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Use Newton's Method (Section 4.7), where needed, to approximate the $x$ -coordinates of the intersections of the curves to at least four decimal places, and then use those approximationsto approximate the area of the region.$$\begin{array}{l}{\text { The region that lies below the curve } y=\sin x \text { and above }} \\ {\text { the line } y=0.2 x, \text { where } x \geq 0}\end{array}$$

$$1.1809$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

Section 1

Area Between Two Curves

Integrals

Integration

Applications of Integration

Area Between Curves

Volume

Arc Length and Surface Area

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Lectures

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we want to find the area of the enclosed region. Of these two equations, the 1st 1 is y equals X sign of X squared, which is in red. And the second is why equals X to the four, which is in blue. And we also are interested in X greater than or equal to zero. So using a graphing calculator, we can see that there are two intersection points, Um, which are at the X equals zero and also X is approximately 0.896 So what we want to find is the area which is enclosed. So this small area right here, So as usual, we're going to determine this area by integrating eso area is equal to integral. We're going from our left most point, which is X equals zero to write most point, which was zero point 896 And we're going to do the top function minus the bottom function. So what we see here is that the top function is X sign of X squared, and the bottom function is X to the force. So we're just going to subtract those x sign of X squared minus X to the fore DX and after solving with this integral will get the approximate area. There is approximate because we we have an approximation for the top people. Intersection point. So that's what it's going to be the approximate area. So let's carry out the integration. The integral of X sign of X squared, uh, is negative half coasts of X squared. And if it's not clear how we do that, you could do a U substitution of U equals X squared. So D'You equals two x d. X. So that's how we get this first anti derivative and the second anti derivative is very clear. It's 1/5 x to the five, and this is again going from X equals 0 to 0.896 So after we plug in everything, we get the approximate answer of 0.37

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