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Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent.

$ \displaystyle \int x^2 \sqrt{1 - x^2}\ dx $

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$\frac{x}{8}\left(2 x^{2}-1\right) \sqrt{1-x^{2}}+\frac{1}{8} \arcsin x+C$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 6

Integration Using Tables and Computer Algebra Systems

Integration Techniques

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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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Use a computer algebra sys…

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Use a computer algebra sy…

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Okay, So this question wants us to compute the following anti derivative using a computer algebra system. Compare that answer to the tabular form and show that they are indeed equivalent. So plugging into a computer algebra system, I was given this answer. But when plugging into a table, you're giving this answer. So let's try expanding the computer's answer and show that that's equal to the table. So all we gotta do is distribute this 1/8 teach term to expand everything out. And I'm just rearranging cause I looks more aesthetically pleasing. If you put the square root on the end and then, as you can see, this is really close. But there's just a factor of X in front of one of the terms, so we can just pull the X out and the other form has the term swapped over. So we'll do that. So all we have to do is pull out and acts from the first center parentheses. So we get to X squared minus one, keeping our square root intact plus 1/8 sign in verse of X plus C and comparing this one last time with our table. We see that the two answers are exactly the same. So our anti derivatives are, after all, equivalent

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