Question

The area of the region bounded by $y-x=2$ and $x^2=y$ is equal to :

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{9}{2}$
• D. $\frac{16}{3}$

Hint:

To find the area between two curves, first find their points of intersection to determine the limits of integration. Then, integrate the difference between the upper curve and the lower curve over that interval.

Answer 1

The Correct Option is C

Given:

Line: y − x = 2
Parabola: x² = y

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Step 1: Express both equations in terms of y

Line: y = x + 2
Parabola: y = x²

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Step 2: Find points of intersection

Set:

x² = x + 2

x² − x − 2 = 0

Using quadratic formula:

x = [1 ± √(1 + 8)] / 2

x = (1 ± 3)/2

x = 2 or x = −1

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Step 3: Find corresponding y-values

For x = 2:

y = 2² = 4

For x = −1:

y = (−1)² = 1

Points of intersection:

(2, 4) and (−1, 1)

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Step 4: Set up the integral

Area = ∫₋₁² [(x + 2) − x²] dx

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Step 5: Evaluate the integral

∫ (x + 2) dx = x²/2 + 2x

∫ x² dx = x³/3

Area = [x²/2 + 2x]₋₁² − [x³/3]₋₁²

First term:

At x = 2 → 2 + 4 = 6
At x = −1 → 1/2 − 2 = −3/2

Difference:

6 − (−3/2) = 15/2

Second term:

At x = 2 → 8/3
At x = −1 → −1/3

Difference:

8/3 − (−1/3) = 3

Final area:

15/2 − 3 = 15/2 − 6/2 = 9/2

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Final Answer:

Area = 9/2