Regions enclosed by curves

Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

𝑦=4𝑥^2, 𝑦=𝑥^2+2

Please show work! :D

Answer 1

find intersection points = integration limits

here's a slightly different problem, using x^4 for one curve instead of x^2 for both

4x^4 =x^2 +2

4x^4-x^2-2 =0

let z=x^2

4z^2-z-2 =0

z=1/8 +/-(1/8)sqr(1+32)

x=+/-sqr(1+/-sqr33)/8)

integrate x^2+2-x^4 dx

=x^3/3 +2x- 4x^5/5

evaluate between above limits = about -.92 to +.92

= (.92)^3)/3 +2(.92) -4(.92)^5)/5 -(-.92)^3)/3 +2(-.92) -4(-.92)^5)/5

= (2/3).92^3+4(.92) +(8/5)(.92^5)

= (2/3 + 8/5).92^3[1+.92^2] +3.68

= (34/15).92^3[1.8464]+ 3.68

= about 3.26 + 3.68

= 6.94

back to the posted problem with 4x^4 and x^2 +2

limits of integration are when 4x^2=x^2 +2

3x^2 = 2

x^2 = 2/3

x = -sqr(2/3) and +sqr(2/3)

integral of x^2+2 -4x^2 dx = integral of 2-3x^2= 2x -x^3

evaluate between the 2 limits of integration, plug in -sqr(2/3) and +sqr(2/3) for x

= 2(sqr(2/3) - (sqr(2/3))^3 -[2(-sqr(2/3) - (-sqr(2/3)^3]

= 2sqr(2/3) - (2/3)sqr(2/3) +2sqr(2/3) -(2/3)sqr(2/3)

=(8/3)sqr(2/3)

=about 2.19932 square units

=about 2.2 square units, rounded off to one decimal place

Answer 2

In terms of the graph, They'll both be parabolas with y=4x² being below the other graph and being skinnier than the other graph. They will intersect when 4x² = x² + 2; this occurs at x = ±√(²/₃). Because there is one graph clearly above the other, we will integrate with respect to x. Also, because both of these functions are even functions (mirror on y-axis), we will integrate from 0 to √(²/₃) and then double that area.

2•∫ (x² + 2 - 4x²) dx --Integrate with the upper function minus the lower function.--

2(2x - x³) --Simplified integration--

2(2√(²/₃) - (²/₃)√(²/₃)) --Plug in upper bound (lower bound x=0 will give 0)--

(⁸/₃)√(²/₃) ≈ 2.177 --Simplify final answer--