Find the areas of the regions enclosed by the lines and curves. y=x^4-4 x^2+4 and y=x^2

Name: Problem 47, Chapter 5: Thomas Calculus
Uploaded: 2020-02-16T08:58:26-08:00
Duration: 10 min 38 s
Channel: Dillon Huddleston
Description: Find the areas of the regions enclosed by the lines and curves. y=x^4-4 x^2+4 and y=x^2

Find the areas of the regions enclosed by the lines and...

Key Concepts

Intersection of Curves

This concept involves setting two function expressions equal to each other and solving for the variable to find the points where the curves cross. It enables the determination of the boundaries of the region whose area is to be calculated.

Definite Integration

Definite integration is the process of calculating the accumulation of a quantity, such as area, over a specific interval. In the context of finding the area between curves, it involves integrating the difference between the functions over the interval determined by their intersection points.

Determining Upper and Lower Functions

Identifying which function lies above the other on the interval of intersection is crucial. This ensures that the integrand, defined as the difference between the upper curve and the lower curve, correctly represents the area between them.

Area Between Curves

The area between curves is found by integrating the vertical distance between the two functions across the interval where they intersect. This concept applies the principles of definite integration to capture the exact enclosed region between the curves.

Transcript

00:01 So for this question, first we consider points of intersection between the two curves.

00:09 Y equal to x to the 4 minus 4x squared plus 4 and y equal to x squared.

00:16 So setting the two to be equal, we have 0 is x to the 4 minus 5 x squared plus 4.

00:30 Now another way we can realize solve that equality is realize that x to the 4 minus 4x squared plus 4 is x squared minus 2 all squared.

00:46 So we set x squared minus 2 all squared to be equal to x squared.

00:52 And this is true when x squared minus 2 is positive minus x.

00:59 Solve the two resulting quadratics in turn.

01:03 So in the case x squared minus 2, well, we want to solve the equation then x squared plus minus x minus 2 is 0.

01:17 And for that, we can use the quadratic formula.

01:24 So here we have b is plus minus 1.

01:36 So our roots will also have plus minus 1.

01:52 And then we had and subtract b squared minus 4 ac, which is 1 minus 4 multiplied by 1, multiplied by minus 2.

02:25 So that's 1 plus 4 times 2.

02:27 It is 9.

02:28 And the square root of that, 9 is 3.

02:33 And we divide by 2a.

02:35 And a is just 1.

02:39 So we divide by 2.

02:41 And so this gives us 4 roots.

02:46 4 by 2 is 2, minus 2 by 2 is minus 1, 2 by 2 is 1, and minus 1 by 3, minus 1, minus 2 by 2, so we have roots minus 2, minus 1, 1 and 2.

03:02 We're in 2 sections occur.

03:05 Now we can see, considering the functions, that the function y is x squared is always a non -negative, and increases increases from zero at the origin.

03:22 On the other hand, y, well, x squared minus 2 all squared will also be non -negative, but we can consider the derivatives of these two functions.

03:38 So, the derivative of y equal to x squared is 2x, and of y equal to x and 4x4 minus 4x4 plus 4 is 4x cubed, minus 8x.

03:51 Now for x near 0, this will be dominated by the minus 8x turn, which is negative, in contrast to the 2x term, which would be positive.

04:07 Again, assuming positive x and the sides of verse from minus x.

04:11 What this implies then is that near the origin, y equals x squared increases faster, then y equals x of 4, minus 4x squared plus 4.

04:24 But actually in this case, it's easier to just know directly that at 0, y equal to 4 minus 4x4 plus 4 is equal to 4, and y equal to x squared is equal to 0.

04:36 So between the roots minus 1 and positive 1, the curve, the quartic, the quartic plot is above the parabola...

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