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Sketch the region enclosed by the given curves and calculate its area.

$ y = 4 - x^2 $, $ y = 0 $

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$$\frac{32}{3}$$

01:37

Frank Lin

01:00

Amrita Bhasin

02:45

Bobby Barnes

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

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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Sketch the region enclosed…

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bounded by the function Y equals four minus x squared and y equals zero. Is the blue shaded region. You see here? The blue curve is the graph of the function uh Y equals four minus X squared. When x zero, you plug zero in for x square. To get four minus zero squared which is four. That gives us this point on the function right here. Uh coordinates zero comma four. And then uh when x is 22 squared is 44 minus 40. So when X is to uh why comes out to equal zero? So we have this point on the graph the curb of our function at two comma zero. To find the area underneath this blue function uh curve and the line Y equals zero. The line Y equals zero is the X axis. Since every point on the X axis um has to coordinate Y equals zero. We're basically looking for this blue region. Now, the area of this blue region, The area underneath the curve uh and above the X axis is equal to the integral from x equals 02, X equals two. Uh Since that's the two values on the X axis that our region is bounded by from X equals 02, X equals two. So the area of this blue region is the integral from 0 to 2 of our function for minus X squared. Uh D X. Now the integral of four is four, X minus. The integral of x square is execute over three. This is a definite integral role. So we are integrating dysfunction between X equals zero and X equals two. So once we integrated it we found the anti derivative we now have to evaluate this expression at two and then subtract this expression evaluated at zero, plugging into everywhere you see X you get four times two -2 Cubed Over three. And then we need to subtract this expression when we plug in zero when we plug zero into this expression uh all of this becomes zero. So you really just be subtracting zero which does help make things a little bit easier. So four times 2 is eight. So tracked two cubed over three is eight thirds and eight minus eight thirds is five and one third. So the area underneath our function and above the X axis is equal to five and one third. Once again, if you want to find uh the area underneath the curve and above the X axis you simply integrate it's a definite integral. You need to know the two values of X that you are integrating between. We integrated between X equals zero and X equals two. Because those were uh when X0, that's when this region starts and when X is two, that's when this region ends. So we integrate from zero to our function for minus x squared dx. To integrate. You find the anti derivative. So the anti derivative of four minus X squared was four X minus X cubed over three. Then you just had to evaluate it at two And you have to evaluate it at zero. So you take this expression when access to And then you subtract the value of this expression when X0. And when we did all that, we got five and one third. So this blue shaded region has an area of 5130.

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