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Evaluate the integral.

$ \displaystyle \int ye^{0.2y} dy $

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04:25

Sky Li

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Siev M.

August 31, 2020

How did you get Y= 5x?

Campbell University

Harvey Mudd College

University of Nottingham

Boston College

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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09:56

Alright. We have a fun inner girl to solve. We could solve it by integration by parts. But this is a perfect one to instead use tabular tabular integration. Tabular integration. So the way it works is you put the function whose derivative would become zero eventually on the left And then we go ahead and take its derivatives. A derivative of Y. Is one. And then we get zero. And then the other function has to be one. We can take anti derivative. So we will take the anti derivative which will look like itself divided by um chain roles. So divided by .2 and then I'll take the anti derivative one more time. I will get E. To the 0.2 Y divided by 0.2 again. So .2 squared. And I can clean this up this by the way is probably easier. If I divide by 1/5. It's really five. Even 0.2 Y. It's a little bit nicer to read. And this then is five squared or 25. It's 0.2. What? Alright, so tabular integration, you do diagonals. This first one's plus the second one is minus. So are integral then equals Y. Times five E. To the 0.2. Why? Minus 25 e. two. 0.2. Why? And we can't forget plus C. And that is our solution. They did kind of clean it up a little bit. So let's try to get in the form they have. If you pull out of 25 then you get y over five E. To the 0.2 Y minus E. 20.2. Why? And um I think the only difference that they have is there still a plus C. I do believe they just wrote that 1/5 as a 0.2. So if you want to look like their official solutions it looks like this. Um Whoops. Um I lost the exponential scope back. Um So we'll get it just like their form Why? Eat at a 0.2 Y -0.2. Why? And I believe now we have the form. Oh they did one more thing, they factored out either factored out to the 0.2 Y. So let's do that. That leaves behind 0.2 y minus one. There we go. Now we have pretty close to the form of the solution already. Hopefully that helped have a fantastic day.

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