Integral Calculus & Differential Equations
Q. Calculate the area under the curve y = 2x + 1 from x = 1 to x = 4.
Solution
The area under the curve is given by ∫(from 1 to 4) (2x + 1) dx = [x^2 + x] from 1 to 4 = (16 + 4) - (1 + 1) = 20.
Correct Answer: A — 15
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Q. Calculate the area under the curve y = x^3 from x = 0 to x = 2.
Solution
The area under the curve is given by ∫(from 0 to 2) x^3 dx = [x^4/4] from 0 to 2 = (16/4) - (0) = 4.
Correct Answer: B — 8
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Q. Find the area between the curves y = x and y = x^2 from x = 0 to x = 1.
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A.
0.5
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B.
1
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C.
0.25
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D.
0.75
Solution
The area between the curves is given by ∫(from 0 to 1) (x - x^2) dx = [x^2/2 - x^3/3] from 0 to 1 = (1/2 - 1/3) = 1/6 = 0.5.
Correct Answer: A — 0.5
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Q. Find the area under the curve y = 3x^2 from x = 1 to x = 2.
Solution
The area under the curve is given by ∫(from 1 to 2) 3x^2 dx = [x^3] from 1 to 2 = (8 - 1) = 7.
Correct Answer: B — 6
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Q. What is the area under the curve y = 1/x from x = 1 to x = 4?
-
A.
ln(4)
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B.
ln(3)
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C.
ln(2)
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D.
ln(1)
Solution
The area under the curve is given by ∫(from 1 to 4) (1/x) dx = [ln(x)] from 1 to 4 = ln(4) - ln(1) = ln(4).
Correct Answer: A — ln(4)
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Q. What is the area under the curve y = 2x^2 + 3 from x = 0 to x = 2?
Solution
The area under the curve is given by ∫(from 0 to 2) (2x^2 + 3) dx = [(2/3)x^3 + 3x] from 0 to 2 = (16/3 + 6) = 10.
Correct Answer: B — 12
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