Find the area bounded by 2 curves?

First, sketch both curves on a same graph [thats highschool math, so surely you can do such sketches].
Then notice your area, your domain.
Then notice that your area goes from x=a to x=b. Find these a & b again via highschool basic algebra.

Write the answers here: a = ____________ b = ______________

Then integrate (top function - bottom function) from x=a to b.

Show your steps here if need be and we can walk you further...

curves intersect at x = [ 1 ±√17 ] / 4 ; y = x - x² is on top...integrate [ top curve - bottom curve ] over [ ( 1 - √17 ) / 4 , ( 1 + √17 ) / 4 ]

do the work yourself

y = x² - 2 ← this is the first curve → C₁ ← blue curve

y = x - x² ← this is the second curve → C₂ ← red curve


Let’s calculate the intersection between the 2 curves.

y = y

x² - 2 = x - x²

2x² - x - 2 = 0

2.[x² - (1/2).x - 1] = 0

x² - (1/2).x - 1 = 0

x² - (1/2).x = 1

x² - (1/2).x + (1/4)² = 1 + (1/4)²

x² - (1/2).x + (1/4)² = 17/16

[x - (1/4)]² = [± (√17)/4]²

x - (1/4) = ± (√17)/4

x = (1/4) ± [(√17)/4]

x = (1 ± √17)/4


x₁ = (1 - √17)/4 ← purple line

x₂ = (1 + √17)/4 ← purple line


area = (from x₁ to x₂) ∫ (C₂ - C₁).dx → because the curve C₂ is over the curve C₁

area = (from x₁ to x₂) ∫ [(x - x²) - (x² - 2)].dx

area = (from x₁ to x₂) ∫ [x - x² - x² + 2].dx

area = (from x₁ to x₂) ∫ [- 2x² + x + 2].dx

area = [- (2/3).x³ + (1/2).x² + 2x] from x₁ to x₂

area = [- (2/3).x₂³ + (1/2).x₂² + 2x₂] - [- (2/3).x₁³ + (1/2).x₁² + 2x₁]

area = - (2/3).x₂³ + (1/2).x₂² + 2x₂ + (2/3).x₁³ - (1/2).x₁² - 2x₁

area = - (2/3).(x₂³ - x₁³) + (1/2).(x₂² - x₁²) + 2.(x₂ - x₁) ← memorize this result


= x₂ - x₁

= [(1 + √17)/4] - [(1 - √17)/4]

= [(1 + √17) - (1 - √17)]/4

= [1 + √17 - 1 + √17]/4

= [2√17]/4

= (√17)/2


= x₂² - x₁²

= [(1 + √17)/4]² - [(1 - √17)/4]²

= [(1 + √17)²/16]² - [(1 - √17)²/16]

= [(1 + √17)² - (1 - √17)²]/16

= [(1 + 2√17 + 17) - (1 - 2√17 + 17)]/16

= [1 + 2√17 + 17 - 1 + 2√17 - 17]/16

= [4√17]/16

= (√17)/4


= x₂³ - x₁³

= [(1 + √17)/4]³ - [(1 - √17)/4]³

= [(1 + √17)³/64] - [(1 - √17)³/64]

= [(1 + √17)³ - (1 - √17)³]/64

= [(1 + √17)².(1 + √17) - (1 - √17)².(1 - √17)]/64

= [(1 + 2√17 + 17).(1 + √17) - (1 - 2√17 + 17).(1 - √17)]/64

= [(18 + 2√17).(1 + √17) - (18 - 2√17).(1 - √17)]/64

= [2.(9 + √17).(1 + √17) - 2.(9 - √17).(1 - √17)]/64

= [(9 + √17).(1 + √17) - (9 - √17).(1 - √17)]/32

= [(9 + 9√17 + √17 + 17) - (9 - 9√17 - √17 + 17)]/32

= [(26 + 10√17) - (26 - 10√17)]/32

= [26 + 10√17 - 26 + 10√17]/32

= [20√17]/32

= (5√17)/8



Restart from the memorized result

area = - (2/3).(x₂³ - x₁³) + (1/2).(x₂² - x₁²) + 2.(x₂ - x₁) → recall: x₂ - x₁ = (√17)/2

area = - (2/3).(x₂³ - x₁³) + (1/2).(x₂² - x₁²) + √17 → recall: x₂² - x₁² = (√17)/4

area = - (2/3).(x₂³ - x₁³) + [(√17)/8] + √17 → recall: x₂³ - x₁³ = (5√17)/8

area = - [(10√17)/24] + [(√17)/8] + √17

area = - [(10√17)/24] + [(3√17)/24] + [(24√17)/24]

area = (- 10√17 + 3√17 + 24√17)/24

area = (17√17)/24

area ≈ 2.92

y = x² - 2
y = x - x²

The area is equal to the area bounded by the x-axis and this curve:

y = (x - x²) - (x² - 2)
y = -2x² + x + 2

x-intercepts:
A([1 - √(17)]/4, 0)
B([1 + √(17)]/4, 0)

AB = √(17)/2

midpoint of AB: (1/4, 0)

Let x = -1/4.
y = -2(1/4)² + (1/4) + 2 = 17/8

middle of arc AB: C(1/4, 17/8)

area(∆ABC)
= 1/2(17/8)[√(17)/2]
= 17√(17)/32

area(parabola section ACB)
= 4/3 area(∆ABC)
= 4/3[17√(17)/32]
= 17√(17)/24

y = x² – 2
y = x – x²

Now we want the area under the second curve (blue) subtracted from that of the first curve (red line). It's complicated by the fact that they are crossing the x axis, where areas become negative.

best way I know to handle this (I have not done this for many decades) is to offset the curves by y = –2 so all the areas are +
y = x² – 2 + 2 = x²
y = x – x² + 2

first, where do the two curves intersect
x² = x – x² + 2
2x² – x – 2 = 0

quadratic equation:
to solve ax² + bx + c = 0
x = [–b ± √(b²–4ac)] / 2a
x = [1 ± √(1+16)] / 4
x = (1/4) ± (1/4)√17
x = 1.281, –0.781

A1 = ∫(x – x² + 2)dx = (1/2)x² – (1/3)x³ + 2x between –0.781 and 1.281
A2 = ∫(x²)dx = (1/3)x³ between –0.781 and 1.281
A = (1/2)x² + 2x between –0.781 and 1.281
A1 = 0.820 + 2.562 + 0.305 – 1.562 = 2.125

but check my arith.

17√17 / 24