Calculus

Application of Integrals

Hard

A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by $y = y(x) = x + \sqrt{2x - 3x^2}$ where 'x' (in meters) is the horizontal distance moved and 'y' (in meters) is the corresponding height. Assuming that his initial point is at origin.

Figure for CUET Mathematics 2022 6 Aug Shift 2 question 44 (Calculus)

Based on the above information answer the Questions

Evaluate $\int_0^{\frac{1}{3}} y \, dx$

$\frac{1}{18} + \frac{2\sqrt{3}\pi}{27}$

$\frac{1}{18} + \sqrt{3}\frac{\pi}{36}$

$\frac{1}{18} + \frac{\sqrt{6}\pi}{9}$

$\frac{1}{18} + \frac{\pi}{36}$

✅ Correct Option: 2

$\int_0^{1/3} x \, dx = \frac{1}{18}$ .

For $\int \sqrt{2x - 3x^2} dx$ , complete the square: $2x - 3x^2 = \frac{1}{3}[1 - (3x-1)^2]$ .

Using substitution $u = 3x - 1$ : $\int_{-1}^{0} \frac{1}{3\sqrt{3}} \sqrt{1 - u^2} du = \frac{1}{3\sqrt{3}} \cdot \frac{\pi}{4} = \frac{\sqrt{3}\pi}{36}$ .

Sum: $\frac{1}{18} + \frac{\sqrt{3}\pi}{36}$ .

Question 41

The value of $x$ for which $\frac{dy}{dx} = 0$ , is

Question 42

$y(x)$ is strictly increasing in the interval

Question 43

The maximum height (in meters) achieved in the first jump is

Question 45

The value of $\frac{dy}{dx}$ when $x = \frac{1}{6}$ , is

Evaluate ∫013y dx\int_0^{\frac{1}{3}} y \, dx∫031​​ydx

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Evaluate $\int_0^{\frac{1}{3}} y \, dx$