Area of the region bounded by the curve x^2 = 4y, x - axis and x = 3 is Option 1: 9/2 sq unit Option 2: 9/4 sq unit Option 3: 9/5 sq unit Option 4: 9/8 sq unit

Answer: 9/4 sq unit. Solution: The curve is y = x^2/4. Area = _0^3 x^2/4 dx = [x^3/12]_0^3 = 27/12 = 9/4 sq units.

Applied Mathematics

Calculus

Easy

Area of the region bounded by the curve $x^2 = 4y$ , $x$ - axis and $x = 3$ is

$\frac{9}{2}$ sq unit

$\frac{9}{4}$ sq unit

$\frac{9}{5}$ sq unit

$\frac{9}{8}$ sq unit

✅ Correct Option: 2

The curve is $y = \frac{x^2}{4}$ .

Area = $\int_0^3 \frac{x^2}{4} dx = \left[\frac{x^3}{12}\right]_0^3 = \frac{27}{12} = \frac{9}{4}$ sq units.

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Let $f : R \to R$ be defined such that $f(x) = 16x^2 - 16x + 12$

(A) Maximum value of $f(x)$ is 8

(B) Minimum value of $f(x)$ is 8

(D) No maximum value of $f(x)$

Choose the correct answer from the options given below :