Question

How to find the area of the region enclosed by the line and curve y=x^4_4x^2+4 and y=x^2

Answer 1

y1 = (x^2 - 2)^2. And. y2 = x^2

We see that the curves are symmetric about y axis. Also y >= 0.

Intersection points are:
x^2 - 2 = x and x^2 - 2 = - x
So x = -1, 2, -2, 1.
Let's find the area for x in [0,1] and [1, 2].
(x^2 - 2})^2 - x^2 is > 0 for x= 1/2 and < 0 for x= 3/2.

Area = 2*(Integral y1-y2 from 0 to 1 + integral y2 - y1 from 1 to 2).

[ x^5/5 - 4/3 x^3 + 4x - x^3/3 ]_0^1 = 38/15.
[ x^3 /3 - x^5/5 + 4/3 x^3 -4 x ]_1^2 = 35/3 - 31/5 -4= 22/15

Total area = 2*4 =8

Answer 2

y = x⁴ - 4x² + 4
y = x²

x² = x⁴ - 4x² + 4
x⁴ - 5x² + 4 = 0
(x² - 1)(x² - 4) = 0
x² = 1 , 4
x = ±1 and ±2
see graph of both functions.
both sides equal area inclosed by functions.
hence, area of whole parts = 2 × area of one part .

area of one part ={(x² -2)² - x²}dx Limit [ 0, 1]
+ {x² - (x² - 2)²}dx Limit [ 1, 2 ]
= 38/15 + 22/15
= 4

hence, area of whole part = 2 × 4 = 8 unit ²