Asked by r
Area bounded by curves y=x^2+4x+2, y=-2x, y=3, x=-1. A little help on how to do this would do. Thank you!
Answers
Answered by
mathhelper
Too many boundaries
I suggest you graph each one using
h tt ps://www.desmos.com/calculator
(delete the 2 spaces at the front of the URL)
to see what I mean.
My interpretation would leave you with a right-angled triangle,
bounded by y = -2x, y = 3, and x = -1
with the parabola not even entering the picture.
Is that what you want?
I suggest you graph each one using
h tt ps://www.desmos.com/calculator
(delete the 2 spaces at the front of the URL)
to see what I mean.
My interpretation would leave you with a right-angled triangle,
bounded by y = -2x, y = 3, and x = -1
with the parabola not even entering the picture.
Is that what you want?
Answered by
oobleck
This is a weird region, and you will have to change boundaries where the parabola intersects y = -2x. That is at (√7-3, 6-2√7)
so, the area is the sum of two integrals
A1 = ∫[-1,√7-3] (3 - (-2x)) dx = 1.0627
A2 = ∫[√7-3,√5-2] (3 - (x^2+4x+2)) dx = 0.71064
Sanity check:
A1 is basically a trapezoid with area (3.3)(0.65)/2 = 1.07
A2 is basically a triangle with area (2.3)(0.586)/2 = 0.67
Looks reasonable to me.
so, the area is the sum of two integrals
A1 = ∫[-1,√7-3] (3 - (-2x)) dx = 1.0627
A2 = ∫[√7-3,√5-2] (3 - (x^2+4x+2)) dx = 0.71064
Sanity check:
A1 is basically a trapezoid with area (3.3)(0.65)/2 = 1.07
A2 is basically a triangle with area (2.3)(0.586)/2 = 0.67
Looks reasonable to me.
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