Area Under Curves
Q. Calculate the area between the curves y = x and y = x^2 from x = 0 to x = 1.
-
A.
0.25
-
B.
0.5
-
C.
0.75
-
D.
1
Solution
The area is given by the integral from 0 to 1 of (x - x^2) dx. This evaluates to [x^2/2 - x^3/3] from 0 to 1 = (1/2 - 1/3) = 1/6 = 0.5.
Correct Answer: B — 0.5
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Q. Calculate the area between the curves y = x^2 and y = 2x from x = 0 to x = 2.
Solution
The area is given by the integral from 0 to 2 of (2x - x^2) dx. This evaluates to [x^2 - x^3/3] from 0 to 2 = (4 - 8/3) = 4/3.
Correct Answer: A — 2
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Q. Calculate the area between the curves y = x^2 and y = 4 from x = 0 to x = 2.
Solution
The area is given by the integral from 0 to 2 of (4 - x^2) dx. This evaluates to [4x - x^3/3] from 0 to 2 = (8 - 8/3) = 16/3.
Correct Answer: A — 4
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Q. Calculate the area under the curve y = cos(x) from x = 0 to x = π/2.
Solution
The area under the curve y = cos(x) from x = 0 to x = π/2 is given by ∫(from 0 to π/2) cos(x) dx = [sin(x)] from 0 to π/2 = 1 - 0 = 1.
Correct Answer: A — 1
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Q. Calculate the area under the curve y = x^2 + 2x from x = 0 to x = 2.
Solution
The area under the curve is given by ∫(from 0 to 2) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 2 = (8/3 + 4) = 20/3.
Correct Answer: B — 6
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Q. Calculate the area under the curve y = x^4 from x = 0 to x = 2.
Solution
The area under the curve y = x^4 from x = 0 to x = 2 is given by ∫(from 0 to 2) x^4 dx = [x^5/5] from 0 to 2 = (32/5) - 0 = 32/5.
Correct Answer: B — 8
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Q. Determine the area between the curves y = x^3 and y = x from x = 0 to x = 1.
-
A.
1/4
-
B.
1/3
-
C.
1/2
-
D.
1/6
Solution
The area is given by the integral from 0 to 1 of (x - x^3) dx. This evaluates to [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.
Correct Answer: A — 1/4
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Q. Determine the area enclosed by the curves y = x^2 and y = 4.
Solution
The area enclosed is found by integrating from -2 to 2: ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 16/3.
Correct Answer: C — 16/3
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Q. Determine the area under the curve y = 1/x from x = 1 to x = 2.
-
A.
ln(2)
-
B.
ln(1)
-
C.
ln(2) - ln(1)
-
D.
ln(2) + ln(1)
Solution
The area under the curve y = 1/x from x = 1 to x = 2 is given by ∫(from 1 to 2) (1/x) dx = [ln(x)] from 1 to 2 = ln(2) - ln(1) = ln(2).
Correct Answer: A — ln(2)
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Q. Determine the area under the curve y = e^x from x = 0 to x = 1.
Solution
The area under the curve y = e^x from 0 to 1 is given by ∫(from 0 to 1) e^x dx = [e^x] from 0 to 1 = e - 1.
Correct Answer: A — e - 1
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Q. Find the area between the curves y = x^2 and y = 4 from x = -2 to x = 2.
Solution
The area between the curves is given by ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = 16/3.
Correct Answer: B — 16/3
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Q. Find the area between the curves y = x^2 and y = 4 from x = 0 to x = 2.
Solution
The area between the curves y = x^2 and y = 4 is given by ∫(from 0 to 2) (4 - x^2) dx = [4x - x^3/3] from 0 to 2 = (8 - 8/3) = 4/3.
Correct Answer: A — 4
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Q. Find the area between the curves y = x^3 and y = x from x = 0 to x = 1.
-
A.
1/4
-
B.
1/3
-
C.
1/2
-
D.
1/6
Solution
The area between the curves is given by ∫(from 0 to 1) (x - x^3) dx = [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.
Correct Answer: B — 1/3
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Q. Find the area under the curve y = e^x from x = 0 to x = 1.
Solution
The area is given by the integral from 0 to 1 of e^x dx. This evaluates to [e^x] from 0 to 1 = e - 1.
Correct Answer: A — e - 1
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Q. Find the area under the curve y = x^2 + 2x from x = 0 to x = 3.
Solution
The area under the curve is given by ∫(from 0 to 3) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 3 = (27/3 + 9) = 18.
Correct Answer: C — 15
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Q. Find the area under the curve y = x^2 from x = 0 to x = 2.
Solution
The area under the curve y = x^2 from 0 to 2 is given by the integral ∫(from 0 to 2) x^2 dx = [x^3/3] from 0 to 2 = (2^3/3) - (0^3/3) = 8/3.
Correct Answer: C — 8/3
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Q. Find the area under the curve y = x^4 from x = 0 to x = 1.
-
A.
1/5
-
B.
1/4
-
C.
1/3
-
D.
1/2
Solution
The area under the curve y = x^4 from 0 to 1 is given by ∫(from 0 to 1) x^4 dx = [x^5/5] from 0 to 1 = 1/5.
Correct Answer: A — 1/5
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Q. Find the area under the curve y = x^4 from x = 0 to x = 2.
Solution
The area is given by the integral from 0 to 2 of x^4 dx. This evaluates to [x^5/5] from 0 to 2 = (32/5) = 16.
Correct Answer: C — 16
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Q. What is the area between the curves y = x^2 and y = 4 from x = -2 to x = 2?
Solution
The area between the curves is given by ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - (8/3)) - (-8 + (8/3)) = 16 - (16/3) = 32/3.
Correct Answer: A — 8
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Q. What is the area under the curve y = 1/x from x = 1 to x = 2?
-
A.
ln(2)
-
B.
1
-
C.
ln(2) - 1
-
D.
0
Solution
The area is given by the integral from 1 to 2 of 1/x dx. This evaluates to [ln(x)] from 1 to 2 = ln(2) - ln(1) = ln(2).
Correct Answer: A — ln(2)
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Q. What is the area under the curve y = cos(x) from x = 0 to x = π/2?
Solution
The area is given by the integral from 0 to π/2 of cos(x) dx. This evaluates to [sin(x)] from 0 to π/2 = 1 - 0 = 1.
Correct Answer: A — 1
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Q. What is the area under the curve y = sin(x) from x = 0 to x = π?
Solution
The area is given by the integral from 0 to π of sin(x) dx. This evaluates to [-cos(x)] from 0 to π = [1 - (-1)] = 2.
Correct Answer: C — π
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Q. What is the area under the curve y = x^3 from x = 1 to x = 2?
Solution
The area under the curve y = x^3 from x = 1 to x = 2 is given by ∫(from 1 to 2) x^3 dx = [x^4/4] from 1 to 2 = (16/4) - (1/4) = 4 - 0.25 = 3.75.
Correct Answer: B — 4
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Q. What is the area under the curve y = x^4 from x = 0 to x = 1?
-
A.
1/5
-
B.
1/4
-
C.
1/3
-
D.
1/2
Solution
The area under the curve y = x^4 from x = 0 to x = 1 is given by ∫(from 0 to 1) x^4 dx = [x^5/5] from 0 to 1 = 1/5.
Correct Answer: A — 1/5
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