going to have minus natural log of one over two, and it's going to be that minus will cancel with that one, giving us plus one fourth. So that's what we get. Now from here, well what I can do is combine this constant with that one. So we have a negative one which is the same thing as negative four over four plus one fourth which would give me negative three fourths. So we're gonna have two natural log of two That's gonna be minus natural log of one over two minus three fourths. And then this right here is about as simple as I can get things. So this right here is gonna be the solution to the problem. And if you wanted to go ahead and type that into a calculator to see what it comes across as a decimal, you should get that this is about 0.64. So this is the result of the problem, and that is how you can find the solution. So this is the process for doing integration by parts where we have bounds that we have to apply to our function once we integrate it. So I hope you found this video helpful, and let me know if you have any questions.

12. Techniques of Integration

Integration by Parts

Calculus

12. Techniques of Integration

Integration by Parts

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Multiple Choice

Evaluate the definite integral.
$the integral from 1 to 2 of x l n x d x$

$negative 0.11$

$0.64$

$0.89$

$1.89$

Verified step by step guidance

Recognize that the integral to evaluate is $ \int_1^2 x \ln(x) \, dx $. This is a definite integral, and the integrand involves a product of $ x $ and $ \ln(x) $.

Use integration by parts to solve the integral. Recall the formula for integration by parts: $ \int u \, dv = uv - \int v \, du $. Choose $ u = \ln(x) $ (since its derivative simplifies) and $ dv = x \, dx $.

Differentiate $ u = \ln(x) $ to find $ du = \frac{1}{x} \, dx $, and integrate $ dv = x \, dx $ to find $ v = \frac{x^2}{2} $.

Substitute into the integration by parts formula: $ \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx $. Simplify the remaining integral.

Evaluate the simplified integral and apply the limits of integration from 1 to 2. Substitute the upper and lower bounds into the resulting expression to compute the definite integral.