Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 11 :

The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq | x + 2 |}$ is equal to [2025]

5
7
24/5
20/3

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(4)

We have, $x^{2} + 4 x + 2 \leq y \leq$ $| x + 2 |$

$\Rightarrow {(x + 2)}^{2} - 2 \leq y \leq$ $| x + 2 |$

$∴$ Required area $= \int_{- 4}^{- 2}$ $| (- x - {2) - (x + 2)}^{2} + 2 |$ $d x + \int_{- 2}^{0}$ $| (x + 2) - {(x + 2)}^{2} + 2 |$ $d x$

$= \int_{- 4}^{- 2}$ $| - x - {(x + 2)}^{2} |$ $d x + \int_{- 2}^{0}$ $| x - {(x + 2)}^{2} + 4 |$ $d x$

$= {[| \frac{- x^{2}}{2} - \frac{{(x + 2)}^{3}}{3} |]}_{- 4}^{- 2} + {[| \frac{x^{2}}{2} - \frac{{(x + 2)}^{3}}{3} + 4 x |]}_{- 2}^{0}$

$= \frac{10}{3} + \frac{10}{3} = \frac{20}{3}$ .

Q 12 :

The area of the region enclosed by the curves $y = e^{x}$ , $y = | e^{x} - 1 |$ and y-axis is: [2025]

$\log_{e} 2$
$1 - \log_{e} 2$
$2 \log_{e} 2 - 1$
$1 + \log_{e} 2$

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(2)

Given: Region is bounded by curves $y = e^{x}$ and $y = | e^{x} - 1 |$ .

The graph is given by

When $e^{x} = | e^{x} - 1 |$

$\Rightarrow e^{x} = - e^{x} + 1$ $[∵ e^{x} - 1 < 0]$

$\Rightarrow 2 e^{x} = 1 \Rightarrow e^{x} = 1 / 2 \Rightarrow x = - \ln (2)$

Now, area is given by.

$A = \int_{- \ln (2)}^{0} [e^{x} - (1 - e^{x})] d x = \int_{- \ln (2)}^{0} (2 e^{x} - 1) d x$

$= {(2 e^{x} - x)}_{- \ln (2)}^{0} = 2 - 1 - \ln (2) = 1 - \ln (2)$ .

Q 13 :

The area of the region ${(x, y) : 0 \leq y \leq 2 | x | + 1, 0 \leq y \leq x^{2} + 1, | x | \leq 3}$ is [2025]

$\frac{17}{3}$
$\frac{32}{3}$
$\frac{64}{3}$
$\frac{80}{3}$

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(3)

Given, the area of region

${(x, y) : 0 \leq y \leq 2 | x | + 1, 0 \leq y \leq x^{2} + 1, | x | \leq 3}$

$∴$ Required area $= 2 \int_{0}^{2} (x^{2} + 1) d x + \int_{2}^{3} (2 x + 1) d x$

$= 2 [{(\frac{x^{3}}{3} + x)}_{0}^{2} + {(\frac{2 x^{2}}{2} + x)}_{2}^{3}]$

$= 2 [\frac{8}{3} + 2] + 2 (6)$

$= 2 [\frac{14}{3}] + 12$

$= \frac{64}{3}$ sq. units.

Q 14 :

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

$2 (\frac{π}{2} - \frac{1}{3})$
$\frac{π}{4} - \frac{1}{3}$
$\frac{π}{2} - \frac{1}{3}$
$\frac{1}{2} (\frac{π}{2} - \frac{1}{3})$

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(3)

We have, $x (1 + y^{2}) = 1$ ... (i)

and $y^{2} = 2 x$ ... (ii)

From equation (i) and (ii), we get

$x (1 + 2 x) = 1 \Rightarrow 2 x^{2} + x - 1 = 0$

$\Rightarrow x = \frac{1}{2}, x = - 1$ (Reject)

From (ii), $y^{2} = 2 (\frac{1}{2}) \Rightarrow y = \pm 1$

$∴$ Required area $= \int_{- 1}^{1} (\frac{1}{1 + y^{2}} - \frac{y^{2}}{2}) d y$

$= (\tan^{- 1} y - \frac{y^{3}}{6}) |_{- 1}^{1} = \frac{π}{2} - \frac{1}{3}$ .

Q 15 :

Let the area of the region ${(x, y) : 2 y \leq x^{2} + 3, y + | x | \leq 3, y \geq | x - 1 |}$ be A. Then 6A is equal to : [2025]

14
16
18
12

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(1)

Required area, A = Rectangle ABDE – Area of region EDC

$\Rightarrow A = 4 - 2 \int_{0}^{1} ((3 - x) - (\frac{x^{2} + 3}{2})) d x$

$\Rightarrow A = 4 - 2 {3 x - \frac{x^{2}}{2} - \frac{x^{3}}{6} - \frac{3}{2} x}_{0}^{1}$

$\Rightarrow A = 4 - 2 {3 - \frac{1}{2} - \frac{1}{6} - \frac{3}{2}} = \frac{7}{3}$

So, 6A = 14.

Q 16 :

Let the area enclosed between the curves $| y | = 1 - x^{2}$ and $x^{2} + y^{2} = 1$ be $α$ . If $9 α = β π + γ; β, γ$ are integers, then the value of $| β - γ |$ equals [2025]

27
15
33
18

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(3)

We have, $c_{1} : | y | = 1 - x^{2}$ and $c_{2} : x^{2} + y^{2} = 1$

$∴$ Required area $= α = 4 [(\begin{matrix} Area of circle in \\ 1^{st} quadrant \end{matrix}) - \int_{0}^{1} (1 - x^{2}) d x]$

$= 4 [(\frac{π (1)}{4}) - {[x - \frac{x^{3}}{3}]}_{0}^{1}]$

$= 4 [\frac{π}{4} - (1 - \frac{1}{3})] = 4 [\frac{π}{4} - \frac{2}{3}]$

$= π - \frac{8}{3}$

Hence, $9 α = 9 π - 24$

On comparing with $9 α = β π + γ$ , we get $β = 9$ and $γ = - 24$

$∴ | β - γ | = | 9 - (- 24) | = 33$ .

Q 17 :

If the area of the region ${(x, y) : | 4 - x^{2} | \leq y \leq x^{2}, y \leq 4, x \geq 0}$ is $(\frac{80 \sqrt{2}}{α} - β), α, β \in N$ , then $α + β$ is equal to _________. [2025]

(22)

Required area $= \int_{\sqrt{2}}^{2} (x^{2} - (4 - x^{2})) d x + (2 \sqrt{2} - 2) \times 4 - \int_{2}^{2 \sqrt{2}} (x^{2} - 4) d x$

$= {[\frac{2 x^{3}}{3} - 4 x]}_{\sqrt{2}}^{2} + 8 \sqrt{2} - 8 - {[\frac{x^{3}}{3} - 4 x]}_{2}^{2 \sqrt{2}}$

$= \frac{40 \sqrt{2}}{3} - 16 = \frac{80 \sqrt{2}}{6} - 6$

$\Rightarrow α = 6, β = 16$

$\Rightarrow α + β = 22$ .

Q 18 :

The area of the region bounded by the curve y = max ${| x |, x | x - 2 |}$ , the x-axis and the lines x = –2 and x = 4 is equal to _________. [2025]

(12)

Required Area = Area of $△$ OAB + Area of region OCDEO

$= \frac{1}{2} \times 2 \times 2 + \int_{0}^{1} (2 x - x^{2}) d x + \int_{1}^{3} x d x + \int_{3}^{4} (x^{2} - 2 x) d x$

$= 2 + \frac{2}{3} + 4 + \frac{16}{3}$

= 12 sq. units.

Q 19 :

If the area of the region ${(x, y) : | x - 5 | \leq y \leq 4 \sqrt{x}}$ is A, then 3A is equal to __________. [2025]

(368)

Area $A = \int_{1}^{25} 4 \sqrt{x} d x - \frac{1}{2} \times 4 \times 4 - \frac{1}{2} \times 20 \times 20$

$= {[4 x^{3 / 2} \times \frac{2}{3}]}_{1}^{25} - 208$

$= \frac{8}{3} (125 - 1) - 208$

$= \frac{368}{3}$ sq. units

$\Rightarrow 3 A = 368$ .

Q 20 :

Let the area of the bounded region ${(x, y) : 0 \leq 9 x \leq y^{2}, y \geq 3 x - 6}$ be A. Then 6A is equal to __________. [2025]

(15)

We have, $0 \leq 9 x \leq y^{2}, y \geq 3 x - 6$

Required area = $A =$ $| \int_{0}^{- 3} \frac{y^{2}}{9} d y + \int_{- 3}^{- 6} (\frac{y + 6}{3}) d y |$

$=$ $| \frac{1}{9} {[\frac{y^{3}}{3}]}_{0}^{- 3} + \frac{1}{3} {(\frac{y^{2}}{2} + 6 y)}_{- 3}^{- 6} |$

$=$ $| \frac{1}{9} [- 9 - 0] + \frac{1}{3} [18 - 36 - \frac{9}{2} + 18] |$

$=$ $| - 1 + \frac{1}{3} [\frac{- 9}{2}] |$

$=$ $| - 1 - \frac{3}{2} |$ $=$ $| \frac{- 5}{2} |$ $= \frac{5}{2} sq . units$

$∴ 6 A = 6 \times \frac{5}{2} = 15$

Q 21 :

Let the function, $f (x) = {\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ a^{2} + b x, & x \geq 1 \end{matrix}$ be differentiable for all $x \in R$ , where a > 1, $b \in R$ . If the area of the region enclosed by y = f(x) and the line y = –20 is $α + β \sqrt{3}, α, β \in Z$ then the value of $α + β$ is __________. [2025]

(34)

Given, f(x) is continuous and differentiable at x = 1.

$f (x) = {\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ a^{2} + b x, & x \geq 1 \end{matrix}$

$\Rightarrow f' (x) = {\begin{matrix} - 6 a x, & x < 1 \\ b, & x \geq 1 \end{matrix}$

L.H.L. at $x = 1 \Rightarrow - 3 a - 2$

R.H.L. at $x = 1 \Rightarrow a^{2} + b$

$∴$ L.H.L. = R.H.L. { $∵$ f(x) is continuous}

$\Rightarrow - 3 a - 2 = a^{2} + b$ ... (i)

L.H.D. at $x = 1 \Rightarrow - 6 a$

R.H.D. at $x = 1 \Rightarrow b$

$∴$ L.H.D. = R.H.D. [ $∵$ f(x) is differentiable]

$\Rightarrow - 6 a = b$ ... (ii)

From (i) and (ii), we get

$- 3 a - 2 = a^{2} - 6 a$

$\Rightarrow a^{2} - 3 a + 2 = 0 \Rightarrow (a - 1) (a - 2) = 0$

$\Rightarrow a = 1 o r a = 2 \Rightarrow a = 2$ ( $∵$ a > 1)

From (ii), b = –12

Now, $f (x) = {\begin{matrix} - 6 x^{2} - 2, & x < 1 \\ 4 - 12 x, & x \geq 1 \end{matrix}$

$∴$ Area of region $= \int_{- \sqrt{3}}^{1} (- 6 x^{2} - 2 + 20) d x + \int_{1}^{2} (4 - 12 x + 20) d x$

$= {(\frac{- 6 x^{3}}{3} - 2 x + 20 x)}_{- \sqrt{3}}^{1} + {(4 x - \frac{12 x^{2}}{2} + 20 x)}_{1}^{2}$

$\Rightarrow 16 + 12 \sqrt{3} + 6 = 22 + 12 \sqrt{3}$

$\Rightarrow 22 + 12 \sqrt{3} = α + β \sqrt{3}$

So, $α = 22 and β = 12$

$∴ α + β = 22 + 12 = 34$

Q 22 :

If the area of the larger portion bounded between the curves $x^{2} + y^{2} = 25$ and $y = | x - 1 |$ is $\frac{1}{4} (b π + c), b, c \in N$ , then b + c is equal to __________. [2025]

(77)

Given, $x^{2} + y^{2} = 25$

$\Rightarrow x^{2} + {(x - 1)}^{2} = 25$ [ $∵$ y = |x –1|]

$\Rightarrow x = 4, - 3$

$∴$ Required area $= 25 π - (\int_{- 3}^{4} (\sqrt{25 - x^{2}} - | x - 1 |) d x)$

$= 25 π - {[\frac{1}{2} x \sqrt{25 - x^{2}} + \frac{25}{2} \sin^{- 1} \frac{x}{5}]}_{- 3}^{4} + \int_{- 3}^{1} (1 - x) d x + \int_{1}^{4} (x - 1) d x$

$= 25 π + \frac{25}{2} - (2 \times 3 + \frac{25}{2} \sin^{- 1} (\frac{4}{5}) + \frac{3}{2} \times 4 + \frac{25}{2} \sin^{- 1} (\frac{3}{5}))$

$= 25 π + \frac{25}{2} - 12 - \frac{25 π}{4}$

$= \frac{75 π}{4} + \frac{1}{2} = \frac{1}{4} (75 π + 2)$

$∴ \frac{1}{4} (b π + c) = \frac{1}{4} (75 π + 2)$

$\Rightarrow b = 75, c = 2$

$\Rightarrow b + c = 77$ .

Q 23 :

The area bounded by the curves $y =$ $| x - 1 |$ $+$ $| x - 2 |$ and $y = 3$ is equal to [2023]

6
3
5
4

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(4)

$Required area is shaded in the given figure, which is a trapezium.$

$∴$ $Area = \frac{1}{2} \times Sum of parallel sides \times Height$

$= \frac{1}{2} \times (A B + C D) \times A M = \frac{1}{2} (1 + 3) \times 2 = 4 sq. units$

Q 24 :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is [2023]

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21

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(2)

$We have,$

$x^{2} \leq y$

$8 - x^{2} \geq y$

$y \leq 7$

$Converting the given inequations into equations, we get x^{2} = y and y = 8 - x^{2}$

$Solving these equations to find their point of intersection$

i.e., $x^{2} = 8 - x^{2}$

$\Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

$∴$ $y = 4$

$Required area = \int_{- 2}^{2} (8 - x^{2} - x^{2}) d x - \int_{- 1}^{1} (8 - x^{2} - 7) d x$

$= \int_{- 2}^{2} (8 - 2 x^{2}) d x - \int_{- 1}^{1} (1 - x^{2}) d x$ $= {[8 x - \frac{2}{3} x^{3}]}_{- 2}^{2} - {[x - \frac{x^{3}}{3}]}_{- 1}^{1}$

$= [(16 - \frac{16}{3}) - (- 16 + \frac{16}{3})] - [\frac{2}{3} + \frac{2}{3}]$

$= 16 - \frac{16}{3} + 16 - \frac{16}{3} - \frac{4}{3}$ $= 20 sq. units$

Q 25 :

Area of the region ${(x, y) : x^{2} + {(y - 2)}^{2} \leq 4, x^{2} \geq 2 y}$ is [2023]

$2 π - \frac{16}{3}$
$π - \frac{8}{3}$
$π + \frac{8}{3}$
$2 π + \frac{16}{3}$

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(1)

$Given that x^{2} + {(y - 2)}^{2} \leq 2^{2} and x^{2} \geq 2 y$

$Convert the given inequalities into equations:$

$x^{2} + {(y - 2)}^{2} = 4 and x^{2} = 2 y$

$Solving circle and parabola simultaneously:$

$2 y + y^{2} - 4 y + 4 = 4,$ $y^{2} - 2 y = 0,$ $y = 0, 2$

$Put y = 2 in x^{2} = 2 y \Rightarrow x = \pm 2 \Rightarrow (2, 2) and (- 2, 2)$

$A_{1} = 2 \times 2 - \frac{1}{4} \cdot π \cdot 2^{2} = 4 - π$

$Required area = 2 [\int_{0}^{2} \frac{x^{2}}{2} d x - (4 - π)]$

$= 2 [\frac{x^{3}}{6} |_{0}^{2} - 4 + π] = 2 [\frac{4}{3} + π - 4] = 2 [π - \frac{8}{3}] = 2 π - \frac{16}{3} sq. units$

Q 26 :

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is [2023]

$\frac{27}{4}$
$\frac{31}{4}$
$\frac{23}{4}$
$\frac{19}{4}$

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(1)

$Given, y = x^{3}$

$Equation of tangent to the curve y = x^{3} at point (- 1, - 1) is$

$y + 1 = 3 (x + 1) \Rightarrow y = 3 x + 2$

$Point of intersection with the curve is (2, 8)$

$So, area = \int_{- 1}^{2} [(3 x + 2) - x^{3}] d x$

$= {[3 \times \frac{x^{2}}{2} + 2 x - \frac{x^{4}}{4}]}_{- 1}^{2}$

$= \frac{27}{4} sq. units$

Q 27 :

The area of the region enclosed by the curve $f (x) = \max {\sin x, \cos x}$ , $- π \leq x \leq π$ and the $x$ -axis is [2023]

$4 (\sqrt{2})$
$2 (\sqrt{2} + 1)$
$2 \sqrt{2} (\sqrt{2} + 1)$
$4$

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(4)

We have $f (x) = \max {\sin x, \cos x}, - π \leq x \leq π$

Now, the area of the region enclosed by the given curve and the $x$ –axis is

$=$ $| \int_{- π}^{- 3 π / 4} \sin x d x |$ $+$ $| \int_{- 3 π / 4}^{- π / 2} \cos x d x |$ $+$ $| \int_{- π / 2}^{π / 4} \cos x d x |$ $+$ $| \int_{π / 4}^{π} \sin x d x |$

$= \frac{- 1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 + 1 + \frac{1}{\sqrt{2}} = 4$

Q 28 :

The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

$\frac{4}{3} (4 \sqrt{2} + 1)$
$\frac{3}{4} (4 \sqrt{2} - 1)$
$\frac{3}{4} (4 \sqrt{2} + 1)$
$\frac{4}{3} (4 \sqrt{2} - 1)$

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(4)

$Required area = 2 (\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y)$

$= 2 ({[\frac{y^{3 / 2}}{3 / 2}]}_{1}^{2} - {[\frac{{(4 - y)}^{3 / 2}}{3 / 2}]}_{2}^{4})$

$= \frac{4}{3} [(2 \sqrt{2} - 1) - (0 - 2 \sqrt{2})] = \frac{4}{3} (4 \sqrt{2} - 1)$

Q 29 :

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is [2023]

$16 \log_{e} 2 - \frac{14}{3}$
$8 \log_{e} 2 - \frac{13}{3}$
$8 \log_{e} 2 + \frac{7}{6}$
$16 \log_{e} 2 + \frac{7}{3}$

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(1)

$Area = \int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x$

$= {(\frac{x^{3}}{3} - x)}_{1}^{2} + 8 {(\log_{e} x)}_{2}^{8} - {(x)}_{2}^{8}$

$= \frac{4}{3} + 8 (2 \log_{e} 2) - 6 = 16 \log_{e} 2 - \frac{14}{3}$

Q 30 :

The area enclosed by the curves $y^{2} + 4 x = 4 and y - 2 x = 2$ is [2023]

$\frac{23}{3}$
$\frac{22}{3}$
$9$
$\frac{25}{3}$

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(3)

Given curves are $y^{2} + 4 x = 4$ and $y - 2 x = 2$

The points of intersection of the given curves are (0, 2) and (- 3, - 4)

$Required area = \int_{- 4}^{2} [(\frac{4 - y^{2}}{4}) - (\frac{y - 2}{2})] d y$

$= \int_{- 4}^{2} \frac{- y^{2} - 2 y + 8}{4} d y = \frac{1}{4} {[- \frac{y^{3}}{3} - y^{2} + 8 y]}_{- 4}^{2}$

$= \frac{1}{4} [(- \frac{8}{3} - 4 + 16) - (\frac{64}{3} - 16 - 32)]$

$= \frac{1}{4} [(\frac{28}{3}) - (- \frac{80}{3})] = \frac{1}{4} (\frac{108}{3}) = \frac{108}{12} = 9 sq. units .$

Q 31 :

Let $A = {(x, y) \in ℝ^{2} : y \geq 0, 2 x \leq y \leq \sqrt{4 - {(x - 1)}^{2}}}$

and $B = {(x, y) \in ℝ \times ℝ : 0 \leq y \leq \min {2 x, \sqrt{4 - {(x - 1)}^{2}}}} .$

Then the ratio of the area of $A$ to the area of $B$ is [2023]

$\frac{π}{π + 1}$
$\frac{π - 1}{π + 1}$
$\frac{π + 1}{π - 1}$
$\frac{π}{π - 1}$

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(2)

Let $A = {(x, y) \in ℝ^{2} : y \geq 0, 2 x \leq y \leq \sqrt{4 - {(x - 1)}^{2}}}$

and $B = {(x, y) \in ℝ \times ℝ : 0 \leq y \leq \min {2 x, \sqrt{4 - {(x - 1)}^{2}}}}$

We have the following diagram $y^{2} + {(x - 1)}^{2} = 4$

$Shaded portion A$

$Area A = area of quadrant of circle - area of (O A B)$

$= \frac{π (4)}{4} - \frac{1}{2} (2) (1) = (π - 1)$ ...(i)

Now, for finding the area of portion B, we have

$Area B = area of (∆ A O B) + area of quadrant of circle$

$= \frac{1}{2} \cdot (1) (2) + π \frac{{(2)}^{2}}{4} = (π + 1)$ ...(ii)

Thus, according to the question, ratio of area A to area

$B = \frac{π - 1}{π + 1}$

Q 32 :

$Let Δ be the area of the region {(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ $Then \frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) is equal to$ [2023]

$\sqrt{3} - \frac{2}{3}$
$2 \sqrt{3} - \frac{2}{3}$
$2 \sqrt{3} - \frac{1}{3}$
$\sqrt{3} - \frac{4}{3}$

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(4)

If $Δ$ be the area of region ${(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ Then we have,

$Area = 2 \int_{1}^{3} 2 \sqrt{x} d x + 2 \int_{3}^{\sqrt{21}} \sqrt{(21 - x^{2})} d x$

$Δ = \frac{8}{3} \cdot (3 \sqrt{3} - 1) + 21 \sin^{- 1} (\frac{2}{\sqrt{7}}) - 6 \sqrt{3}$

$\Rightarrow [Δ - 21 \sin^{- 1} (\frac{2}{\sqrt{7}})] = \frac{2 \sqrt{3} - \frac{8}{3}}{2}$

$\Rightarrow \frac{1}{2} [Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}] = \sqrt{3} - \frac{4}{3}$

Q 33 :

The area of the region $A = {(x, y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{π}{2}}$ is [2023]

$\sqrt{5} - 2 \sqrt{2} + 1$
$\sqrt{5} + 2 \sqrt{2} - 4.5$
$1 - \frac{3}{\sqrt{2}} + \frac{4}{\sqrt{5}}$
$\frac{3}{\sqrt{5}} - \frac{3}{\sqrt{2}} + 1$

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(1)

The area of the region, $A = {(x, y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{π}{2}}$

$\Rightarrow$ $| \cos x - \sin x |$ $\leq y \leq \sin x$

For finding the intersecting point we must have

$\cos x - \sin x = \sin x$

$\Rightarrow \tan x = \frac{1}{2}$

Let $ϕ = \tan^{- 1} \frac{1}{2}$

$So, \tan ϕ = \frac{1}{2}, \sin ϕ = \frac{1}{\sqrt{5}}, \cos ϕ = \frac{2}{\sqrt{5}}$

$Area = \int_{ϕ}^{π / 2} (\sin x - | \cos x - \sin x |) d x$

$= \int_{ϕ}^{π / 4} (\sin x - (\cos x - \sin x)) d x + \int_{π / 4}^{π / 2} (\sin x - (\sin x - \cos x)) d x$

$= \int_{ϕ}^{π / 4} (2 \sin x - \cos x) d x + \int_{π / 4}^{π / 2} \cos x d x$

$= {[- 2 \cos x - \sin x]}_{ϕ}^{π / 4} + {[\sin x]}_{π / 4}^{π / 2}$

$= \sqrt{5} - 2 \sqrt{2} + 1$

Q 34 :

$Let q be the maximum integral value of p in [0,10] for which the roots of the equation x^{2} - p x + \frac{5}{4} p = 0 are rational.$ Then the area of the region
${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is [2023]

$164$
$243$
$\frac{125}{3}$
$25$

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(2)

Given equation is $x^{2} - p x + \frac{5}{4} p = 0$

$D = p^{2} - 5 p,$ which is a perfect square when $p = 9$ .

$∴$ $q = 9$

$Required area = \int_{0}^{9} {(x - 9)}^{2} d x$

$= \int_{0}^{9} (x^{2} - 18 x + 81) d x$ $= {[\frac{x^{3}}{3} - 18 \times \frac{x^{2}}{2} + 81 x]}_{0}^{9}$

$= 243 sq. units.$

Q 35 :

If the area of region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ is equal to $\frac{n + 2}{n + 1} - \frac{π}{n - 1},$ then the natural number $n$ is equal to _______ . [2023]

(5)

Given, $x^{2} + y^{2} - 2 y \geq 0; x^{2} - 2 y \leq 0; x \geq y$

$Area of segment O A B = \frac{90 °}{360 °} \times π {(1)}^{2} - \frac{1}{2} \times 1 \times 1$

$= (\frac{π}{4} - \frac{1}{2})$

Hence, required area $= \int_{0}^{2} (x - \frac{x^{2}}{2}) d x - (\frac{π}{4} - \frac{1}{2})$

$= {(\frac{x^{2}}{2} - \frac{x^{3}}{6})}_{0}^{2} - (\frac{π - 2}{4}) = 2 - \frac{4}{3} - \frac{π}{4} + \frac{1}{2} = \frac{7}{6} - \frac{π}{4}$

Now, $\frac{n + 2}{n + 1} - \frac{π}{n - 1} = \frac{7}{6} - \frac{π}{4}$

For $n = 5$ , $\frac{n + 2}{n + 1} - \frac{π}{n - 1} = \frac{7}{6} - \frac{π}{4}$ $∴$ $n = 5$

Q 36 :

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f (x) = \min {x^{2} + \frac{3}{4}, 1 + [x]},$ where $[x]$ denotes the greatest integer $\leq x$ , be $A$ . Then the value of $12 A$ is____________ . [2023]

(17)

Q 37 :

Let $y = p (x)$ be the parabola passing through the points $(- 1, 0), (0, 1)$ and $(1, 0)$ . If the area of the region

${(x, y) : {(x + 1)}^{2} + {(y - 1)}^{2} \leq 1, y \leq p (x)}$ is $A$ , then $12 (π - 4 A)$ is equal to ______ . [2023]

(16)

There can be infinite parabola through given points.

In question, it must be given that axis of parabola is parallel to $y$ -axis.

Equation of parabola passing through (-1, 0), (0, 1) and (1, 0) is $x^{2} = - (y - 1)$ ...(i)

$∴$ $Required area, A = \int_{- 1}^{0} [(1 - x^{2}) - (1 - \sqrt{1 - {(x + 1)}^{2}})] d x$

$= \int_{- 1}^{0} (- x^{2} + \sqrt{1 - {(x + 1)}^{2}}) d x$

$= {[\frac{- x^{3}}{3} + \frac{x + 1}{2} \sqrt{1 - {(x + 1)}^{2}} + \frac{1}{2} \sin^{- 1} (\frac{x + 1}{1})]}_{- 1}^{0}$

$= \frac{1}{2} (\frac{π}{2}) - \frac{1}{3} = \frac{π}{4} - \frac{1}{3}$

$∴$ $12 (π - 4 A) = 12 (π - 4 (\frac{π}{4} - \frac{1}{3})) = 12 (\frac{4}{3}) = 16$

Q 38 :

If the area of the region ${(x, y) : | x^{2} - 2 | \leq y \leq x}$ is A, then $6 A + 16 \sqrt{2}$ is equal to _______ . [2023]

(27)

$We have,$

$Region = {(x, y) : | x^{2} - 2 | \leq y \leq x}$

The graph of the region is as shown in figure.

$Required area, A = \int_{1}^{\sqrt{2}} [x - {- (x^{2} - 2)}] d x + \int_{\sqrt{2}}^{2} {x - (x^{2} - 2)} d x$

$= \int_{1}^{\sqrt{2}} (x^{2} + x - 2) d x + \int_{\sqrt{2}}^{2} (- x^{2} + x + 2) d x$

$= {(\frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x)}_{1}^{\sqrt{2}} + {(\frac{- x^{3}}{3} + \frac{x^{2}}{2} + 2 x)}_{\sqrt{2}}^{2}$

$= (\frac{2 \sqrt{2}}{3} + 1 - 2 \sqrt{2}) - (\frac{1}{3} + \frac{1}{2} - 2)$ $+ (\frac{- 8}{3} + 2 + 4) - (\frac{- 2 \sqrt{2}}{3} + 1 + 2 \sqrt{2})$

$= \frac{27 - 16 \sqrt{2}}{6}$ $∴$ $6 A + 16 \sqrt{2} = 27 - 16 \sqrt{2} + 16 \sqrt{2} = 27$

Q 39 :

If A is the area in the first quadrant enclosed by the curve $C : 2 x^{2} - y + 1 = 0$ , the tangent to C at the point (1, 3) and the line $x + y = 1$ , then the value of 60A is ______ . [2023]

(16)

$Given, C : 2 x^{2} - y + 1 = 0$

$\Rightarrow 4 x - \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = 4 x \Rightarrow {[\frac{d y}{d x}]}_{(1, 3)} = 4$

$Equation of tangent to the curve C is$

$y - 3 = 4 (x - 1)$

$\Rightarrow y - 4 x + 1 = 0 ...(i)$

$Also, given line is x + y = 1 ...(ii)$

$Solving (i) and (ii), we get x = 0.4, y = 0.6$

$Given curve and equation (ii) intersect each other at (0, 1) and (- 0.5, 1.5) .$

$Required area,$

$A = \int_{0}^{1} (2 x^{2} + 1) d x - area of ∆ E O A - area of ∆ B E D + area of ∆ E D C$

$= {[\frac{2 x^{3}}{3} + x]}_{0}^{1} - \frac{1}{2} \times 1 \times 1 - \frac{1}{2} (1 - \frac{1}{4}) \times 3 + \frac{1}{2} \times (1 - \frac{1}{4}) \times 0.6$

$= (\frac{2}{3} + 1) - \frac{1}{2} - \frac{9}{8} + \frac{9}{40} = \frac{16}{60} sq. units$

$∴$ $60 A = \frac{16}{60} \times 60 = 16$

Q 40 :

If the area bounded by the curve $2 y^{2} = 3 x,$ lines $x + y = 3, y = 0$ and outside the circle ${(x - 3)}^{2} + y^{2} = 2$ is A, then $4 (π + 4 A)$ is equal to _____ . [2023]

(42)

$Given, y^{2} = \frac{3 x}{2}, x + y = 3, y = 0, and {(x - 3)}^{2} + y^{2} = 2$

$\Rightarrow 2 y^{2} = 3 (3 - y)$

$\Rightarrow 2 y^{2} + 3 y - 9 = 0 \Rightarrow 2 y^{2} - 3 y + 6 y - 9 = 0$

$\Rightarrow (2 y - 3) (y + 3) = 0 \Rightarrow y = \frac{3}{2}, - 2$

$Required area, A = \int_{0}^{\frac{3}{2}} ((3 - y) - \frac{2 y^{2}}{3}) d y - \frac{π}{8} (2)$

$= {(3 y - \frac{y^{2}}{2} - \frac{2 y^{3}}{9})}_{0}^{3 / 2} - \frac{π}{4}$ $= 3 \times \frac{3}{2} - \frac{9}{8} - \frac{2}{9} (\frac{27}{8}) - \frac{π}{4}$

$= \frac{9}{2} - \frac{9}{8} - \frac{3}{4} - \frac{π}{4}$

$∴$ $4 A + π = 4 [\frac{9}{2} - \frac{9}{8} - \frac{3}{4}] = \frac{21}{2} = 10.50$

$∴$ $4 (4 A + π) = 42$

Topic Question Set

Q 11 :

The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq | x + 2 |}$ is equal to [2025]

Q 12 :

The area of the region enclosed by the curves $y = e^{x}$ , $y = | e^{x} - 1 |$ and y-axis is: [2025]

Q 13 :

The area of the region ${(x, y) : 0 \leq y \leq 2 | x | + 1, 0 \leq y \leq x^{2} + 1, | x | \leq 3}$ is [2025]

Q 14 :

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

Q 15 :

Let the area of the region ${(x, y) : 2 y \leq x^{2} + 3, y + | x | \leq 3, y \geq | x - 1 |}$ be A. Then 6A is equal to : [2025]

Q 16 :

Let the area enclosed between the curves $| y | = 1 - x^{2}$ and $x^{2} + y^{2} = 1$ be $α$ . If $9 α = β π + γ; β, γ$ are integers, then the value of $| β - γ |$ equals [2025]

Q 17 :

If the area of the region ${(x, y) : | 4 - x^{2} | \leq y \leq x^{2}, y \leq 4, x \geq 0}$ is $(\frac{80 \sqrt{2}}{α} - β), α, β \in N$ , then $α + β$ is equal to _________. [2025]

Q 18 :

The area of the region bounded by the curve y = max ${| x |, x | x - 2 |}$ , the x-axis and the lines x = –2 and x = 4 is equal to _________. [2025]

Q 19 :

If the area of the region ${(x, y) : | x - 5 | \leq y \leq 4 \sqrt{x}}$ is A, then 3A is equal to __________. [2025]

Q 20 :

Let the area of the bounded region ${(x, y) : 0 \leq 9 x \leq y^{2}, y \geq 3 x - 6}$ be A. Then 6A is equal to __________. [2025]

Q 22 :

If the area of the larger portion bounded between the curves $x^{2} + y^{2} = 25$ and $y = | x - 1 |$ is $\frac{1}{4} (b π + c), b, c \in N$ , then b + c is equal to __________. [2025]

Q 23 :

The area bounded by the curves $y =$ $| x - 1 |$ $+$ $| x - 2 |$ and $y = 3$ is equal to [2023]

Q 24 :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is [2023]

Q 25 :

Area of the region ${(x, y) : x^{2} + {(y - 2)}^{2} \leq 4, x^{2} \geq 2 y}$ is [2023]

Q 26 :

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is [2023]

Q 27 :

The area of the region enclosed by the curve $f (x) = \max {\sin x, \cos x}$ , $- π \leq x \leq π$ and the $x$ -axis is [2023]

Q 28 :

The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

Q 29 :

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is [2023]

Q 30 :

The area enclosed by the curves $y^{2} + 4 x = 4 and y - 2 x = 2$ is [2023]

Q 31 :

Let $A = {(x, y) \in ℝ^{2} : y \geq 0, 2 x \leq y \leq \sqrt{4 - {(x - 1)}^{2}}}$

and $B = {(x, y) \in ℝ \times ℝ : 0 \leq y \leq \min {2 x, \sqrt{4 - {(x - 1)}^{2}}}} .$

Then the ratio of the area of $A$ to the area of $B$ is [2023]

Q 32 :

$Let Δ be the area of the region {(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ $Then \frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) is equal to$ [2023]

Q 33 :

The area of the region $A = {(x, y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{π}{2}}$ is [2023]

Q 34 :

$Let q be the maximum integral value of p in [0,10] for which the roots of the equation x^{2} - p x + \frac{5}{4} p = 0 are rational.$ Then the area of the region
${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is [2023]

Q 35 :

If the area of region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ is equal to $\frac{n + 2}{n + 1} - \frac{π}{n - 1},$ then the natural number $n$ is equal to _______ . [2023]

Q 36 :

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f (x) = \min {x^{2} + \frac{3}{4}, 1 + [x]},$ where $[x]$ denotes the greatest integer $\leq x$ , be $A$ . Then the value of $12 A$ is____________ . [2023]

(17)

Q 37 :

Let $y = p (x)$ be the parabola passing through the points $(- 1, 0), (0, 1)$ and $(1, 0)$ . If the area of the region

${(x, y) : {(x + 1)}^{2} + {(y - 1)}^{2} \leq 1, y \leq p (x)}$ is $A$ , then $12 (π - 4 A)$ is equal to ______ . [2023]

Q 38 :

If the area of the region ${(x, y) : | x^{2} - 2 | \leq y \leq x}$ is A, then $6 A + 16 \sqrt{2}$ is equal to _______ . [2023]

Q 39 :

If A is the area in the first quadrant enclosed by the curve $C : 2 x^{2} - y + 1 = 0$ , the tangent to C at the point (1, 3) and the line $x + y = 1$ , then the value of 60A is ______ . [2023]

Q 40 :

If the area bounded by the curve $2 y^{2} = 3 x,$ lines $x + y = 3, y = 0$ and outside the circle ${(x - 3)}^{2} + y^{2} = 2$ is A, then $4 (π + 4 A)$ is equal to _____ . [2023]

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Topic Question Set

Q 11 : The area of the region {(x,y):x2+4x+2≤y≤|x+2|} is equal to [2025]

Q 12 : The area of the region enclosed by the curves y=ex, y=|ex–1| and y-axis is: [2025]

Q 13 : The area of the region {(x,y):0≤y≤2|x|+1, 0≤y≤x2+1, |x|≤3} is [2025]

Q 14 : The area of the region bounded by the curves x(1+y2)=1 and y2=2x is : [2025]

Q 15 : Let the area of the region {(x,y):2y≤x2+3, y+|x|≤3, y≥|x–1|} be A. Then 6A is equal to : [2025]

Q 16 : Let the area enclosed between the curves |y|=1–x2 and x2+y2=1 be α. If 9α=βπ+γ; β, γ are integers, then the value of |β–γ| equals [2025]

Q 17 : If the area of the region {(x,y):|4–x2|≤y≤x2, y≤4, x≥0} is (802α–β), α, β∈N, then α+β is equal to _________. [2025]

Q 18 : The area of the region bounded by the curve y = max {|x|,x|x–2|}, the x-axis and the lines x = –2 and x = 4 is equal to _________. [2025]

Q 19 : If the area of the region {(x,y):|x–5|≤y≤4x} is A, then 3A is equal to __________. [2025]

Q 20 : Let the area of the bounded region {(x,y):0≤9x≤y2, y≥3x–6} be A. Then 6A is equal to __________. [2025]

Q 21 : Let the function, f(x)={–3ax2–2,x<1a2+bx,x≥1 be differentiable for all x∈R, where a > 1, b∈R. If the area of the region enclosed by y = f(x) and the line y = –20 is α+β3, α, β∈Z then the value of α+β is __________. [2025]

Q 22 : If the area of the larger portion bounded between the curves x2+y2=25 and y=|x–1| is 14(bπ+c), b, c∈N, then b + c is equal to __________. [2025]

Q 23 : The area bounded by the curves y=|x-1|+|x-2| and y=3 is equal to [2023]

Q 24 : The area of the region {(x,y):x2≤y≤8-x2, y≤7} is [2023]

Q 25 : Area of the region {(x,y):x2+(y-2)2≤4, x2≥2y} is [2023]

Q 26 : The area of the region enclosed by the curve y=x3 and its tangent at the point (-1,-1) is [2023]

Q 27 : The area of the region enclosed by the curve f(x)=max{sinx,cosx}, -π≤x≤π and the x-axis is [2023]

Q 28 : The area of the region {(x,y):x2≤y≤|x2-4|,y≥1} is [2023]

Q 29 : The area of the region given by {(x,y):xy≤8, 1≤y≤x2} is [2023]

Q 30 : The area enclosed by the curves y2+4x=4 and y-2x=2 is [2023]

Q 31 : Let A={(x,y)∈ℝ2:y≥0, 2x≤y≤ 4-(x-1)2 } and B={(x,y)∈ℝ×ℝ:0≤y≤min{2x, 4-(x-1)2 }}. Then the ratio of the area of A to the area of B is [2023]

Q 32 : Let Δ be the area of the region {(x,y)∈ℝ2:x2+y2≤21, y2≤4x, x≥1}. Then 12(Δ-21sin-127) is equal to [2023]

Q 33 : The area of the region A={(x,y):|cosx-sinx|≤y≤sinx, 0≤x≤π2} is [2023]

Q 34 : Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2-px+54p=0 are rational. Then the area of the region {(x,y):0≤y≤(x-q)2, 0≤x≤q} is [2023]

Q 35 : If the area of region S={(x,y): 2y-y2≤x2≤2y, x≥y} is equal to n+2n+1-πn-1, then the natural number n is equal to _______ . [2023]

Q 36 : Let the area enclosed by the lines x+y=2, y=0, x=0 and the curve f(x)=min{x2+34, 1+[x]}, where [x] denotes the greatest integer ≤x, be A. Then the value of 12A is____________ . [2023]

(17)

Q 37 : Let y=p(x) be the parabola passing through the points (-1,0),(0,1) and (1,0). If the area of the region {(x,y):(x+1)2+(y-1)2≤1, y≤p(x)} is A, then 12(π-4A) is equal to ______ . [2023]

Q 38 : If the area of the region {(x,y):|x2-2|≤y≤x} is A, then 6A+162 is equal to _______ . [2023]

Q 39 : If A is the area in the first quadrant enclosed by the curve C: 2x2-y+1=0, the tangent to C at the point (1, 3) and the line x+y=1, then the value of 60A is ______ . [2023]

Q 40 : If the area bounded by the curve 2y2=3x, lines x+y=3, y=0 and outside the circle (x-3)2+y2=2 is A, then 4(π+4A) is equal to _____ . [2023]

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Q 11 :

The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq | x + 2 |}$ is equal to [2025]

Q 12 :

The area of the region enclosed by the curves $y = e^{x}$ , $y = | e^{x} - 1 |$ and y-axis is: [2025]

Q 13 :

The area of the region ${(x, y) : 0 \leq y \leq 2 | x | + 1, 0 \leq y \leq x^{2} + 1, | x | \leq 3}$ is [2025]

Q 14 :

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

Q 15 :

Let the area of the region ${(x, y) : 2 y \leq x^{2} + 3, y + | x | \leq 3, y \geq | x - 1 |}$ be A. Then 6A is equal to : [2025]

Q 16 :

Let the area enclosed between the curves $| y | = 1 - x^{2}$ and $x^{2} + y^{2} = 1$ be $α$ . If $9 α = β π + γ; β, γ$ are integers, then the value of $| β - γ |$ equals [2025]

Q 17 :

If the area of the region ${(x, y) : | 4 - x^{2} | \leq y \leq x^{2}, y \leq 4, x \geq 0}$ is $(\frac{80 \sqrt{2}}{α} - β), α, β \in N$ , then $α + β$ is equal to _________. [2025]

Q 18 :

The area of the region bounded by the curve y = max ${| x |, x | x - 2 |}$ , the x-axis and the lines x = –2 and x = 4 is equal to _________. [2025]

Q 19 :

If the area of the region ${(x, y) : | x - 5 | \leq y \leq 4 \sqrt{x}}$ is A, then 3A is equal to __________. [2025]

Q 20 :

Let the area of the bounded region ${(x, y) : 0 \leq 9 x \leq y^{2}, y \geq 3 x - 6}$ be A. Then 6A is equal to __________. [2025]

Q 22 :

If the area of the larger portion bounded between the curves $x^{2} + y^{2} = 25$ and $y = | x - 1 |$ is $\frac{1}{4} (b π + c), b, c \in N$ , then b + c is equal to __________. [2025]

Q 23 :

The area bounded by the curves $y =$ $| x - 1 |$ $+$ $| x - 2 |$ and $y = 3$ is equal to [2023]

Q 24 :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is [2023]

Q 25 :

Area of the region ${(x, y) : x^{2} + {(y - 2)}^{2} \leq 4, x^{2} \geq 2 y}$ is [2023]

Q 26 :

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is [2023]

Q 27 :

The area of the region enclosed by the curve $f (x) = \max {\sin x, \cos x}$ , $- π \leq x \leq π$ and the $x$ -axis is [2023]

Q 28 :

The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

Q 29 :

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is [2023]

Q 30 :

The area enclosed by the curves $y^{2} + 4 x = 4 and y - 2 x = 2$ is [2023]

Q 31 :

Let $A = {(x, y) \in ℝ^{2} : y \geq 0, 2 x \leq y \leq \sqrt{4 - {(x - 1)}^{2}}}$

and $B = {(x, y) \in ℝ \times ℝ : 0 \leq y \leq \min {2 x, \sqrt{4 - {(x - 1)}^{2}}}} .$

Then the ratio of the area of $A$ to the area of $B$ is [2023]

Q 32 :

$Let Δ be the area of the region {(x, y) \in ℝ^{2} : x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1} .$ $Then \frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) is equal to$ [2023]

Q 33 :

The area of the region $A = {(x, y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{π}{2}}$ is [2023]

Q 34 :

$Let q be the maximum integral value of p in [0,10] for which the roots of the equation x^{2} - p x + \frac{5}{4} p = 0 are rational.$ Then the area of the region
${(x, y) : 0 \leq y \leq {(x - q)}^{2}, 0 \leq x \leq q}$ is [2023]

Q 35 :

If the area of region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ is equal to $\frac{n + 2}{n + 1} - \frac{π}{n - 1},$ then the natural number $n$ is equal to _______ . [2023]

Q 36 :

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f (x) = \min {x^{2} + \frac{3}{4}, 1 + [x]},$ where $[x]$ denotes the greatest integer $\leq x$ , be $A$ . Then the value of $12 A$ is____________ . [2023]

Q 37 :

Let $y = p (x)$ be the parabola passing through the points $(- 1, 0), (0, 1)$ and $(1, 0)$ . If the area of the region

${(x, y) : {(x + 1)}^{2} + {(y - 1)}^{2} \leq 1, y \leq p (x)}$ is $A$ , then $12 (π - 4 A)$ is equal to ______ . [2023]

Q 38 :

If the area of the region ${(x, y) : | x^{2} - 2 | \leq y \leq x}$ is A, then $6 A + 16 \sqrt{2}$ is equal to _______ . [2023]

Q 39 :

If A is the area in the first quadrant enclosed by the curve $C : 2 x^{2} - y + 1 = 0$ , the tangent to C at the point (1, 3) and the line $x + y = 1$ , then the value of 60A is ______ . [2023]

Q 40 :

If the area bounded by the curve $2 y^{2} = 3 x,$ lines $x + y = 3, y = 0$ and outside the circle ${(x - 3)}^{2} + y^{2} = 2$ is A, then $4 (π + 4 A)$ is equal to _____ . [2023]