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Determine the area of the region between the given curves.$$x-2 y^{2}=1 \text { and } x+2 y=5.$$

9

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 8

Applications of the Definite Integral

Integrals

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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Find the area of the regio…

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Determine the area of the …

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Sketch and find the area o…

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This question is testing you on your ability to find the area of the region between the curves. Mhm. So the two curves given to us are y equals negative X squared plus six X minus five. And why goes to x minus five. So our first step is to determine where the two curves intersect because we're not giving up bounds to find the area for. So we can do this just by setting the two equations to each other and solving it out. So this because negative skirt is four X. Because zero fires cancel out. I can divide both sides by negative one. And then yeah this is plus and then this becomes like sorry, set the internet to Mexico zero and X equals four. So are bound to be set up Who zero through four. Next week had to visualize the graft the graph to see where or to see which graph is on top of the other. So the line to x minus five goes this but this is a negative five and then negative. The problem negative X squared plus six X minus five also intersects at negative five. So, and then they intersect that uh four or X equals four, speaking for. Mhm. All right. So we can see by this past that we want to find this area right here. So because our arrival is on top, we can set up this by having the problem first and then subtracting two X minus five from it. D. X. He's become a zero before negative X squared. Mine is four. Thanks. D. X. If you solve this uh integral out, we should get an answer that is 32/3 and that's your final answer.

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