Lesson 1.08e
1.08e Areas under and between curves Quiz: OCR Maths, Unit 8
20 questions
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Lesson 1.08e, Areas under and between curves: 20 multiple choice questions for the OCR Maths (H240), Unit 8: Integration, written with Revision Ninja.
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The 20 questions
-
How is the physical area found for a region lying entirely below the x-axis?
- Reciprocal of integral
- Unmodified integral
- Integral squared
- Modulus of integral
-
Which expression gives the area between upper curve f(x) and lower curve g(x) on [a, b]?
- Integral of (f x g)
- Integral of (f + g)
- Integral of (g - f)
- Integral of (f - g)
-
Find the area under y = x^2 from x = 0 to x = 2.
- 4
- 8/3
- 2
- 16/3
-
Find the area between y = 4 - x^2 and the x-axis from x = -2 to x = 2.
- 64/3
- 16/3
- 32/3
- 8
-
Find the area between y = x and y = x^2 for 0 <= x <= 1.
- 1/3
- 1/6
- 1/2
- 1/12
-
Find the area between y = 2x + 3 and y = x^2.
- 9
- 32/3
- 16/3
- 10
-
Find the area under y = sin x from x = 0 to x = pi.
- 0
- pi
- 1
- 2
-
Find the area between y = x(x - 2) and the x-axis for 0 <= x <= 2.
- 8/3
- 2/3
- 4/3
- -4/3
-
Find the area between y = e^x, the x-axis, x = 0 and x = 1.
- e
- 1
- e + 1
- e - 1
-
Find the area between y = 1/x, the x-axis, x = 1 and x = e.
- 1/e
- 1
- e - 1
- e
-
Find the area between y = x^3 - x and the x-axis for -1 <= x <= 1.
- 0
- 1
- 1/2
- 1/4
-
Find the area between y = x and y = x^3 for 0 <= x <= 1.
- 1/4
- 1/3
- 1/6
- 1/2
-
A curve is given by x = t^2 and y = t for 0 <= t <= 1. What is the area under it?
- 2/3
- 1
- 1/3
- 1/2
-
If the area under y = kx from x = 0 to x = 2 is 6, what is k?
- 6
- 3
- 2
- 12
-
Find the area under y = x^2 - 4 from x = 0 to x = 2 (as a positive value).
- -16/3
- 16/3
- 4
- 8/3
-
Find the area between y = sqrt(x) and y = x^2 for 0 <= x <= 1.
- 1/3
- 1/6
- 2/3
- 1/2
-
Why does integrating a curve below the x-axis result in a negative value?
- x-values are negative
- y-values are negative
- Limits are negative
- Gradients are negative
-
Find the area between y = 2 and y = x^2.
- 4 sqrt(2)
- 4 sqrt(2)/3
- 2 sqrt(2)
- 8 sqrt(2)/3
-
Find the area under y = cos x from x = 0 to x = pi/2.
- 2
- 1
- 0
- pi/2
-
Find the area under y = 3 from x = 1 to x = 4.
- 3
- 12
- 9
- 6
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