JAMIA 2024 — Mathematics PYQ

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Question

JAMIA 2024 — Mathematics PYQ

JAMIA | Mathematics | 2024

The area bounded by the curves $y^{2} = 4 x$ and $y = x$ is equal to

Choose the correct answer:

Explanation

Step 1: Intersection Points

Pehle hum curves ko solve karke limits nikaalte hain:

y^2 = 4x \quad \text{and} \quad y = x

Substitute $y = x$ into $y^2 = 4x$ :

x^2 = 4x

x^2 - 4x = 0

x(x - 4) = 0

Limits are $x = 0$ to $x = 4$ .

Step 2: Integral Setup

Area formula: $A = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$

A = \int_{0}^{4} (\sqrt{4x} - x) \, dx

A = \int_{0}^{4} (2x^{1/2} - x) \, dx

Step 3: Calculation

A = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{4}

A = \left[ \frac{4}{3}x^{3/2} - \frac{x^2}{2} \right]_{0}^{4}

Upper limit ( $x=4$ ) put karte hain:

A = \left( \frac{4}{3}(4)^{3/2} - \frac{4^2}{2} \right) - 0

A = \frac{4}{3}(8) - \frac{16}{2}

A = \frac{32}{3} - 8

A = \frac{32 - 24}{3}

A = \frac{8}{3}

Final Answer:

Area = \frac{8}{3} \text{ sq. units}