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The area, in square unit, bounded by the curves y = x3, y = x2 and the ordinates x = 1, x = 2 is
- a)17/12
- b)12/13
- c)2/7
- d)7/2
Correct answer is option 'A'. Can you explain this answer?
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The area, in square unit, bounded by the curves y = x3, y = x2 and th...
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The area, in square unit, bounded by the curves y = x3, y = x2 and th...
Given curves and bounds:
The given curves are y = x^3, y = x^2, and the ordinates x = 1, x = 2.
Finding the points of intersection:
To find the points of intersection, we need to solve for x when y = x^3 and y = x^2.
Setting x^3 = x^2, we get x = 0 and x = 1 as the points of intersection.
Setting up the integral:
To find the area bounded by the curves, we need to set up the integral. The area can be calculated as the difference between the curves x^3 and x^2 within the bounds of x = 1 and x = 2.
Integrating the function:
Integrating the function f(x) = x^3 - x^2 gives us F(x) = (1/4)x^4 - (1/3)x^3.
Finding the area:
Substitute the bounds x = 2 and x = 1 into the antiderivative F(x) and find the difference.
Area = F(2) - F(1) = [(1/4)(16) - (1/3)(8)] - [(1/4)(1) - (1/3)(1)] = (4 - 8/3) - (1/4 - 1/3) = 12/3 - 8/3 = 4/3 = 16/12 = 17/12.
Therefore, the area bounded by the given curves and ordinates is 17/12 square units.
The given curves are y = x^3, y = x^2, and the ordinates x = 1, x = 2.
Finding the points of intersection:
To find the points of intersection, we need to solve for x when y = x^3 and y = x^2.
Setting x^3 = x^2, we get x = 0 and x = 1 as the points of intersection.
Setting up the integral:
To find the area bounded by the curves, we need to set up the integral. The area can be calculated as the difference between the curves x^3 and x^2 within the bounds of x = 1 and x = 2.
Integrating the function:
Integrating the function f(x) = x^3 - x^2 gives us F(x) = (1/4)x^4 - (1/3)x^3.
Finding the area:
Substitute the bounds x = 2 and x = 1 into the antiderivative F(x) and find the difference.
Area = F(2) - F(1) = [(1/4)(16) - (1/3)(8)] - [(1/4)(1) - (1/3)(1)] = (4 - 8/3) - (1/4 - 1/3) = 12/3 - 8/3 = 4/3 = 16/12 = 17/12.
Therefore, the area bounded by the given curves and ordinates is 17/12 square units.
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The area, in square unit, bounded by the curves y = x3, y = x2 and the ordinates x = 1, x = 2 isa)17/12b)12/13c)2/7d)7/2Correct answer is option 'A'. Can you explain this answer?
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