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The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then the area bounded by the curve, the x-axis and the line x = 1 is
- a)5/6
- b)6/5
- c)1/6
- d)6
Correct answer is option 'A'. Can you explain this answer?
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The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. I...
Solution:
Given that the slope of the tangent to the curve is 2x - 1. Let's integrate this expression to find the equation of the curve.
Integration of the slope of the tangent gives us:
∫(2x - 1) dx = x^2 - x + C
Since the curve passes through the point (1, 2), we can substitute these values into the equation to find the value of C.
2 = 1^2 - 1 + C
2 = 1 - 1 + C
2 = C
So, the equation of the curve is y = x^2 - x + 2.
To find the area bounded by the curve, the x-axis, and the line x = 1, we need to find the definite integral of the curve between the limits x = 0 and x = 1.
Area = ∫[0,1] (x^2 - x + 2) dx
Using the power rule of integration, we can find the antiderivative of the curve:
= [x^3/3 - x^2/2 + 2x] from 0 to 1
Substituting the limits into the antiderivative expression:
= [(1^3/3 - 1^2/2 + 2(1)) - (0^3/3 - 0^2/2 + 2(0))]
Simplifying the expression:
= [(1/3 - 1/2 + 2) - (0 - 0 + 0)]
= [(1/3 - 1/2 + 2)]
= [(2/6 - 3/6 + 12/6)]
= [11/6]
Therefore, the area bounded by the curve, the x-axis, and the line x = 1 is 11/6.
The correct answer is option A) 5/6.
Given that the slope of the tangent to the curve is 2x - 1. Let's integrate this expression to find the equation of the curve.
Integration of the slope of the tangent gives us:
∫(2x - 1) dx = x^2 - x + C
Since the curve passes through the point (1, 2), we can substitute these values into the equation to find the value of C.
2 = 1^2 - 1 + C
2 = 1 - 1 + C
2 = C
So, the equation of the curve is y = x^2 - x + 2.
To find the area bounded by the curve, the x-axis, and the line x = 1, we need to find the definite integral of the curve between the limits x = 0 and x = 1.
Area = ∫[0,1] (x^2 - x + 2) dx
Using the power rule of integration, we can find the antiderivative of the curve:
= [x^3/3 - x^2/2 + 2x] from 0 to 1
Substituting the limits into the antiderivative expression:
= [(1^3/3 - 1^2/2 + 2(1)) - (0^3/3 - 0^2/2 + 2(0))]
Simplifying the expression:
= [(1/3 - 1/2 + 2) - (0 - 0 + 0)]
= [(1/3 - 1/2 + 2)]
= [(2/6 - 3/6 + 12/6)]
= [11/6]
Therefore, the area bounded by the curve, the x-axis, and the line x = 1 is 11/6.
The correct answer is option A) 5/6.
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The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then the area bounded by the curve, the x-axis and the line x = 1 isa)5/6b)6/5c)1/6d)6Correct answer is option 'A'. Can you explain this answer?
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