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The area bounded by the curve x2 = 4y and the straight line x = 4y − 2 is
- a)8/9 sq. unit
- b)9/8 sq. unit
- c)4/3 sq. unit
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The area bounded by the curve x2= 4yand the straight line x = 4y &minu...
We have,
x2 = 4y
x = 4y − 2
On solving both the equations simultaneously, we get the points of intersection
Required area
x2 = 4y
x = 4y − 2
On solving both the equations simultaneously, we get the points of intersection
Required area
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Community Answer
The area bounded by the curve x2= 4yand the straight line x = 4y &minu...
To find the area bounded by the curve x^2 = 4y and the straight line x = 4y, we need to find the points of intersection between the curve and the line.
First, let's solve the equation x^2 = 4y for y:
x^2 = 4y
y = x^2/4
Next, let's solve the equation x = 4y for y:
x = 4y
y = x/4
Now, we can find the points of intersection by setting the two equations equal to each other:
x^2/4 = x/4
x^2 = x
x^2 - x = 0
x(x - 1) = 0
From this, we find two possible values for x: x = 0 and x = 1.
Substituting these values into the equation y = x^2/4, we find the corresponding y-values:
For x = 0:
y = (0^2)/4 = 0
For x = 1:
y = (1^2)/4 = 1/4
So, the points of intersection are (0, 0) and (1, 1/4).
To find the area between the curve and the line, we need to integrate the difference between the two equations with respect to x. Since the curve is above the line in the given interval, the integral will be:
∫[0, 1] [(x^2/4) - (x/4)] dx
= ∫[0, 1] [(x^2 - x)/4] dx
= (1/4) ∫[0, 1] (x^2 - x) dx
= (1/4) [ (x^3/3) - (x^2/2) ] [0, 1]
= (1/4) [ (1/3) - (1/2) ]
= (1/4) [ (2/6) - (3/6) ]
= (1/4) (-1/6)
= -1/24
Therefore, the area bounded by the curve x^2 = 4y and the straight line x = 4y is -1/24 square units.
First, let's solve the equation x^2 = 4y for y:
x^2 = 4y
y = x^2/4
Next, let's solve the equation x = 4y for y:
x = 4y
y = x/4
Now, we can find the points of intersection by setting the two equations equal to each other:
x^2/4 = x/4
x^2 = x
x^2 - x = 0
x(x - 1) = 0
From this, we find two possible values for x: x = 0 and x = 1.
Substituting these values into the equation y = x^2/4, we find the corresponding y-values:
For x = 0:
y = (0^2)/4 = 0
For x = 1:
y = (1^2)/4 = 1/4
So, the points of intersection are (0, 0) and (1, 1/4).
To find the area between the curve and the line, we need to integrate the difference between the two equations with respect to x. Since the curve is above the line in the given interval, the integral will be:
∫[0, 1] [(x^2/4) - (x/4)] dx
= ∫[0, 1] [(x^2 - x)/4] dx
= (1/4) ∫[0, 1] (x^2 - x) dx
= (1/4) [ (x^3/3) - (x^2/2) ] [0, 1]
= (1/4) [ (1/3) - (1/2) ]
= (1/4) [ (2/6) - (3/6) ]
= (1/4) (-1/6)
= -1/24
Therefore, the area bounded by the curve x^2 = 4y and the straight line x = 4y is -1/24 square units.
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The area bounded by the curve x2= 4yand the straight line x = 4y − 2isa)8/9 sq. unitb)9/8 sq. unitc)4/3 sq. unitd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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