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

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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2

3

4

(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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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]

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

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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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(164)

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

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

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

$∴$ $12 A = 164 sq. units$

Topic Question Set

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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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Topic Question Set

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]

0%

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]

0%

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]

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

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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]

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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]

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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]