The area bounded by the curve y = x^2 between x = 0 and x = in the first quadrant is : Option 1: 0 Option 2: (^3)/3 Option 3: 2 Option 4: (4^3)/3

Answer: (^3)/3. Solution: We need the area under y = x^2, above the x-axis, from x = 0 to x = . Whenever we need the area under a curve, we use definite integration — think of it as summing up infinitely thin vertical strips between the curve and the x-axis. Setting up the integral: Area = _0^() x^2 dx Since both x and y = x^2 are non-negative between 0 and, we are entirely in the first quadrant. Applying the power rule x^n dx = (x^(n+1))/(n+1) + C : _0^() x^2 dx = [x^3/3]_0^() Substituting the upper and lower limits: = (()^3)/3 - ((0)^3)/3 = (^3)/3 - 0 = (^3)/3 Don't confuse this with area under y = x or y = x which also commonly use 0 to as limits. Here the curve is y = x^2, a basic polynomial, so we simply apply the power rule — no trigonometry involved.

Applied Mathematics

Calculus

Easy

The area bounded by the curve $y = x^2$ between $x = 0$ and $x = \pi$ in the first quadrant is :

$\frac{\pi^3}{3}$

$\frac{4\pi^3}{3}$

✅ Correct Option: 2

We need the area under $y = x^2$ , above the $x$ -axis, from $x = 0$ to $x = \pi$ .

Whenever we need the area under a curve, we use definite integration — think of it as summing up infinitely thin vertical strips between the curve and the $x$ -axis.

Setting up the integral:

$\text{Area} = \int_0^{\pi} x^2 \, dx$

Since both $x$ and $y = x^2$ are non-negative between $0$ and $\pi$ , we are entirely in the first quadrant.

Applying the power rule $\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + C$ :

$\int_0^{\pi} x^2 \, dx = \left[\dfrac{x^3}{3}\right]_0^{\pi}$

Substituting the upper and lower limits:

$= \dfrac{(\pi)^3}{3} - \dfrac{(0)^3}{3}$

$= \dfrac{\pi^3}{3} - 0$

$= \boxed{\dfrac{\pi^3}{3}}$

Don't confuse this with area under $y = \sin x$ or $y = \cos x$ which also commonly use $0$ to $\pi$ as limits. Here the curve is $y = x^2$ , a basic polynomial, so we simply apply the power rule — no trigonometry involved.

2026: 30 May Shift 2

Let C = C(x) be the total cost of producing x units of a product and AC is the average cost and MC is the marginal cost, then

Match List-I with List-II

List-I	List-II
(A) AC decreases with x	(I) when MC = AC
(B) AC increases with x	(II) when MC < AC
(C) AC doesn't change with x	(III) when MC = 0
(D) C is constant.	(IV) when AC < MC

Choose the correct answer from the options given below:

2023: 11 Aug Shift 1

The order and degree Of the differential equation $\frac{d^2 y}{dx^2} + 2 e^{-x} = 0$ , respectively are

2026: 29 May Shift 2

The marginal cost (MC) of producing x units of an item in a factory is given by $MC=0.006x^2-0.001x+4$ .

If the fixed cost is ₹1000, then the cost of producing 200 units will be :

2026: 14 May Shift 2

The second order derivative of the function $y=\log(4x^3+6)$ with respect to 'x' will be:

2026: 14 May Shift 2

If $R(x)$ is the total revenue received and $C(x)$ is the total cost incurred in the production of x units of a commodity, then the profit function $P(x)$

2023: 23 May Shift 3

For a manufacturer, total cost function is given by $C = \frac{x^2}{25} + 2x$ . Which of the following are correct ?

A. 2.6 is the marginal cost when 5 units are produced.

B. $\frac{2x}{25} + 2$ is the marginal cost function.

C. $\frac{x}{25} + 2$ is the average cost function.

D. 2.4 is the marginal cost when 5 units are produced.

E. $\frac{x}{25} + 1$ is the average cost function.

Choose the correct answer from the options given below:

The area bounded by the curve y=x2y = x^2y=x2 between x=0x = 0x=0 and x=πx = \pix=π in the first quadrant is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The area bounded by the curve $y = x^2$ between $x = 0$ and $x = \pi$ in the first quadrant is :