This one is a little tricky. First graph all functions/lines.
Then find region that is bounded by all functions/lines.
The region I found is shaded in link below:
http://www.flickr.com/photos/56185495@N0…
A = ∫₀¹ (3e^x - (3 - x²)) dx + ∫₁² (3e^x - 2x) dx
A = ∫₀¹ (3e^x - 3 + x²) dx + ∫₁² (3e^x - 2x) dx
A = (3e^x - 3x + x³/3)|₀¹ + (3e^x - x²)|₁²
A = [(3e - 3 + 1/3) - (3 - 0 + 0)] + [(3e² - 4) - (3e - 1)]
A = (3e - 17/3) + (3e² - 3e - 3)
A = 3e² - 26/3
A ≈ 13.5005
Mαthmφm
Then find region that is bounded by all functions/lines.
The region I found is shaded in link below:
http://www.flickr.com/photos/56185495@N0…
A = ∫₀¹ (3e^x - (3 - x²)) dx + ∫₁² (3e^x - 2x) dx
A = ∫₀¹ (3e^x - 3 + x²) dx + ∫₁² (3e^x - 2x) dx
A = (3e^x - 3x + x³/3)|₀¹ + (3e^x - x²)|₁²
A = [(3e - 3 + 1/3) - (3 - 0 + 0)] + [(3e² - 4) - (3e - 1)]
A = (3e - 17/3) + (3e² - 3e - 3)
A = 3e² - 26/3
A ≈ 13.5005
Mαthmφm
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A diagram shows that there is only one region bounded by all four of these graphs so I presume it is the area of that region that is required.
The region is to the left of x = 2, above both y = 3 - x^2 and y = 2x and below y = 3e^x.
You need to find where y = 3 - x^2 and y = 2x cross by solving
3 - x^2 = 2x ----> x^2 + 2x - 3 = 0 ----> (x + 3)(x - 1) = 0 ----> x = 1 or -3.
Only the result x = 1 matters. it shows that the required area is
INT [0, 2] 3e^x dx - INT [1, 2] 2x dx - INT [0, 1] 3 - x^2 dx
Can you finish from there?
The region is to the left of x = 2, above both y = 3 - x^2 and y = 2x and below y = 3e^x.
You need to find where y = 3 - x^2 and y = 2x cross by solving
3 - x^2 = 2x ----> x^2 + 2x - 3 = 0 ----> (x + 3)(x - 1) = 0 ----> x = 1 or -3.
Only the result x = 1 matters. it shows that the required area is
INT [0, 2] 3e^x dx - INT [1, 2] 2x dx - INT [0, 1] 3 - x^2 dx
Can you finish from there?
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There are many regions bounded by these curves and lines. Can you be a little more specific?