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]

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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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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]

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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]

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

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[IMAGE 29]--------------------------------

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]

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

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

[IMAGE 30]------------------------------------------------

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

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

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

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

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[IMAGE 31]

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

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

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

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

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

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]

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184
154
172

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[IMAGE 32]----------------------------

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

[IMAGE 33]----------------------------------

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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[IMAGE 34]-------------------------------------

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

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

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

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

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

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

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

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

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]

0%

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]

0%

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]

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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]

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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]

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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]

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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]