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The area bounded by the parabolas y = (x+1)2 and y = (x−1)2 and the line y = (1/4) is equal to
- a)4/3
- b)1/3
- c)1/6
- d)4
Correct answer is option 'B'. Can you explain this answer?
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The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and t...
To find the area bounded by the parabolas y = (x - 1)^2 and y = (x + 1)^2, we need to find the points of intersection first.
Setting the two equations equal to each other, we have:
(x - 1)^2 = (x + 1)^2
Expanding both sides, we get:
x^2 - 2x + 1 = x^2 + 2x + 1
Simplifying, we have:
-4x = 4
Dividing both sides by -4, we get:
x = -1
So the two parabolas intersect at x = -1.
Now we can find the y-values of the intersection point by substituting x = -1 into either equation:
y = (-1 - 1)^2 = (-2)^2 = 4
Therefore, the two parabolas intersect at the point (-1, 4).
To find the area bounded by the parabolas, we need to find the area between the two curves from x = -1 to x = 1.
Integrating the upper curve (y = (x + 1)^2) and subtracting the integral of the lower curve (y = (x - 1)^2):
∫[(x + 1)^2 - (x - 1)^2] dx from x = -1 to x = 1
Expanding the expressions:
∫[(x^2 + 2x + 1) - (x^2 - 2x + 1)] dx from x = -1 to x = 1
Simplifying:
∫(4x) dx from x = -1 to x = 1
Integrating:
2x^2 from x = -1 to x = 1
Substituting the limits:
2(1)^2 - 2(-1)^2
2(1) - 2(1)
2 - 2
0
Therefore, the area bounded by the parabolas y = (x - 1)^2 and y = (x + 1)^2 is 0.
Setting the two equations equal to each other, we have:
(x - 1)^2 = (x + 1)^2
Expanding both sides, we get:
x^2 - 2x + 1 = x^2 + 2x + 1
Simplifying, we have:
-4x = 4
Dividing both sides by -4, we get:
x = -1
So the two parabolas intersect at x = -1.
Now we can find the y-values of the intersection point by substituting x = -1 into either equation:
y = (-1 - 1)^2 = (-2)^2 = 4
Therefore, the two parabolas intersect at the point (-1, 4).
To find the area bounded by the parabolas, we need to find the area between the two curves from x = -1 to x = 1.
Integrating the upper curve (y = (x + 1)^2) and subtracting the integral of the lower curve (y = (x - 1)^2):
∫[(x + 1)^2 - (x - 1)^2] dx from x = -1 to x = 1
Expanding the expressions:
∫[(x^2 + 2x + 1) - (x^2 - 2x + 1)] dx from x = -1 to x = 1
Simplifying:
∫(4x) dx from x = -1 to x = 1
Integrating:
2x^2 from x = -1 to x = 1
Substituting the limits:
2(1)^2 - 2(-1)^2
2(1) - 2(1)
2 - 2
0
Therefore, the area bounded by the parabolas y = (x - 1)^2 and y = (x + 1)^2 is 0.
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The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer?
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