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

Q 41 :

Twice the area (in sq. units) of the region ${(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y$ and $y \leq 1 + \sqrt{x}}$ is

(5)

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

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

$= (1 + \frac{2}{3}) + (6 - 2 - 3 + \frac{1}{2}) - (\frac{8}{12})$

$= \frac{5}{3} + \frac{3}{2} - \frac{2}{3}$

$= 1 + \frac{3}{2}$

$= \frac{5}{2} sq. unit$

$∴$ $2 I = 5$

Q 42 :

Area of the region bounded by $y = x$ and $y = x^{2}$ is _____

$\frac{1}{4}$
$\frac{1}{3}$
$\frac{1}{5}$
$\frac{1}{6}$

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4

(4)

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

Q 43 :

The area of the region ${(x, y) : y \leq π - | x |, y \leq | x \sin x |, y \geq 0}$ is: [2026]

$1 + \frac{π^{2}}{8}$
$2 + \frac{π^{2}}{4}$
$\frac{π^{2}}{8} - 1$
$4 + \frac{π^{2}}{2}$

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3

4

(2)

We have,

${(x, y) : y \leq π - | x |, y \leq | x \sin x |, y \geq 0}$

Required area

$= 2 [\int_{0}^{π / 2} (x \sin x) d x + \frac{1}{2} \times π / 2 \times π / 2]$

$= 2 [x (- \cos x) |_{0}^{π / 2} + \int_{0}^{π / 2} \cos x d x] + \frac{π^{2}}{4}$

$= 2 + \frac{π^{2}}{4} sq. units$

Q 44 :

The area of the region bounded by the curves $x + 3 y^{2} = 0$ and $x + 4 y^{2} = 1$ is equal to : [2026]

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{4}{3}$
$\frac{5}{3}$

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3

4

(3)

We have, $x + 3 y^{2} = 0$

$\Rightarrow y^{2} = - \frac{x}{3}$ $...(i)$

and $x + 4 y^{2} = 1$

$\Rightarrow 4 y^{2} = 1 - x$

$\Rightarrow y^{2} = - \frac{1}{4} (x - 1)$ $...(ii)$

Solving (i) & (ii), we get

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

$\Rightarrow 4 x = 3 x - 3 \Rightarrow x = - 3 \Rightarrow y = \pm 1$

Area bounded by curves

$=$ $| \int_{- 1}^{1} (1 - 4 y^{2}) - (- 3 y^{2}) |$ $d y$

$= \int_{- 1}^{1} (1 - y^{2}) d y = 2 \int_{0}^{1} (1 - y^{2}) d y$

$= 2 {[y - \frac{y^{3}}{3}]}_{0}^{1} = 2 [1 - \frac{1}{3}] = \frac{4}{3} sq. units$

Q 45 :

The area of the region $R = {(x, y) : x y \leq 27, 1 \leq y \leq x^{2}}$ is equal to: [2026]

$78 \log_{e} 3 - \frac{52}{3}$
$54 \log_{e} 3 - \frac{52}{3}$
$54 \log_{e} 3 - \frac{26}{3}$
$54 \log_{e} 3 + \frac{26}{3}$

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4

(2)

Required area=

$\int_{1}^{3} (x^{2} - 1) d x + \int_{3}^{27} (\frac{27}{x} - 1) d x$

$= {[\frac{x^{3}}{3} - x]}_{1}^{3} + {[27 \ln x - x]}_{3}^{27}$

$= (9 - 3) - (\frac{1}{3} - 1) + (27 \ln 27 - 27) - (27 \ln 3 - 3)$

$= 6 + \frac{2}{3} + 27 (\ln 27 - \ln 3) - 24$

$= \frac{20}{3} + 27 \ln 9 - 24$

$= \frac{20 - 72}{3} + 54 \ln 3$

$= 54 \ln 3 - \frac{52}{3} sq. units$

Q 46 :

The area of the region ${(x, y) : 0 \leq y \leq 6 - x, y^{2} \geq 4 x - 3, x \geq 0}$ is : [2026]

8
9
12
15

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4

(2)

Required area

$= \int_{0}^{3} \frac{y^{2} + 3}{4} d y + \int_{3}^{6} (6 - y) d y$

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

$= \frac{1}{4} [9 + 9] + [(36 - 18) - (18 - \frac{9}{2})]$

$= \frac{18}{4} + 18 - \frac{27}{2} = \frac{9}{2} - \frac{27}{2} + 18 = - 9 + 18$

$= 9 sq. units$

Q 47 :

Let $e$ be the base of natural logarithm and let $f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$ and $g : {1, e, e^{2}, e^{3}} \to {1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $ϕ (x) = {[f^{- 1} {g^{- 1} (\frac{1}{2})}]}^{x}$ , then the area of the region $R {(x, y) : x^{2} \leq y \leq ϕ (x), 0 \leq x \leq 1}$ is : [2026]

$\frac{3 - \log_{e} (2)}{3 \log_{e} (2)}$
$\frac{1}{3 \log_{e} (2)}$
$3 + \log_{e} (2)$
$\frac{3 + \log_{e} (2)}{2 + \log_{e} (3)}$

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2

3

4

(1)

$f : {1, 2, 3, 4} \to {1, e, e^{2}, e^{3}}$

$∵$ $f is strictly decreasing$

$∴$ $f (1) = e^{3}, f (2) = e^{2}, f (3) = e, f (4) = 1$

$g : {1, e, e^{2}, e^{3}} \to {1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}}$

$∵$ $g is strictly increasing$

$∴$ $g (1) = \frac{1}{4}, g (e) = \frac{1}{3}, g (e^{2}) = \frac{1}{2}$ $and g (e^{3}) = 1$

$∵$ $ϕ (x) = {[f^{- 1} (g^{- 1} (\frac{1}{2}))]}^{x}$

$∵$ $g (e^{2}) = \frac{1}{2} \Rightarrow g^{- 1} (\frac{1}{2}) = e^{2}$

$and f (2) = e^{2} \Rightarrow f^{- 1} (e^{2}) = 2$

$∴$ $ϕ (x) = 2^{x}$

$Now, y = 2^{x}, and y = x^{2} from x = 0 to x = 1$

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

$= {(\frac{2^{x}}{\log_{e} 2} - \frac{x^{3}}{3})}_{0}^{1} = \frac{1}{\log_{e} 2} - \frac{1}{3}$

$= \frac{3 - \log_{e} 2}{3 \log_{e} 2}$

Q 48 :

The area of the region ${(x, y) : x^{2} - 8 x \leq y \leq - x}$ is : [2026]

$\frac{343}{6}$
$\frac{637}{6}$
$\frac{437}{6}$
$\frac{523}{6}$

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4

(1)

Given, ${(x, y) : x^{2} - 8 x \leq y \leq - x}$

The given region contains a parabola i.e., $y = x^{2} - 8 x$ and a straight line i.e., $y = - x$

Solving the above two equations, we get

$x^{2} - 8 x = - x$

$\Rightarrow x^{2} - 7 x = 0$

$\Rightarrow x = 0; x = 7$

$Required area = \int_{0}^{7} (- x - (x^{2} - 8 x)) d x$

$= \int_{0}^{7} (7 x - x^{2}) d x = {(\frac{7 x^{2}}{2} - \frac{x^{3}}{3})}_{0}^{7}$

$= \frac{343}{2} - \frac{343}{3} = \frac{343}{6}$

Q 49 :

Let $f : (1, \infty) \to R$ be a function defined as $f (x) = \frac{x - 1}{x + 1}$ . Let $f^{i + 1} (x) = f (f^{i} (x)), i = 1, 2, \dots, 25$ , where $f^{1} (x) = f (x)$ .

If $g (x) + f^{26} (x) = 0, x \in (1, \infty)$ , then the area of the region bounded by the curves $y = g (x), 2 y = 2 x - 3$ , $y = 0$ and $x = 4$ is : [2026]

$\frac{1}{8} + \log_{e} 2$
$\frac{1}{4} + \log_{e} 2$
$\frac{5}{6} + 3 \log_{e} 2$
$\frac{5}{6} + \log_{e} 2$

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4

(1)

$f (x) = \frac{x - 1}{x + 1}$

$f^{2} (x) = f (f (x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{- 1}{x}$

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

$f^{4} (x) = f (f^{3} (x)) = \frac{\frac{x + 1}{1 - x} - 1}{\frac{x + 1}{1 - x} + 1} = x$

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

$This is periodic with period 4.$

$For f^{26} (x) we find remainder of \frac{26}{4} which is 2, so$

$f^{26} (x) = f^{2} (x) = \frac{- 1}{x}$

$Now, g (x) = - f^{26} (x) = \frac{1}{x}$

$Region is bounded by g (x) = \frac{1}{x}; 2 y = 2 x - 3$

$y = 0 and x = 4$

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

$= {(\frac{x^{2}}{2} - \frac{3 x}{2})}_{3 / 2}^{2} + {(\ln x)}_{2}^{4} = \frac{1}{8} + \ln 2$

Q 50 :

If the area of the region bounded by $16 x^{2} - 9 y^{2} = 144$ and $8 x - 3 y = 24$ is $A$ , then $3 (A + 6 \log_{e} (3))$ is equal to _______. [2026]

(24)

$Given, C_{1} : 16 x^{2} - 9 y^{2} = 144$

$\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

$and line L_{1} : 8 x - 3 y = 24$

$\Rightarrow y = \frac{8 x - 24}{3}$

$For point of intersection,$

$\frac{x^{2}}{9} - \frac{{(8 x - 24)}^{2}}{16 \times 9} = 1 \Rightarrow x^{2} - 8 x + 15 = 0$

$\Rightarrow (x - 3) (x - 5) = 0 \Rightarrow x = 3 and x = 5$

$∴$ $A = \int_{3}^{5} (y_{line} - y_{hyperbola}) d x$

$= \int_{3}^{5} (\frac{8 x - 24}{3} - \frac{4}{3} \sqrt{x^{2} - 9}) d x$

$= \frac{4}{3} \int_{3}^{5} (2 x - 6 - \sqrt{x^{2} - 9}) d x$

$= \frac{4}{3} {(x^{2} - 6 x - \frac{x}{2} \sqrt{x^{2} - 9} + \frac{9}{2} \ln | x + \sqrt{x^{2} - 9} |)}_{3}^{5}$

$= \frac{4}{3} [(25 - 30 - \frac{5}{2} \sqrt{25 - 9} + \frac{9}{2} \ln | 5 + \sqrt{16} |) - (9 - 18 + \frac{9}{2} \ln 3)]$

$= \frac{4}{3} [- 6 + \frac{9}{2} \ln | 9 | - \frac{9}{2} \ln 3]$

$= \frac{4}{3} (- 6 + 9 \ln 3 - \frac{9}{2} \ln 3)$

$\Rightarrow A = \frac{4}{3} (- 6 + \frac{9}{2} \ln 3)$

$\Rightarrow A = - 8 + 6 \ln 3 \Rightarrow$ $| A |$ $= 8 - 6 \ln 3$

$∴$ $3 (A + 6 \ln 3) = 3 (8 - 6 \ln 3 + 6 \ln 3) = 24$

Q 51 :

Let $f : R \to R$ be a function such that $f (x) + 3 f (\frac{π}{2} - x) = \sin x, x \in R .$

Let the maximum value of $f$ on $R$ be $α$ . If the area of the region bounded by the curves $g (x) = x^{2}$ and $h (x) = β x^{3}, β > 0$ , is $α^{2}$ , then $30 β^{3}$ is equal to ______ [2026]

(16)

We have,

$f (x) + 3 f (\frac{π}{2} - x) = \sin x$ $...(i)$

Put $x \to \frac{π}{2} - x$ ,

$f (\frac{π}{2} - x) + 3 f (x) = \cos x$ $...(ii)$

From (i) and (ii), we get

$f (x) = \frac{1}{8} [3 \cos x - \sin x]$

$f {(x)}_{\max} = \frac{\sqrt{10}}{8} = α$

Intersection points of $g (x)$ and $h (x)$ are $(0, 0)$ and $(\frac{1}{β}, \frac{1}{β^{2}})$

Required area

$=$ $| \int_{0}^{1 / β} (x^{2} - β x^{3}) d x |$ $= \frac{1}{12 β^{3}} = α^{2} (Given)$

Thus, $30 β^{3} = 16$

Topic Question Set

Q 41 :

Twice the area (in sq. units) of the region ${(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y$ and $y \leq 1 + \sqrt{x}}$ is

Q 42 :

Area of the region bounded by $y = x$ and $y = x^{2}$ is _____

Q 43 :

The area of the region ${(x, y) : y \leq π - | x |, y \leq | x \sin x |, y \geq 0}$ is: [2026]

Q 44 :

The area of the region bounded by the curves $x + 3 y^{2} = 0$ and $x + 4 y^{2} = 1$ is equal to : [2026]

Q 45 :

The area of the region $R = {(x, y) : x y \leq 27, 1 \leq y \leq x^{2}}$ is equal to: [2026]

Q 46 :

The area of the region ${(x, y) : 0 \leq y \leq 6 - x, y^{2} \geq 4 x - 3, x \geq 0}$ is : [2026]

Q 48 :

The area of the region ${(x, y) : x^{2} - 8 x \leq y \leq - x}$ is : [2026]

Q 50 :

If the area of the region bounded by $16 x^{2} - 9 y^{2} = 144$ and $8 x - 3 y = 24$ is $A$ , then $3 (A + 6 \log_{e} (3))$ is equal to _______. [2026]

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Topic Question Set

Q 41 : Twice the area (in sq. units) of the region {(x,y):x≥0, x+y≤3, x2≤4y and y≤1+x} is

Q 42 : Area of the region bounded by y=x and y=x2 is _____

Q 43 : The area of the region {(x,y):y≤π-|x|,y≤|xsinx|,y≥0} is: [2026]

Q 44 : The area of the region bounded by the curves x+3y2=0 and x+4y2=1 is equal to : [2026]

Q 45 : The area of the region R={(x,y):xy≤27,1≤y≤x2} is equal to: [2026]

Q 46 : The area of the region {(x,y):0≤y≤6-x,y2≥4x-3,x≥0} is : [2026]

Q 48 : The area of the region {(x,y):x2-8x≤y≤-x} is : [2026]

Q 49 : Let f:(1,∞)→R be a function defined as f(x)=x-1x+1. Let fi+1(x)=f(fi(x)),i=1,2,…,25, where f1(x)=f(x). If g(x)+f26(x)=0, x∈(1,∞), then the area of the region bounded by the curves y=g(x), 2y=2x-3, y=0 and x=4 is : [2026]

Q 50 : If the area of the region bounded by 16x2-9y2=144 and 8x-3y=24 is A, then 3(A+6loge(3)) is equal to _______. [2026]

Q 51 : Let f:R→R be a function such that f(x)+3f(π2-x)=sinx, x∈R. Let the maximum value of f on R be α. If the area of the region bounded by the curves g(x)=x2 and h(x)=βx3, β>0, is α2, then 30β3 is equal to ______ [2026]

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Q 41 :

Twice the area (in sq. units) of the region ${(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y$ and $y \leq 1 + \sqrt{x}}$ is

Q 42 :

Area of the region bounded by $y = x$ and $y = x^{2}$ is _____

Q 43 :

The area of the region ${(x, y) : y \leq π - | x |, y \leq | x \sin x |, y \geq 0}$ is: [2026]

Q 44 :

The area of the region bounded by the curves $x + 3 y^{2} = 0$ and $x + 4 y^{2} = 1$ is equal to : [2026]

Q 45 :

The area of the region $R = {(x, y) : x y \leq 27, 1 \leq y \leq x^{2}}$ is equal to: [2026]

Q 46 :

The area of the region ${(x, y) : 0 \leq y \leq 6 - x, y^{2} \geq 4 x - 3, x \geq 0}$ is : [2026]

Q 48 :

The area of the region ${(x, y) : x^{2} - 8 x \leq y \leq - x}$ is : [2026]

Q 50 :

If the area of the region bounded by $16 x^{2} - 9 y^{2} = 144$ and $8 x - 3 y = 24$ is $A$ , then $3 (A + 6 \log_{e} (3))$ is equal to _______. [2026]