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

Q 31 :
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

(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 32 :
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 33 :
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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4

(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 34 :
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 35 :
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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4

(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 36 :
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 37 :
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]

0%

(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 38 :
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]

0%

(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 39 :
If the area of the region ${(x, y) : | x - 5 | \leq y \leq 4 \sqrt{x}}$ is A, then 3A is equal to __________. [2025]

0%

(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 40 :
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]

0%

(15)

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

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

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

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

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

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

$∴ 6 A = 6 \times \frac{5}{2} = 15$

Topic Question Set

Q 31 :
The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq | x + 2 |}$ is equal to [2025]

Q 32 :
The area of the region enclosed by the curves $y = e^{x}$ , $y = | e^{x} - 1 |$ and y-axis is: [2025]

Q 33 :
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 34 :
The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

Q 35 :
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 36 :
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 37 :
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]

0%

Q 38 :
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]

0%

Q 39 :
If the area of the region ${(x, y) : | x - 5 | \leq y \leq 4 \sqrt{x}}$ is A, then 3A is equal to __________. [2025]

0%

Q 40 :
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]

0%

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Topic Question Set

Q 31 : The area of the region {(x,y):x2+4x+2≤y≤|x+2|} is equal to [2025]

Q 32 : The area of the region enclosed by the curves y=ex, y=|ex–1| and y-axis is: [2025]

Q 33 : The area of the region {(x,y):0≤y≤2|x|+1, 0≤y≤x2+1, |x|≤3} is [2025]

Q 34 : The area of the region bounded by the curves x(1+y2)=1 and y2=2x is : [2025]

Q 35 : 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 36 : 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 37 : 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]

0%

Q 38 : 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]

0%

Q 39 : If the area of the region {(x,y):|x–5|≤y≤4x} is A, then 3A is equal to __________. [2025]

0%

Q 40 : Let the area of the bounded region {(x,y):0≤9x≤y2, y≥3x–6} be A. Then 6A is equal to __________. [2025]

0%

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Q 31 :
The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq | x + 2 |}$ is equal to [2025]

Q 32 :
The area of the region enclosed by the curves $y = e^{x}$ , $y = | e^{x} - 1 |$ and y-axis is: [2025]

Q 33 :
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 34 :
The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

Q 35 :
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 36 :
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 37 :
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 38 :
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 39 :
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 40 :
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]