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Find the area of the region under the given curve from 1 to 2.

$ y = \dfrac{x^2 + 1}{3x - x^2} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Baylor University

University of Nottingham

Idaho State University

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.

03:26

Find the area of the regio…

04:49

02:04

01:28

03:44

this problem is asking for the integral from one to two of this function. And let me rewrite this denominator by factoring chips just three minus X. That's the area under the curve from one to two. So here the numerator and denominator of the same degree. They're both degree too. So we should along the vision here. So go to the side and do long division because we always should do this before a partial fractions. It's minus three X and then we have three. X Plus one is our remainder. So using this long division, I can rewrite this integral negative one and then three X plus one x three minus x t x And then we'LL go out and do partial fraction to composition on this fraction here using what the author calls Case one. So here we have a three minus X. So it's good and multiply both sides of this by distant dominator here on the left that on the right, so much factor on the X and we see that b minus a must be three and three a must be won. So we get that moves. So here we have one equals three soft ray and then used in this equation. Here we have B minus a third is three or weaken right that is nine over three. So soft for be there. Just add that one over three over. So we have our A and B. Now let's just go ahead and plug those in Now. We could replace this fraction here with the partial fraction. And let's write that on the next page. Once a two, we have the negative one from a long division and then for a partial fractions. That's our area. Now let's go ahead and evaluate these inner rules. If this last one. If this is bothering you with the three minus X, feel free to do it. Use up here. You could do U equals three minus six and to uni equals negative DX. That should work. So let's integrate these three. The first one just is minus X for the next one. Natural log. Absolute value X and for the last one, ten over three with a minus natural log. Absolute value three minus X, and you could see the negative is coming from the use of. And don't forget our UN points wanted to it's good and plug those in one at a time. Ellen. One there. We know that zero Ellen one. That's also zero minus ten over three. Ellen, too, And then just go out and simplify this eleven over three. Natural onto. That's our answer.

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