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

Q 1 :

If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

2
- 1
1
0

1

2

3

4

(2)

Area of region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1} = \int_{1}^{2} (\frac{1}{x} - \frac{a}{x^{2}}) d x$

$\Rightarrow (\ln 2) - \frac{1}{7} = {[\ln x + \frac{a}{x}]}_{1}^{2} = \ln 2 + \frac{a}{2} - (\ln (1 + a))$

$= \ln 2 - \frac{a}{2} \Rightarrow \frac{a}{2} = \frac{1}{7} \Rightarrow a = \frac{2}{7}$

$So, 7 a - 3 = \frac{2}{7} \times 7 - 3 = - 1$

Q 2 :

The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is [2024]

$\sqrt{3} π$
$\sqrt{3} π + \frac{3}{4}$
$\sqrt{3} π - \frac{3}{4}$
$\sqrt{3} π + 1$

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(1)

We have, $x^{2} + 3 y^{2} = 18$

$\Rightarrow \frac{x^{2}}{18} + \frac{y^{2}}{6} = 1$

$\Rightarrow \frac{x^{2}}{(3 \sqrt{2})^{2}} + \frac{y^{2}}{(\sqrt{6})^{2}} = 1$

For point of intersection of ellipse and line $y = x,$ we have, $x^{2} + 3 x^{2} = 18$

$\Rightarrow x^{2} = \frac{18}{4} \Rightarrow x^{2} = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}}$

Required area $= \int_{0}^{3 / \sqrt{2}} x d x + \int_{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18 - x^{2}}{3}} d x$

$= {[\frac{x^{2}}{2}]}_{0}^{3 / \sqrt{2}} + \frac{1}{\sqrt{3}} {[\frac{x}{2} \sqrt{18 - x^{2}} + 9 \sin^{- 1} \frac{x}{3 \sqrt{2}}]}_{3 / \sqrt{2}}^{3 \sqrt{2}}$

$= \frac{9}{4} + \frac{1}{\sqrt{3}} [9 \sin^{- 1} (1) - \frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}} - 9 \sin^{- 1} (\frac{1}{2})]$

$= \frac{9}{4} + \frac{1}{\sqrt{3}} [\frac{9 π}{2} - \frac{9 \sqrt{3}}{4} - \frac{9 π}{6}] = \frac{1}{\sqrt{3}} 3 π = \sqrt{3} π$

Q 3 :

Three points $O (0, 0), P (a, a^{2}), Q (- b, b^{2}), a > 0, b > 0,$ are on the parabola $y = x^{2} .$ Let $S_{1}$ be the area of the region bounded by the line PQ and the parabola, and $S_{2}$ be the area of the triangle $O P Q .$ If the minimum value of $\frac{S_{1}}{S_{2}}$ is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

(7)

Equation of line $P Q$ is, $y - a^{2} = \frac{b^{2} - a^{2}}{- b - a} (x - a)$

$\Rightarrow y = x (a - b) + a b$

$∴ S_{1} = \int_{- b}^{a} (x (a - b) + a b - x^{2}) d x$

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

$= (a - b) \frac{(a^{2} - b^{2})}{2} + a b (a + b) - \frac{(a^{3} + b^{3})}{3}$

$\Rightarrow S_{1} = \frac{1}{6} {(a + b)}^{3}$ ...(i)

Also, $S_{2} = \frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ a & a^{2} & 1 \\ - b & b^{2} & 1 \end{matrix} |$ $= \frac{1}{2} a b (a + b)$

$Now, \frac{S_{1}}{S_{2}} = \frac{{(a + b)}^{3} / 6}{a b (a + b) / 2} = \frac{1}{3} \frac{{(a + b)}^{2}}{a b}$

$= \frac{a^{2} + b^{2} + 2 a b}{3 a b} = \frac{a}{3 b} + \frac{b}{3 a} + \frac{2}{3} = \frac{1}{3} (\frac{a}{b} + \frac{b}{a} + 2)$

$Now, \frac{a}{b} + \frac{1}{a / b} \geq 2$

$∴ Minimum value of \frac{S_{1}}{S_{2}} = \frac{4}{3}$

$So, m + n = 4 + 3 = 7$

Q 4 :

The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]

(8)

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

Point of intersection

$y (k y - 2 (1 - \frac{2 y}{k})) = 0$

$\Rightarrow k y^{2} - 2 (y - \frac{2 y^{2}}{k}) = 0$

$\Rightarrow y = 0 and k y = 2 (1 - \frac{2 y}{k}) \Rightarrow k y + \frac{4 y}{k} = 2$

$y = \frac{2}{k + \frac{4}{k}} = \frac{2 k}{k^{2} + 4}$ $∴ A = \int_{0}^{\frac{2 k}{k^{2} + 4}} ((y - \frac{k y^{2}}{2}) - (\frac{2 y^{2}}{k})) \cdot d y$

$= {[\frac{y^{2}}{2} - (\frac{k}{2} + \frac{2}{k}) \cdot \frac{y^{3}}{3}]}_{0}^{\frac{2 k}{k^{2} + 4}}$

$= {(\frac{2 k}{k^{2} + 4})}^{2} [\frac{1}{2} - \frac{k^{2} + 4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^{2} + 4}] = \frac{1}{6} \times 4 \times {(\frac{1}{k + \frac{4}{k}})}^{2}$

$A.M. \geq G.M. \Rightarrow (\frac{k + \frac{4}{k}}{2}) \geq 2 \Rightarrow k + \frac{4}{k} \geq 4$

$So, area is maximum when k = \frac{4}{k} \Rightarrow k = 2, - 2$

$∴ Sum of squares of all possible values of k$

$= 2^{2} + {(- 2)}^{2} = 8$

Q 5 :

One of the points of intersection of the curves $y = 1 + 3 x - 2 x^{2}$ and $y = \frac{1}{x}$ is $(\frac{1}{2}, 2) .$ Let the area of the region enclosed by these curves be $\frac{1}{24} (l \sqrt{5} + m) - n \log_{e} (1 + \sqrt{5}),$ where $l, m, n \in N .$ Then $l + m + n$ is equal to [2024]

32
30
29
31

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(2)

Given curves are $y = 1 + 3 x - 2 x^{2} and y = \frac{1}{x}$

On solving, $2 x^{3} - 3 x^{2} - x + 1 = 0$

$\Rightarrow$ $(2 x - 1) (x^{2} - x - 1) = 0 \Rightarrow x = \frac{1}{2}, x = \frac{1 \pm \sqrt{5}}{2}$

Required Area = $\int_{\frac{1}{2}}^{\frac{1 + \sqrt{5}}{2}} (1 + 3 x - 2 x^{2} - \frac{1}{x}) d x$

$= {[x + \frac{3 x^{2}}{2} - \frac{2 x^{3}}{3} - \ln x]}_{\frac{1}{2}}^{\frac{1 + \sqrt{5}}{2}}$

$= [\frac{\sqrt{5} + 1}{2} + \frac{3}{8} {(\sqrt{5} + 1)}^{2} - \frac{1}{12} {(\sqrt{5} + 1)}^{3} - \ln (\frac{\sqrt{5} + 1}{2})] - (\frac{1}{2} + \frac{3}{8} - \frac{1}{12} - \ln \frac{1}{2})$

$= \frac{1}{24} [12 (\sqrt{5} + 1) + 9 {(\sqrt{5} + 1)}^{2} - 2 {(\sqrt{5} + 1)}^{3} - 12 - 9 + 2]$ $- (\ln \frac{\sqrt{5} + 1}{2} + \ln 2)$

$= \frac{1}{24} [12 (\sqrt{5} + 1) + 9 (6 + 2 \sqrt{5}) - 2 (5 \sqrt{5} + 1 + 3 \sqrt{5} (\sqrt{5} + 1) - 19)] - \ln (\frac{\sqrt{5} + 1}{2} \times 2)$

$= \frac{1}{24} [14 \sqrt{5} + 15] - \ln (\sqrt{5} + 1)$

$= \frac{1}{24} (l \sqrt{5} + m) - n \log_{e} (1 + \sqrt{5})$ [Given]

$∴$ $l = 14,$ $m = 15$ and $n = 1$

Hence, $l + m + n = 30$

Q 6 :

The area (in sq. units) of the region described by ${(x, y) : y^{2} \leq 2 x, and y \geq 4 x - 1}$ is [2024]

$\frac{11}{12}$
$\frac{11}{32}$
$\frac{9}{32}$
$\frac{8}{9}$

1

2

3

4

(3)

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

$= {[\frac{y^{2}}{8} + \frac{y}{4} - \frac{y^{3}}{6}]}_{- \frac{1}{2}}^{1}$

$= (\frac{1}{8} + \frac{1}{4} - \frac{1}{6}) - (\frac{1}{32} - \frac{1}{8} + \frac{1}{48}) = \frac{9}{32}$

Q 7 :

The area enclosed between the curves $y = x$ $| x |$ and $y = x -$ $| x |$ is: [2024]

$\frac{8}{3}$
$\frac{2}{3}$
$\frac{4}{3}$
$1$

1

2

3

4

(3)

$y = x$ $| x |$ $= {\begin{matrix} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{matrix}$

$y = x -$ $| x |$ $= {\begin{matrix} 0, & x \geq 0 \\ 2 x, & x < 0 \end{matrix}$

$∴$ $Required Area = \int_{- 2}^{0} (- x^{2} - 2 x) d x$

$= {[- \frac{x^{3}}{3} - x^{2}]}_{- 2}^{0}$

$= \frac{- 8}{3} + 4 = \frac{4}{3}$

Q 8 :

Let the area of the region enclosed by the curves $y = 3 x, 2 y = 27 - 3 x and y = 3 x - x \sqrt{x}$ be $A .$ Then 10A is equal to [2024]

162
184
154
172

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2

3

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(1)

Required area,

$A = \int_{0}^{3} (3 x - (3 x - x \sqrt{x})) d x + \int_{3}^{9} \frac{27 - 3 x}{2} - (3 x - x \sqrt{x}) d x$

$= \int_{0}^{3} x^{3 / 2} d x + \int_{3}^{9} \frac{27}{2} - \frac{9 x}{2} + x^{3 / 2} d x$

$= \frac{2}{5} {[x^{5 / 2}]}_{0}^{3} + \frac{27}{2} {[x]}_{3}^{9} - \frac{9}{4} {[x^{2}]}_{3}^{9} + \frac{2}{5} {[x^{5 / 2}]}_{3}^{9}$

$= \frac{2}{5} 3^{5 / 2} + \frac{27}{2} \times 6 - \frac{9}{4} \times 72 + \frac{2}{5} [9^{5 / 2} - 3^{5 / 2}]$

$= 81 - 162 + \frac{2}{5} \cdot 3^{5} = \frac{81}{5}$

So, $10 A = 10 \times \frac{81}{5} = 162$

Q 9 :

The area of the region in the first quadrant inside the circle $x^{2} + y^{2} = 8$ and outside the parabola $y^{2} = 2 x$ is equal to: [2024]

$\frac{π}{2} - \frac{2}{3}$
$\frac{π}{2} - \frac{1}{3}$
$π - \frac{1}{3}$
$π - \frac{2}{3}$

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(4)

We have, $x^{2} + y^{2} = 8 and y^{2} = 2 x$

Required area = area of the shaded region

$= \int_{0}^{2} \frac{y^{2}}{2} d y + \int_{2}^{2 \sqrt{2}} \sqrt{8 - y^{2}} d y$

$= \frac{1}{6} {[y^{3}]}_{0}^{2} + {[\frac{y}{2} \sqrt{8 - y^{2}} + 4 \sin^{- 1} (\frac{y}{2 \sqrt{2}})]}_{2}^{2 \sqrt{2}}$

$= \frac{4}{3} + [2 π - 2 - π] = π - \frac{2}{3}$

Q 10 :

The parabola $y^{2} = 4 x$ divides the area of the circle $x^{2} + y^{2} = 5$ in two parts. The area of the smaller part is equal to : [2024]

$\frac{1}{3} + 5 \sin^{- 1} (\frac{2}{\sqrt{5}})$
$\frac{2}{3} + \sqrt{5} \sin^{- 1} (\frac{2}{\sqrt{5}})$
$\frac{2}{3} + 5 \sin^{- 1} (\frac{2}{\sqrt{5}})$
$\frac{1}{3} + \sqrt{5} \sin^{- 1} (\frac{2}{\sqrt{5}})$

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(3)

The points of intersection of $y^{2} = 4 x$ and $x^{2} + y^{2} = 5$ are (1, 2) and (1, -2).

$∴$ $Required Area = 2 {Area of OACO + Area of CABC}$

$= 2 [\int_{0}^{1} 2 \sqrt{x} d x + \int_{1}^{\sqrt{5}} \sqrt{5 - x^{2}} d x]$

$= 2 {[| \frac{4}{3} x^{\frac{3}{2}} |_{0}^{1} + (\frac{1}{2} x \sqrt{5 - x^{2}} + \frac{5}{2} \sin^{- 1} \frac{x}{\sqrt{5}})]}_{1}^{\sqrt{5}}$

$= 2 [(\frac{4}{3} - 0) + (0 + \frac{5 π}{4}) - (1 + \frac{5}{2} \sin^{- 1} \frac{1}{\sqrt{5}})]$

$= 2 [\frac{1}{3} + \frac{5 π}{4} - \frac{5}{2} \sin^{- 1} \frac{1}{\sqrt{5}}] = \frac{2}{3} + 5 [\frac{π}{2} - \sin^{- 1} \frac{1}{\sqrt{5}}]$

$= \frac{2}{3} + 5 \cos^{- 1} (\frac{1}{\sqrt{5}})$ $[∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$

$= \frac{2}{3} + 5 \sin^{- 1} (\frac{2}{\sqrt{5}})$

Q 11 :

The area enclosed by the curves $x y + 4 y = 16$ and $x + y = 6$ is equal to: [2024]

$28 - 30 \log_{e} 2$
$32 - 30 \log_{e} 2$
$30 - 28 \log_{e} 2$
$30 - 32 \log_{e} 2$

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(4)

We have, $y = \frac{16}{x + 4} and x + y = 6 \Rightarrow y = 6 - x$

For intersecting points:

$6 - x = \frac{16}{x + 4} \Rightarrow (6 - x) (x + 4) = 16$

$\Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow x^{2} - 4 x + 2 x - 8 = 0$

$\Rightarrow x (x - 4) + 2 (x - 4) = 0$

$\Rightarrow (x + 2) (x - 4) = 0 \Rightarrow x = - 2, 4$

$∴$ $y = 8, when x = - 2 and y = 2, when x = 4$

So, $Area = \int_{- 2}^{4} ((6 - x) - \frac{16}{x + 4}) d x = {[6 x - \frac{x^{2}}{2} - 16 \log_{e} (x + 4)]}_{- 2}^{4}$

$= 24 - 8 - 16 \log_{e} 8 + 12 + 2 + 16 \log_{e} 2 = 30 - 16 \log_{e} (\frac{8}{2})$

$= 30 - 16 \log_{e} 2^{2} = 30 - 32 \log_{e} 2$

Q 12 :

The area (in square units) of the region bounded by the parabola $y^{2} = 4 (x - 2)$ and the line $y = 2 x - 8,$ is [2024]

9
7
8
6

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2

3

4

(1)

Parabola : $y^{2} = 4 (x - 2)$ ...(i)

and line : $y = 2 x - 8$ ...(ii)

$\Rightarrow 4 {(x - 4)}^{2} = 4 (x - 2)$

$\Rightarrow x^{2} - 9 x + 18 = 0$

$\Rightarrow x = 3, 6$

From (ii), we get

$\Rightarrow y = - 2, 4$

Points of intersection of (i) and (ii) are $(3, - 2)$ and $(6, 4)$

$∴ Required Area = \int_{- 2}^{4} [(\frac{y + 8}{2}) - (\frac{y^{2}}{4} + 2)] d y$

$= {[\frac{y^{2}}{4} - \frac{y^{3}}{12} + 2 y]}_{- 2}^{4}$

$= 9 sq. units$

Q 13 :

The area of the region

${(x, y) : y^{2} \leq 4 x, x < 4, \frac{x y (x - 1) (x - 2)}{(x - 3) (x - 4)} > 0, x \neq 3}$ is [2024]

$\frac{32}{3}$
$\frac{16}{3}$
$\frac{8}{3}$
$\frac{64}{3}$

1

2

3

4

(1)

The area of region $x y \frac{(x - 1) (x - 2)}{(x - 3) (x - 4)}, y^{2} - 4 x \leq 0$

$\Rightarrow y x (x - 1) (x - 2) (x - 3) (x - 4) < 0$

Case 1: $y > 0$

So, $x (x - 1) (x - 2) (x - 3) (x - 4) < 0$

$\Rightarrow x \in (0, 1) \cup (2, 3)$

Case 2: $y < 0$

So, $x (x - 1) (x - 2) (x - 3) (x - 4) > 0 \Rightarrow x \in (1, 2) \cup (3, 4)$

$Area = \int_{0}^{4} 2 \sqrt{x} d x = \frac{2}{3} \times 16 = \frac{32}{3}$

Q 14 :

The area of the region enclosed by the parabolas $y = 4 x - x^{2}$ and $3 y = {(x - 4)}^{2}$ is equal to [2024]

6
4
$\frac{14}{3}$
$\frac{32}{9}$

1

2

3

4

(1)

Given, $y = 4 x - x^{2}$ ...(i)

$3 y = {(x - 4)}^{2}$ ...(ii)

From (i),

$3 y = 3 (4 x - x^{2}) = 12 x - 3 x^{2}$

From (ii),

${(x - 4)}^{2} = 12 x - 3 x^{2}$

$\Rightarrow x^{2} - 8 x + 16 = 12 x - 3 x^{2} \Rightarrow x^{2} - 5 x + 4 = 0$

$\Rightarrow (x - 1) (x - 4) = 0 \Rightarrow x = 1, 4$

Points of intersection are $(1, 3)$ and $(4, 0)$

Required area = $\int_{1}^{4} [(4 x - x^{2}) - \frac{{(x - 4)}^{2}}{3}] d x$

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

$= [(32 - \frac{64}{3} - \frac{1}{3} (0)) - (2 - \frac{1}{3} - \frac{1}{3} (\frac{- 27}{3}))]$

$= 30 - \frac{64}{3} + \frac{1}{3} - 3 = 27 - 21 = 6 sq. units$

Q 15 :

The area of the region enclosed by the parabolas $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$ is ___________ [2024]

(72)

Given, $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$

Both curves passes through the origin

For the points of intersection, we have $x^{2} - 5 x = 7 x - x^{2}$

$\Rightarrow 2 x^{2} = 12 x$

$\Rightarrow 2 x (x - 6) = 0$

$\Rightarrow x = 0, 6$

$∴$ Area bounded by the curves $= \int_{0}^{6} [(7 x - x^{2}) - (x^{2} - 5 x)] d x$

$= \int_{0}^{6} (7 x - x^{2} - x^{2} + 5 x) d x = \int_{0}^{6} (12 x - 2 x^{2}) d x$

$= 12 (\frac{6^{2}}{2}) - \frac{2}{3} {(6)}^{3}$

$= 72 sq. units$

Q 16 :

Let the area of the region enclosed by the curve $y = \min {\sin x, \cos x}$ and the $x$ -axis between $x = - π$ to $x = π$ be $A .$ Then $A^{2}$ is equal to ______ . [2024]

(16)

Required area, $A = \int_{- π}^{\frac{- 3 π}{4}} - \cos x d x + \int_{\frac{- 3 π}{4}}^{0} - \sin x d x + \int_{0}^{\frac{π}{4}} \sin x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos x d x + \int_{\frac{π}{2}}^{π} - \cos x d x$

$= {[- \sin x]}_{- π}^{- \frac{3 π}{4}} + {[\cos x]}_{- \frac{3 π}{4}}^{0} - {[\cos x]}_{0}^{\frac{π}{4}} + {[\sin x]}_{\frac{π}{4}}^{\frac{π}{2}} - {[\sin x]}_{\frac{π}{2}}^{π}$

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

$∴$ $A^{2} = 16$

Q 17 :

Let the area of the region ${(x, y) : x - 2 y + 4 \geq 0, x + 2 y^{2} \geq 0, x + 4 y^{2} \leq 8, y \geq 0}$ be $\frac{m}{n},$ where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to _____ . [2024]

(119)

Given, $x - 2 y + 4 = 0$ ...(i)

$x + 2 y^{2} = 0$ ...(ii)

$x + 4 y^{2} = 8$ ...(iii)

The point intersection of (i) and (iii) are $(- 1, 1.5)$ and $(- 8, - 2) .$

The point of intersection of (i) and (ii) are $(- 2, 1)$ and $(- 8, - 2) .$

$∴$ Required area

$\int_{0}^{1} [(8 - 4 y^{2}) - (- 2 y^{2})] d y + \int_{1}^{3 / 2} [(8 - 4 y^{2}) - (2 y - 4)] d y$

$= {[8 y - \frac{2 y^{3}}{3}]}_{0}^{1} + {[12 y - y^{2} - \frac{4}{3} y^{3}]}_{1}^{3 / 2}$

$= (8 - \frac{2}{3}) + (12 (\frac{3}{2}) - {(\frac{3}{2})}^{2} - \frac{4}{3} {(\frac{3}{2})}^{3}) - (12 - 1 - \frac{4}{3})$

$= \frac{22}{3} + 18 - \frac{9}{4} - \frac{9}{2} - \frac{29}{3} = \frac{107}{12} = \frac{m}{n}$

$∴ m + n = 119$

Q 18 :

If the area of the region ${(x, y) : 0 \leq y \leq \min {2 x, 6 x - x^{2}}}$ is $A,$ then 12A is equal to ____________. [2024]

(304)

Region ${(x, y) : 0 \leq y \leq \min (2 x, 6 x - x^{2})}$

$∴$ Let A be the area of the required region.

$A = 2 \int_{0}^{4} x d x + \int_{4}^{6} (6 x - x^{2}) d x$

$= {[x^{2}]}_{0}^{4} + {[3 x^{2} - \frac{x^{3}}{3}]}_{4}^{6}$

$= 16 + 3 (36 - 16) - \frac{1}{3} (216 - 64)$

$= 16 + 60 - \frac{152}{3} = \frac{228 - 152}{3} = \frac{76}{3} \Rightarrow A = \frac{76}{3} sq. units$

$∴$ $12 A = 12 \times \frac{76}{3} = 4 \times 76 = 304$

Q 19 :

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 169$ which is below the line $5 x - y = 13$ is $\frac{π α}{2 β} - \frac{65}{2} + \frac{α}{β} \sin^{- 1} (\frac{12}{13}),$ where $α, β$ are coprime numbers. Then $α + β$ is equal to _______ . [2024]

(171)

$∴ Required Area = \int_{- 13}^{12} (\sqrt{169 - y^{2}} - \frac{y + 13}{5}) d y$

$= {[\frac{1}{2} y \sqrt{169 - y^{2}} + \frac{169}{2} \sin^{- 1} \frac{y}{13}]}_{- 13}^{12} - {[\frac{y^{2}}{10} + \frac{13}{5} y]}_{- 13}^{12}$

$= [\frac{1}{2} \times 12 \times 5 + \frac{169}{2} \sin^{- 1} \frac{12}{13}] - [0 + \frac{169}{2} (- \frac{π}{2})] - [\frac{144}{10} + \frac{13}{5} \times 12] + [\frac{169}{10} - \frac{169}{5}]$

$= \frac{169}{2} (\frac{π}{2}) + \frac{169}{2} \sin^{- 1} (\frac{12}{13}) - \frac{65}{2}$

$∴$ $α = 169 and β = 2$

$Thus, α + β = 169 + 2 = 171$

Q 20 :

Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min {x^{2} + 2, 2 x + 2}}$ be $A .$ Then 12A is equal to ___________ . [2024]

(164)

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

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

$\Rightarrow A = \frac{8}{3} + 4 + (9 + 6 - 4 - 4) = \frac{41}{3}$

$∴$ $12 A = 164 sq. units$

Q 21 :

The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

(5)

Given, equation of parabola

${(y - 2)}^{2} = (x - 1)$ ...(i)

and line $x - 2 y + 4 = 0$ ...(ii)

From (ii), we have

$x - 2 (y - 2) = 0 \Rightarrow y - 2 = \frac{x}{2}$

Putting in (i), we get ${(\frac{x}{2})}^{2} = (x - 1)$

$\Rightarrow x^{2} - 4 x + 4 = 0$

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

From (ii), we have, $y = 3$

$∴$ Intersection point of (i) and (ii) is (2,3).

$∴$ $Required area = \int_{0}^{3} ({(y - 2)}^{2} + 1) d y - Area of triangle$

$= \int_{0}^{3} (y^{2} - 4 y + 5) d y - \frac{1}{2} \times 2 \times 1$

$= {[\frac{y^{3}}{3} - 2 y^{2} + 5 y]}_{0}^{3} - 1 = (9 - 18 + 15) - 1 = 6 - 1 = 5 sq. units$

Q 22 :

The area of the region ${(x, y) : | x - y | \leq y \leq 4 \sqrt{x}}$ is [2025]

512
$\frac{2048}{3}$
$\frac{1024}{3}$
$\frac{512}{3}$

1

2

3

4

(3)

We have, $| x - y |$ $\leq y \leq 4 \sqrt{x}$

Now, $y =$ $| x - y |$ $\Rightarrow y^{2} = {(x - y)}^{2}$

$\Rightarrow y = \frac{x}{2} and x = 0$

Required Area $= \int_{0}^{64} (4 \sqrt{x} - \frac{x}{2}) d x$

$= {[\frac{4 x^{3 / 2}}{3 / 2} - \frac{x^{2}}{4}]}_{0}^{64} = \frac{8}{3} \cdot 8^{3} - \frac{64^{2}}{4} = 64^{2} (\frac{1}{12}) = \frac{1024}{3}$ .

Q 23 :

Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

$\frac{1}{2}$
$\sqrt{2}$
$\sqrt{5}$
2

1

2

3

4

(1)

We have, $y = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$\Rightarrow \frac{d y}{d x} = - 2 + e^{- x} \cdot e^{x} f (x) + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$= - 2 + y + y + 2 x - 1$

$\Rightarrow \frac{d y}{d x} - 2 y = 2 x - 3$

This is a linear differential equation, so we have

$y e^{- 2 x} = \int (2 x - 3) \cdot e^{- 2 x} d x + c$ [ $∵ I . F . = e^{\int - 2 d x} = e^{- 2 x}$ ]

$\Rightarrow y e^{- 2 x} = - \frac{(2 x - 3)}{2} e^{- 2 x} + \int e^{- 2 x} \cdot d x + c$

$\Rightarrow y e^{- 2 x} = \frac{- (2 x - 3)}{2} e^{- 2 x} - \frac{e^{- 2 x}}{2} + c$

$∵ f (0) = 1 ∴ c = 1 - \frac{3}{2} + \frac{1}{2} = 0$

$\Rightarrow y = - \frac{(2 x - 3)}{2} - \frac{1}{2} \Rightarrow x + y = 1$

So, area $= \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$ .

Q 24 :

A line passing through the point A(–2, 0), touches the parabola $P : y^{2} = x - 2$ at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

$\frac{7}{3}$
2
3
$\frac{8}{3}$

1

2

3

4

(4)

Given parabola $P : y^{2} = x - 2$ and line is passing through A(–2, 0)

Tangent is y = m(x + 2)

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

$\Rightarrow m^{2} x^{2} + (4 m^{2} - 1) x + (4 m^{2} + 2) = 0$

As D = 0

$\Rightarrow {(4 m^{2} - 1)}^{2} = 4 m^{2} (4 m^{2} + 2) \Rightarrow m = \frac{1}{4}$

So, tangent is $y = \frac{1}{4} (x + 2)$

Point of tangency is (6, 2)

$∴$ Area of region $= \int_{0}^{2} (y^{2} + 2) - (4 y - 2) d y$

$= \frac{8}{3} - 8 + 8 = \frac{8}{3}$ sq. units.

Q 25 :

If the area of the region bounded by the curves $y = 4 - \frac{x^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals [2025]

240
220
210
250

1

2

3

4

(4)

Given, $y = 4 - \frac{x^{2}}{4}$ ... (i)

and $y = \frac{x - 4}{2}$ ... (ii)

From (i) & (ii), we get

x = –6, 4 and y = –5, 0

Thus, point of intersection of (i) & (ii) are (–6, 5) and (4, 0).

Required Area $= \int_{- 6}^{4} {(4 - \frac{x^{2}}{4}) - (\frac{x - 4}{2})} d x$

$\Rightarrow α = \int_{- 6}^{4} {- \frac{x^{2}}{4} - \frac{x}{2} + 6} d x$

$α =$ $- \frac{x^{3}}{12} - \frac{x^{2}}{4} + 6 x |_{- 6}^{4}$

$= - (\frac{64}{12} + \frac{216}{12}) - (\frac{16}{4} - \frac{36}{4}) + 6 (4 + 6)$

$= \frac{125}{3}$

$∴ 6 α = 250$ .

Q 26 :

If the area of the region ${(x, y) : 1 + x^{2} \leq y \leq \min {x + 7, 11 - 3 x}}$ is A, then 3A is equal to [2025]

47
49
50
46

1

2

3

4

(3)

$A = \int_{- 2}^{1} (x + 7 - x^{2} - 1) d x + \int_{1}^{2} (11 - 3 x - x^{2} - 1) d x$

$= {[\frac{x^{2}}{2} + 6 x - \frac{x^{3}}{3}]}_{- 2}^{1} + {[10 x - \frac{3 x^{2}}{2} - \frac{x^{3}}{3}]}_{1}^{2}$

$= \frac{50}{3} \Rightarrow 3 A = 50$ .

Q 27 :

The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

$3 π + 8$
$3 π - 8$
$6 π - 16$
$6 π - 8$

1

2

3

4

(3)

We have, ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ , a circle with centre $(2 \sqrt{3}, 0)$ and radius $\sqrt{12}$ and a parabola $y^{2} = 2 \sqrt{3} x$ .

$∴$ Required Area = $\frac{π (12)}{2} - 2 \int_{0}^{2 \sqrt{3}} {(2 \sqrt{3})}^{\frac{1}{2}} x^{1 / 2} d x$

$= 6 π - 2 {(2 \sqrt{3})}^{1 / 2} {[\frac{x^{3 / 2}}{3} \cdot 2]}_{0}^{2 \sqrt{3}}$

$= 6 π - \frac{4}{3} {(2 \sqrt{3})}^{1 / 2} {(2 \sqrt{3})}^{3 / 2}$

$= 6 π - \frac{4}{3} {(2 \sqrt{3})}^{2} = 6 π - 16$ .

Q 28 :

Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

$e^{4} + 1$
$e^{2} - 1$
$e^{4} - 1$
$e^{2} + 1$

1

2

3

4

(2)

We have, f(x + y) = f(x)f(y)

Let $f (x) = e^{λ x}$

Differentiating w.r.t. x, we get

$f' (x) = λ \cdot e^{λ x}$

Put x = 0, we get

$f' (0) = λ$ [ $∵ f' (0) = 4 a$ ]

$\Rightarrow λ = 4 a$

So, $f (x) = e^{4 a x}$

Now, $f'' (x) - 3 a f' (x) - f (x) = 0$

$\Rightarrow \frac{d}{d x} (λ e^{λ x}) - 3 a (λ e^{λ x}) - e^{λ x} = 0$

$\Rightarrow λ^{2} - 3 a λ - 1 = 0$

$\Rightarrow 16 a^{2} - 12 a^{2} - 1 = 0 \Rightarrow 4 a^{2} = 1$

$\Rightarrow a = \frac{1}{2}$ [ $∵ a > 0$ ]

$∴ f (x) = e^{2 x}$

Now, $f (x) = f (a x) = e^{x}$

$∴$ Area of region $\int_{0}^{2} e^{x} d x = {[e^{x}]}_{0}^{2} = e^{2} - 1$ .

Q 29 :

The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

5
4/3
8/3
8

1

2

3

4

(3)

The given curves are $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$

Points of intersection are (2, 0) and (0, 4).

$∴$ Required Area $= \int_{0}^{2} (\sqrt{16 - 8 x} - (x^{2} - 4 x + 4)) d x$

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

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

$= \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$ sq. units.

Q 30 :

If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

6
8
7
5

1

2

3

4

(4)

Required area $= a + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$

$= a + {[a x + e^{x} + e^{- x}]}_{0}^{1}$

$\Rightarrow 2 a + e + e^{- 1} - 2 = e + 8 + \frac{1}{e}$

$\Rightarrow 2 a - 2 = 8$

$\Rightarrow 2 a = 10 \Rightarrow a = 5$ .

Topic Question Set

Q 1 :

If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

Q 2 :

The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is [2024]

Q 4 :

The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]

Q 5 :

One of the points of intersection of the curves $y = 1 + 3 x - 2 x^{2}$ and $y = \frac{1}{x}$ is $(\frac{1}{2}, 2) .$ Let the area of the region enclosed by these curves be $\frac{1}{24} (l \sqrt{5} + m) - n \log_{e} (1 + \sqrt{5}),$ where $l, m, n \in N .$ Then $l + m + n$ is equal to [2024]

Q 6 :

The area (in sq. units) of the region described by ${(x, y) : y^{2} \leq 2 x, and y \geq 4 x - 1}$ is [2024]

Q 7 :

The area enclosed between the curves $y = x$ $| x |$ and $y = x -$ $| x |$ is: [2024]

Q 8 :

Let the area of the region enclosed by the curves $y = 3 x, 2 y = 27 - 3 x and y = 3 x - x \sqrt{x}$ be $A .$ Then 10A is equal to [2024]

Q 9 :

The area of the region in the first quadrant inside the circle $x^{2} + y^{2} = 8$ and outside the parabola $y^{2} = 2 x$ is equal to: [2024]

Q 10 :

The parabola $y^{2} = 4 x$ divides the area of the circle $x^{2} + y^{2} = 5$ in two parts. The area of the smaller part is equal to : [2024]

Q 11 :

The area enclosed by the curves $x y + 4 y = 16$ and $x + y = 6$ is equal to: [2024]

Q 12 :

The area (in square units) of the region bounded by the parabola $y^{2} = 4 (x - 2)$ and the line $y = 2 x - 8,$ is [2024]

Q 13 :

The area of the region

${(x, y) : y^{2} \leq 4 x, x < 4, \frac{x y (x - 1) (x - 2)}{(x - 3) (x - 4)} > 0, x \neq 3}$ is [2024]

Q 14 :

The area of the region enclosed by the parabolas $y = 4 x - x^{2}$ and $3 y = {(x - 4)}^{2}$ is equal to [2024]

Q 15 :

The area of the region enclosed by the parabolas $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$ is ___________ [2024]

Q 16 :

Let the area of the region enclosed by the curve $y = \min {\sin x, \cos x}$ and the $x$ -axis between $x = - π$ to $x = π$ be $A .$ Then $A^{2}$ is equal to ______ . [2024]

Q 17 :

Let the area of the region ${(x, y) : x - 2 y + 4 \geq 0, x + 2 y^{2} \geq 0, x + 4 y^{2} \leq 8, y \geq 0}$ be $\frac{m}{n},$ where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to _____ . [2024]

Q 18 :

If the area of the region ${(x, y) : 0 \leq y \leq \min {2 x, 6 x - x^{2}}}$ is $A,$ then 12A is equal to ____________. [2024]

Q 19 :

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 169$ which is below the line $5 x - y = 13$ is $\frac{π α}{2 β} - \frac{65}{2} + \frac{α}{β} \sin^{- 1} (\frac{12}{13}),$ where $α, β$ are coprime numbers. Then $α + β$ is equal to _______ . [2024]

Q 20 :

Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min {x^{2} + 2, 2 x + 2}}$ be $A .$ Then 12A is equal to ___________ . [2024]

Q 21 :

The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

Q 22 :

The area of the region ${(x, y) : | x - y | \leq y \leq 4 \sqrt{x}}$ is [2025]

Q 23 :

Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Q 24 :

A line passing through the point A(–2, 0), touches the parabola $P : y^{2} = x - 2$ at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

Q 25 :

If the area of the region bounded by the curves $y = 4 - \frac{x^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals [2025]

Q 26 :

If the area of the region ${(x, y) : 1 + x^{2} \leq y \leq \min {x + 7, 11 - 3 x}}$ is A, then 3A is equal to [2025]

Q 27 :

The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

Q 28 :

Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

Q 29 :

The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

Q 30 :

If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

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Topic Question Set

Q 1 : If the area of the region {(x,y):ax2≤y≤1x,1≤x≤2,0<a<1} is (loge2)-17 then the value of 7a-3 is equal to : [2024]

Q 2 : The area (in square units) of the region enclosed by the ellipse x2+3y2=18 in the first quadrant below the line y=x is [2024]

Q 3 : Three points O(0,0),P(a,a2),Q(-b,b2),a>0,b>0, are on the parabola y=x2. Let S1 be the area of the region bounded by the line PQ and the parabola, and S2 be the area of the triangle OPQ. If the minimum value of S1S2 is mn,gcd(m,n)=1, then m+n is equal to _______ . [2024]

Q 4 : The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y2=kx and ky2=2(y-x) is maximum, is equal to _______ . [2024]

Q 5 : One of the points of intersection of the curves y=1+3x-2x2 and y=1x is (12,2). Let the area of the region enclosed by these curves be 124(l5+m)-nloge(1+5), where l,m,n∈N. Then l+m+n is equal to [2024]

Q 6 : The area (in sq. units) of the region described by {(x,y):y2≤2x, and y≥4x-1} is [2024]

Q 7 : The area enclosed between the curves y=x|x| and y=x-|x| is: [2024]

Q 8 : Let the area of the region enclosed by the curves y=3x, 2y=27-3x and y=3x-xx be A. Then 10A is equal to [2024]

Q 9 : The area of the region in the first quadrant inside the circle x2+y2=8 and outside the parabola y2=2x is equal to: [2024]

Q 10 : The parabola y2=4x divides the area of the circle x2+y2=5 in two parts. The area of the smaller part is equal to : [2024]

Q 11 : The area enclosed by the curves xy+4y=16 and x+y=6 is equal to: [2024]

Q 12 : The area (in square units) of the region bounded by the parabola y2=4(x-2) and the line y=2x-8, is [2024]

Q 13 : The area of the region {(x,y):y2≤4x, x<4, xy(x-1)(x-2)(x-3)(x-4)>0, x≠3} is [2024]

Q 14 : The area of the region enclosed by the parabolas y=4x-x2 and 3y=(x-4)2 is equal to [2024]

Q 15 : The area of the region enclosed by the parabolas y=x2-5x and y=7x-x2 is ___________ [2024]

Q 16 : Let the area of the region enclosed by the curve y=min{sinx,cosx} and the x-axis between x=-π to x=π be A. Then A2 is equal to ______ . [2024]

Q 17 : Let the area of the region {(x,y):x-2y+4≥0, x+2y2≥0, x+4y2≤8, y≥0} be mn, where m and n are coprime numbers. Then m+n is equal to _____ . [2024]

Q 18 : If the area of the region {(x,y):0≤y≤min{2x,6x-x2}} is A, then 12A is equal to ____________. [2024]

Q 19 : The area (in sq. units) of the part of the circle x2+y2=169 which is below the line 5x-y=13 is πα2β-652+αβsin-1(1213), where α,β are coprime numbers. Then α+β is equal to _______ . [2024]

Q 20 : Let the area of the region {(x,y):0≤x≤3,0≤y≤min{x2+2,2x+2}} be A. Then 12A is equal to ___________ . [2024]

Q 21 : The area of the region enclosed by the parabola (y-2)2=x-1, the line x-2y+4=0 and the positive coordinate axes is ____________. [2024]

Q 22 : The area of the region {(x,y):|x–y|≤y≤4x} is [2025]

Q 23 : Let f:[0,∞)→R be a differentiable function such that f(x)=1–2x+∫0xex–tf(t)dt for all x∈[0,∞). Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Q 24 : A line passing through the point A(–2, 0), touches the parabola P : y2=x–2 at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

Q 25 : If the area of the region bounded by the curves y=4–x24 and y=x–42 is equal to α, then 6α equals [2025]

Q 26 : If the area of the region {(x,y):1+x2≤y≤min {x+7,11–3x}} is A, then 3A is equal to [2025]

Q 27 : The area of the region, inside the circle (x–23)2+y2=12 and outside the parabola y2=23x is : [2025]

Q 28 : Let f : R→R be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, y∈R. If f'(0)=4a and f satisfies f''(x)=3af'(x)–f(x)=0, a > 0, then the area of the region R={(x,y) | 0≤y≤f(ax), 0≤x≤2} is : [2025]

Q 29 : The area of the region enclosed by the curves y=x2–4x+4 and y2=16–8x is : [2025]

Q 30 : If the area of the region {(x,y):–1≤x≤1, 0≤y≤a+e|x|–e–x, a>0} is e2+8e+1e, then the value of a is : [2025]

Want Us to Email you About Special Offers & Updates?

Q 1 :

If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

Q 2 :

The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is [2024]

Q 4 :

The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]

Q 6 :

The area (in sq. units) of the region described by ${(x, y) : y^{2} \leq 2 x, and y \geq 4 x - 1}$ is [2024]

Q 7 :

The area enclosed between the curves $y = x$ $| x |$ and $y = x -$ $| x |$ is: [2024]

Q 8 :

Let the area of the region enclosed by the curves $y = 3 x, 2 y = 27 - 3 x and y = 3 x - x \sqrt{x}$ be $A .$ Then 10A is equal to [2024]

Q 9 :

The area of the region in the first quadrant inside the circle $x^{2} + y^{2} = 8$ and outside the parabola $y^{2} = 2 x$ is equal to: [2024]

Q 10 :

The parabola $y^{2} = 4 x$ divides the area of the circle $x^{2} + y^{2} = 5$ in two parts. The area of the smaller part is equal to : [2024]

Q 11 :

The area enclosed by the curves $x y + 4 y = 16$ and $x + y = 6$ is equal to: [2024]

Q 12 :

The area (in square units) of the region bounded by the parabola $y^{2} = 4 (x - 2)$ and the line $y = 2 x - 8,$ is [2024]

Q 13 :

The area of the region

${(x, y) : y^{2} \leq 4 x, x < 4, \frac{x y (x - 1) (x - 2)}{(x - 3) (x - 4)} > 0, x \neq 3}$ is [2024]

Q 14 :

The area of the region enclosed by the parabolas $y = 4 x - x^{2}$ and $3 y = {(x - 4)}^{2}$ is equal to [2024]

Q 15 :

The area of the region enclosed by the parabolas $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$ is ___________ [2024]

Q 16 :

Let the area of the region enclosed by the curve $y = \min {\sin x, \cos x}$ and the $x$ -axis between $x = - π$ to $x = π$ be $A .$ Then $A^{2}$ is equal to ______ . [2024]

Q 17 :

Let the area of the region ${(x, y) : x - 2 y + 4 \geq 0, x + 2 y^{2} \geq 0, x + 4 y^{2} \leq 8, y \geq 0}$ be $\frac{m}{n},$ where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to _____ . [2024]

Q 18 :

If the area of the region ${(x, y) : 0 \leq y \leq \min {2 x, 6 x - x^{2}}}$ is $A,$ then 12A is equal to ____________. [2024]

Q 19 :

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 169$ which is below the line $5 x - y = 13$ is $\frac{π α}{2 β} - \frac{65}{2} + \frac{α}{β} \sin^{- 1} (\frac{12}{13}),$ where $α, β$ are coprime numbers. Then $α + β$ is equal to _______ . [2024]

Q 20 :

Let the area of the region ${(x, y) : 0 \leq x \leq 3, 0 \leq y \leq \min {x^{2} + 2, 2 x + 2}}$ be $A .$ Then 12A is equal to ___________ . [2024]

Q 21 :

The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

Q 22 :

The area of the region ${(x, y) : | x - y | \leq y \leq 4 \sqrt{x}}$ is [2025]

Q 23 :

Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Q 24 :

A line passing through the point A(–2, 0), touches the parabola $P : y^{2} = x - 2$ at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

Q 25 :

If the area of the region bounded by the curves $y = 4 - \frac{x^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals [2025]

Q 26 :

If the area of the region ${(x, y) : 1 + x^{2} \leq y \leq \min {x + 7, 11 - 3 x}}$ is A, then 3A is equal to [2025]

Q 27 :

The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

Q 28 :

Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

Q 29 :

The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

Q 30 :

If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]