The area bounded by the curve x^2 = 4y and the line x = 4y - 2 is : Option 1: 9/8 Option 2: 9/4 Option 3: 9/2 Option 4: 9/7

Answer: 9/8. Solution: Intersection: x^2 = x + 2 x^2 - x - 2 = 0 x = -1, 2. Area = _(-1)^2((x+2)/4 - x^2/4)dx = 1/4[x^2/2 + 2x - x^3/3]_(-1)^2 = 1/4 * 9/2 = 9/8.

Calculus

Application of Integrals

Medium

The area bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$ is :

$\frac{9}{8}$

$\frac{9}{4}$

$\frac{9}{2}$

$\frac{9}{7}$

✅ Correct Option: 1

Intersection: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow x = -1, 2$ . Area $= \int_{-1}^{2}\left(\frac{x+2}{4} - \frac{x^2}{4}\right)dx = \frac{1}{4}\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} = \frac{1}{4} \cdot \frac{9}{2} = \frac{9}{8}$ .

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The area bounded by the curve x2=4yx^2 = 4yx2=4y and the line x=4y−2x = 4y - 2x=4y−2 is :

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CUET Mathematics Past Year Questions (Topic-Wise)

The area bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$ is :