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The area bounded by the curve y = 4x - x2 and x-axis is
- a)30/7 sq. units
- b)31/7 sq. units
- c)32/3 sq. units
- d)34/3 sq. units
Correct answer is option 'C'. Can you explain this answer?
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The area bounded by the curve y = 4x - x2 and x-axis isa)30/7 sq. unit...
Understanding the Problem:
The problem asks us to find the area bounded by the curve y = 4x - x^2 and x-axis.
Solution:
Step 1: Find the x-intercepts:
To find the x-intercepts, we set y = 0 in the given equation:
0 = 4x - x^2
0 = x(4 - x)
x = 0 or x = 4
Step 2: Calculate the area:
To find the area bounded by the curve and the x-axis, we need to find the integral of the curve function with respect to x:
Area = ∫[0,4] (4x - x^2) dx
Area = [2x^2 - (x^3)/3] from 0 to 4
Area = [2(4)^2 - ((4)^3)/3] - [2(0)^2 - ((0)^3)/3]
Area = [32 - 64/3] - [0 - 0]
Area = 96/3 - 64/3
Area = 32/3 sq. units
Therefore, the area bounded by the curve y = 4x - x^2 and x-axis is 32/3 sq. units, which corresponds to option 'c'.
The problem asks us to find the area bounded by the curve y = 4x - x^2 and x-axis.
Solution:
Step 1: Find the x-intercepts:
To find the x-intercepts, we set y = 0 in the given equation:
0 = 4x - x^2
0 = x(4 - x)
x = 0 or x = 4
Step 2: Calculate the area:
To find the area bounded by the curve and the x-axis, we need to find the integral of the curve function with respect to x:
Area = ∫[0,4] (4x - x^2) dx
Area = [2x^2 - (x^3)/3] from 0 to 4
Area = [2(4)^2 - ((4)^3)/3] - [2(0)^2 - ((0)^3)/3]
Area = [32 - 64/3] - [0 - 0]
Area = 96/3 - 64/3
Area = 32/3 sq. units
Therefore, the area bounded by the curve y = 4x - x^2 and x-axis is 32/3 sq. units, which corresponds to option 'c'.
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The area bounded by the curve y = 4x - x2 and x-axis isa)30/7 sq. unitsb)31/7 sq. unitsc)32/3 sq. unitsd)34/3 sq. unitsCorrect answer is option 'C'. Can you explain this answer?
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