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

Q 21 :

Let A be the area bounded by the curve $y = x$ $| x - 3 |$ , the $x$ -axis and the ordinates $x = - 1$ and $x = 2$ . Then 12A is equal to _____. [2023]

(62)

$y = x$ $| x - 3 |$ $= x (3 - x), when - 1 \leq x \leq 2$

$Required area, A =$ $| \int_{- 1}^{0} (3 x - x^{2}) d x |$ $+ \int_{0}^{2} (3 x - x^{2}) d x$

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

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

$= \frac{11}{6} + \frac{10}{3}$

$Now 12 A = 12 (\frac{11}{6} + \frac{10}{3}) = 22 + 40 = 62$

Q 22 :

If the area of the region bounded by the curves $y^{2} - 2 y = - x$ and $x + y = 0$ is A, then 8A is equal to _________ . [2023]

(36)

Required area $= \int_{0}^{3} [(2 y - y^{2}) - (- y)] d y$

$= \int_{0}^{3} (3 y - y^{2}) d y$

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

$= \frac{27}{2} - \frac{27}{3} = \frac{9}{2} sq. units$

$∴$ $8 A = 8 \times \frac{9}{2} = 36$

Q 23 :

If the area enclosed by the parabolas $P_{1} : 2 y = 5 x^{2}$ and $P_{2} : x^{2} - y + 6 = 0$ is equal to the area enclosed by $P_{1}$ and $y = α x, α > 0$ , then $α^{3}$ is equal to _______ . [2023]

(600)

Given, $P_{1} : 2 y = 5 x^{2}, P_{2} : x^{2} - y + 6 = 0 and y = α x$

The point of intersection of $P_{1}$ and $P_{2}$ is given by

$2 y = 5 (y - 6) \Rightarrow 2 y = 5 y - 30 \Rightarrow y = 10$

Put $y = 10$ in $P_{1}$ , we get $2 (10) = 5 x^{2}$

$\Rightarrow 5 x^{2} = 20 \Rightarrow x = \pm 2$

Thus $P_{1}$ and $P_{2}$ intersect at $(- 2, 10)$ and $(2, 10)$ .

$Area = 2 \int_{0}^{2} (x^{2} + 6 - \frac{5 x^{2}}{2}) d x$ $= 2 {[\frac{x^{3}}{3} + 6 x - \frac{5 x^{3}}{6}]}_{0}^{2}$

$= 2 [(\frac{8}{3} + 12 - 5 \times \frac{8}{6}) - 0] = 2 (8) = 16 sq. units$

Now, the point of intersection of $2 y = 5 x^{2}$ and $y = α x$ , is given by

$\Rightarrow 2 α x = 5 x^{2} \Rightarrow x = \frac{2 α}{5}$

$Area = \int_{0}^{2 α / 5} (α x - \frac{5 x^{2}}{2}) d x = {[\frac{α x^{2}}{2} - \frac{5 x^{3}}{6}]}_{0}^{2 α / 5}$

$= \frac{α}{2} (\frac{4 α^{2}}{25}) - \frac{5}{6} (\frac{8 α^{3}}{125}) = \frac{2 α^{3}}{25} - \frac{4 α^{3}}{75} = \frac{2 α^{3}}{75}$

$\frac{2 α^{3}}{75} = 16 [Given]$

$\Rightarrow α^{3} = 75 \times 8 = 600$

Q 24 :

Let $α$ be the area of the larger region bounded by the curve $y^{2} = 8 x$ and the lines $y = x$ and $x = 2$ , which lies in the first quadrant. Then the value of $3 α$ is equal to _____ . [2023]

(22)

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

The shaded region has the largest area. So,

$Area = \int_{2}^{8} (2 \sqrt{2} \sqrt{x} - x) d x$

$α = 2 \sqrt{2} \int_{2}^{8} x^{1 / 2} d x - \int_{2}^{8} x d x$

$= 2 \sqrt{2} \times \frac{2}{3} {[x^{3 / 2}]}_{2}^{8} - {[\frac{x^{2}}{2}]}_{2}^{8}$

$= \frac{4 \sqrt{2}}{3} [16 \sqrt{2} - 2 \sqrt{2}] - \frac{1}{2} [64 - 4] = \frac{112}{3} - 30 = \frac{22}{3}$

$Now 3 α = 22$

Q 25 :

Let A be the area of the region ${(x, y) : y \geq x^{2}, y \geq {(1 - x)}^{2}, y \leq 2 x (1 - x)}$ . Then 540A is equal to ___________ . [2023]

(25)

Region ${(x, y) : y \geq x^{2}, y \geq {(1 - x)}^{2}, y \leq 2 x (1 - x)}$

$y = x^{2}$ $\dots (i)$

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

$y = 2 x (1 - x)$ $\dots (i i i)$

Solving (ii) and (iii), we get

$x = \frac{1}{3} and x = 1$

$∴$ $Required area A = 2 \int_{1 / 3}^{1 / 2} [2 x - 2 x^{2} - {(1 - x)}^{2}] d x$

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

$= 2 \int_{1 / 3}^{1 / 2} (- 3 x^{2} + 4 x - 1) d x$ $= {[- x^{3} + 2 x^{2} - x]}_{1 / 3}^{1 / 2} = \frac{5}{108}$

$∴$ $540 A = 540 \times \frac{5}{108} = 25$

Q 26 :

Let for $x \in R,$ $f (x) = \frac{x + | x |}{2}$ and $g (x) =$ ${\begin{matrix} x, & x < 0 \\ x^{2}, & x \geq 0 \end{matrix}$ . Then area bounded by the curve $y = (f o g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to _______ . [2023]

(72)

Here, $f (x) = \frac{x + | x |}{2} =$ ${\begin{matrix} x, & x \geq 0 \\ 0, & x < 0 \end{matrix}$

$g (x) =$ ${\begin{matrix} x^{2}, & x \geq 0 \\ x, & x < 0 \end{matrix}$

$f o g (x) = f (g (x)) =$ ${\begin{matrix} g (x), & g (x) \geq 0 \\ 0, & g (x) < 0 \end{matrix}$ $=$ ${\begin{matrix} x^{2}, & x \geq 0 \\ 0, & x < 0 \end{matrix}$

$Required area = A_{1} + A_{2}$

$= \int_{0}^{3} (\frac{x + 15}{2} - x^{2}) d x + \frac{1}{2} \times \frac{15}{2} \times 15$

$= {[\frac{x^{2}}{4} + \frac{15 x}{2} - \frac{x^{3}}{3}]}_{0}^{3} + \frac{225}{4} = \frac{9}{4} + \frac{45}{2} - 9 + \frac{225}{4} = \frac{288}{4} = 72$

Q 27 :

Let the area of the region ${(x, y) : | 2 x - 1 | \leq y \leq | x^{2} - x |, 0 \leq x \leq 1}$ be A. Then ${(6 A + 11)}^{2}$ is equal to __________. [2023]

(125)

$y \geq$ $| 2 x - 1 |$ $,$ $y \leq$ $| x^{2} - x |$

Both curves are symmetric about $x = \frac{1}{2}$

Hence

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

$A = 2 \int_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}} (- x^{2} + 3 x - 1) d x = 2 {[\frac{- x^{3}}{3} + \frac{3}{2} x^{2} - x]}_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}}$

$= \frac{5 \sqrt{5} - 11}{6}$

$On solving, 6 A + 11 = 5 \sqrt{5}, {(6 A + 11)}^{2} = 125$

Q 28 :

Let the area of the region bounded by the curve $y = \max {\sin x, \cos x},$ lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be A. Then $A + A^{2}$ is equal to ________ . [2026]

(12)

$A = \int_{0}^{π / 4} \cos x d x + \int_{π / 4}^{π} \sin x d x + \int_{π}^{5 π / 4} - \sin x d x + \int_{5 π / 4}^{3 π / 2} - \cos x d x$

$A = {(\sin x)}_{0}^{π / 4} + {(\cos x)}_{π / 4}^{π} + {(\cos x)}_{π}^{5 π / 4} + (\sin x)_{5 π / 4}^{3 π / 2}$

$A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 3$

$A^{2} + A = 12$

Q 29 :

The area of the region inside the ellipse $x^{2} + 4 y^{2} = 4$ and outside the region bounded by the curves $y = | x | - 1 and y = 1 - | x |$ is [2026]

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

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(2)

$Required area = area of ellipse - shaded area$

$= π \times 2 \times 1 - 4 (\frac{1}{2} \times 1 \times 1) = 2 π - 2$

Q 30 :

Let $A_{1}$ be the bounded area enclosed by the curves $y = x^{2} + 2,$ $x + y = 8$ and $y$ -axis that lies in the first quadrant. Let $A_{2}$ be the bounded area enclosed by the curves $y = x^{2} + 2$ , $y^{2} = x$ , $x = 2$ , and $y$ -axis that lies in the first quadrant. Then $A_{1} - A_{2}$ is equal to [2026]

$\frac{2}{3} (\sqrt{2} + 1)$
$\frac{2}{3} (4 \sqrt{2} + 1)$
$\frac{2}{3} (3 \sqrt{2} + 1)$
$\frac{2}{3} (2 \sqrt{2} + 1)$

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(4)

$A_{1} = \int_{0}^{2} ((8 - x) - (x^{2} + 2)) d x$

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

$A_{1} {(6 x - \frac{x^{2}}{2} - \frac{x^{3}}{3})}_{0}^{2}$ $= 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$

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

$A_{2} = {(\frac{x^{3}}{3} + 2 x)}_{0}^{2} - \frac{4 \sqrt{2}}{3}$

$A_{2} = \frac{8}{3} + 4 - \frac{4 \sqrt{2}}{3} = \frac{20}{3} - \frac{4 \sqrt{2}}{3}$

$A_{1} - A_{2} = \frac{2}{3} + \frac{4 \sqrt{2}}{3}$

Q 31 :

Let the line $x = - 1$ divide the area of the region ${(x, y) : 1 + x^{2} \leq y \leq 3 - x}$ in the ratio $m : n$ , $\gcd (m, n) = 1$ . Then $m + n$ is equal to [2026]

26
25
28
27

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3

4

(4)

$\frac{m}{n} = \frac{\int_{- 1}^{1} [(3 - x) - (1 + x^{2})] d x}{\int_{- 2}^{1} [(3 - x) - (1 + x^{2})] d x} = \frac{20}{7}$

$∴$ $m + n = 20 + 7$

$= 27$

Q 32 :

Let $P_{1} : y = 4 x^{2} and 𝑃_{2} : 𝑦 = 𝑥^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1} and P_{2}$ is six times the area of the bounded region enclosed between the line $y = α x, α > 0$ and $P_{1}$ , then $α$ is equal to : [2026]

15
12
8
6

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(2)

$Area bounded between P_{1} & P_{2} is$

$\int_{- 3}^{3} ((x^{2} + 27) - (4 x^{2})) d x$

$(P.O.I. of P_{1} & P_{2} is x = \pm 3)$

$= 2 \int_{0}^{3} (27 - 3 x^{2}) d x = 2 {[27 x - x^{3}]}_{0}^{3}$

$= 2 [81 - 27] = 108$

$∴$ $Area bounded between P_{1} & L is 18 sq. units$

$(Area between x^{2} = 4 a y & line x = m y is \frac{8 a^{2}}{3 m^{3}})$

$∴$ $Area between x^{2} = \frac{y}{4} & x = \frac{y}{α} is$

$\frac{8 {(\frac{1}{16})}^{2}}{3 {(\frac{1}{α})}^{3}} = 18$

$\Rightarrow \frac{\frac{8}{16 \cdot 16}}{\frac{3}{α^{3}}} = 18 \Rightarrow α^{3} = 2^{6} \cdot 3^{3}$

$\Rightarrow α = 12$

Q 33 :

Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y =$ $| α x - 5 |$ $-$ $| 1 - α x |$ $+ α x^{2} .$

Then $(f (0) + f (1))$ is equal to [2026]

9
12
7
14

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(3)

$at α = 0 \Rightarrow f (0)$

$x = 0, x = 1, y^{2} = x$

$y =$ $| 0 \cdot x - 5 |$ $-$ $| 1 - 0 \cdot x |$ $+ 0 \cdot x^{2}$

$y = 4$

$A_{1} = \int_{0}^{1} (4 - \sqrt{x}) d x$

$\begin{matrix} = \end{matrix}$ $4 x - \frac{x^{3 / 2}}{3 / 2} |_{0}^{1}$

$= 4 - \frac{2}{3} (1) = \frac{10}{3}$

$at α = 1 \Rightarrow f (1)$

$x = 0, x = 1, y^{2} = x$

$y =$ $| x - 5 | - | 1 - x | + x^{2}, x \in (0, 1)$

$y = 5 - x - (1 - x) + x^{2}$

$y = 4 + x^{2}$

$A_{2} = \int_{0}^{1} [(4 + x^{2}) - \sqrt{x}] d x$

$=$ $4 x + \frac{x^{3}}{3} - \frac{x^{\frac{3}{2}}}{\frac{3}{2}} |_{0}^{1}$

$= 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$

$| f (0) + f (1) |$ $=$ $| A_{1} + A_{2} |$ $=$ $| \frac{10}{3} + \frac{11}{3} |$ $=$ $| \frac{21}{3} |$ $= 7$

$option (3)$

Q 34 :

The area of the region enclosed between the circles $x^{2} + y^{2} = 4$ and $x^{2} + {(y - 2)}^{2} = 4$ is: [2026]

$\frac{2}{3} (4 π - 3 \sqrt{3})$
$\frac{4}{3} (2 π - \sqrt{3})$
$\frac{4}{3} (2 π - 3 \sqrt{3})$
$\frac{2}{3} (2 π - 3 \sqrt{3})$

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(1)

$A = 2 \int_{0}^{\sqrt{3}} [\sqrt{4 - x^{2}} - (2 - \sqrt{4 - x^{2}})] d x$

$= 2 \int_{0}^{\sqrt{3}} (2 \sqrt{4 - x^{2}} - 2) d x$

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

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

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

$= \frac{8 π}{3} - 2 \sqrt{3} (Sq. units)$

Q 35 :

If the area of the region ${(x, y) : 1 - 2 x \leq y \leq 4 - x^{2}, x \geq 0, y \geq 0}$ is $\frac{α}{β}, α, β \in ℕ, \gcd (α, β) = 1$ , then the value of $(α + β)$ is: [2026]

91
73
85
67

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(2)

$Required area = \frac{2}{3} \times 8 - \frac{1}{2} \times \frac{1}{2} \times 1$

$= \frac{16}{3} - \frac{1}{4} = \frac{61}{12} = \frac{α}{β}$

$\Rightarrow α + β = 73$

Q 36 :

The area of the region $A = {(x, y) : 4 x^{2} + y^{2} \leq 8 and y^{2} \leq 4 x}$ is: [2026]

$\frac{π}{2} + 2$
$\frac{π}{2} + \frac{1}{3}$
$π + \frac{2}{3}$
$π + 4$

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4

(3)

$A = \int_{0}^{2} 2 \sqrt{x} d x + 2 \int_{1}^{\sqrt{2}} \sqrt{8 - 4 x^{2}} d x$

$= \frac{8}{3} [x^{3 / 2}] \int_{0}^{1} + 4 \int_{1}^{\sqrt{2}} \sqrt{2 - x^{2}} d x$

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

$= \frac{8}{3} + 2 [2 \times \frac{π}{2} - 1 - 2 \times \frac{π}{4}]$

$= \frac{8}{3} + 2 π - 2 - π$

$= π + \frac{2}{3} sq. units$

Q 37 :

The area of the region $R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}$ is [2026]

$\frac{1}{3} (49 \log_{e} (2) - 15)$
$\frac{2}{3} (20 \log_{e} (2) + 9)$
$\frac{2}{3} (24 \log_{e} (2) - 7)$
$\frac{1}{3} (40 \log_{e} (2) + 27)$

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(3)

$A = \int_{1}^{2} (x^{2} - 1) d x + \int_{2}^{8} (\frac{8}{x} - 1) d x$

$A = 8 \log_{e} 4 - \frac{14}{3} = 16 \log_{e} 2 - \frac{14}{3}$

$= \frac{2}{3} (24 \log_{e} 2 - 7)$

Q 38 :

The area of smaller region bounded by the curve $9 x^{2} + 4 y^{2} - 36 x + 16 y + 16 = 0$ and the line $3 x + 2 y = 8$ is

$\frac{3}{2} (π + 2)$
$3 (π - 2)$
$\frac{3}{4} (π - 2)$
$\frac{3}{2} (π - 2)$

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(4)

$\frac{π}{4} a b - \frac{1}{2} a b$

where a, b are length of semi minor and major axis.

Q 39 :

Let S be the region bounded by the curves $y = x^{3}$ and $y^{2} = x .$ The curve $y = 2$ $| x |$ divides S into two regions of areas $R_{1}$ and $R_{2} .$ If $\max {R_{1}, R_{2}} = R_{2}$ ,then $\frac{R_{2}}{R_{1}} =$

(19)

$= \int_{0}^{1} \sqrt{x} d x - \frac{3}{4}$

$= {[\frac{2 x^{3 / 2}}{3} - \frac{x^{4}}{4}]}_{1}^{0} = \frac{5}{12}$

$R_{1} = \int_{0}^{1 / 4} (\sqrt{x} - 2 x) d x$

$= {[\frac{2 x^{3 / 2}}{3} - x^{2}]}_{0}^{1 / 4} = \frac{1}{48}$

$∴$ $R_{2} = \frac{19}{48}$

$So, \frac{R_{2}}{R_{1}} = 19$

Q 40 :

The area of the region bounded by the curve $x = 2 y + 3$ and the lines $y = 1, y = - 1$ is:

4 sq. units
$\frac{3}{2}$ sq. units
6 sq. units
8 sq. units

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4

(3)

Ans. 6 sq. units

Explanation :

Required area, $A = \int_{- 1}^{1} (2 y + 3) d y$

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

$= {[y^{2} + 3 y]}_{- 1}^{1}$

$= (1 + 3) - (1 - 3)$

$= 4 - (- 2)$

$= 6 sq. units.$

Q 41 :

The area of the region bounded by the curves $y =$ $| x - 2 |$ $,$ $x = 1,$ $x = 3$ and the $X$ -axis is:

4 sq. units
2 sq. units
3 sq. units
1 sq. unit

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(4)

Ans. 1 sq. unit

Explanation:

Required area, $A = \int_{1}^{3}$ $| x - 2 |$ $d x$

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

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

$= \frac{1}{2} + \frac{1}{2} = 1 sq. unit.$

Q 42 :

The area bounded by the curve $y = x$ , $X$ -axis and ordinates $x = - 1$ to $x = 2$ is:

0
$\frac{1}{2}$ sq. unit
$\frac{3}{2}$ sq. units
$\frac{5}{2}$ sq. units

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(4)

Ans. $\frac{5}{2}$ sq. units

Explanation :

Bounded area, $=$ $| \int_{- 1}^{0} x d x |$ $+$ $| \int_{0}^{2} x d x |$

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

Q 43 :

The area of the region bounded by the line $2 y + x = 8$ , $X$ -axis and the lines $x = 2$ and $x = 4$ is:

5 sq. units
7 sq. units
9 sq. units
10 sq. units

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(1)

Ans. 5 sq. units

Explanation:

Required area = Area of the shaded region

$= \int_{x = 2}^{4} y d x$

$= \int_{2}^{4} \frac{1}{2} (8 - x) d x$

$= \frac{1}{2} {[8 x - \frac{x^{2}}{2}]}_{2}^{4}$

$= \frac{1}{2} [(8 \cdot (4) - \frac{4^{2}}{2}) - (8 \cdot (2) - \frac{2^{2}}{2})]$

$= 5 sq. unit.$

Q 44 :

The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2, x = 3$ is:

$\frac{7}{2}$ sq. units
$\frac{9}{2}$ sq. units
$\frac{11}{2}$ sq. units
$\frac{13}{2}$ sq. units

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(1)

Ans. $\frac{7}{2}$ sq. units

Explanation :

Required area, $A = \int_{2}^{3} (x + 1) d x = {[\frac{x^{2}}{2} + x]}_{2}^{3}$

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

$= [\frac{5}{2} + 1] = \frac{7}{2} sq. units.$

Q 45 :

Area of the region in the first quadrant enclosed by the $X$ -axis, the line $y = x$ and the circle $x^{2} + y^{2} = 32$ is:

$16 π$ sq. units
$4 π$ sq. units
$32 π$ sq. units
$24 π$ sq. units

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(2)

Ans. $4 π$ sq. units

Explanation:

We have, area enclosed by $X$ -axis i.e., $y = 0$ , $y = x$ and the circle $x^{2} + y^{2} = 32$ in first quadrant.

Since, $x^{2} + {(x)}^{2} = 32$ $[∵ y = x]$

$\Rightarrow$ $2 x^{2} = 32$

$\Rightarrow$ $x = \pm 4$

So, the intersection point of circle $x^{2} + y^{2} = 32$ and line $y = x$ are (4, 4) or $(- 4, - 4)$ .

and $x^{2} + y^{2} = {(4 \sqrt{2})}^{2}$

Since, $y = 0$

$∴$ $x^{2} + {(0)}^{2} = 32$

$\Rightarrow$ $x = \pm 4 \sqrt{2}$

So, the circle intersects the $X$ -axis at $(\pm 4 \sqrt{2}, 0)$ .

Area of shaded region

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

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

$= \frac{16}{2} + [\frac{4 \sqrt{2}}{2} \cdot 0 + 16 \sin^{- 1} (\frac{4 \sqrt{2}}{4 \sqrt{2}}) - \frac{4}{2} \sqrt{{(4 \sqrt{2})}^{2} - 16} - 16 \sin^{- 1} (\frac{4}{4 \sqrt{2}})]$

$= 8 + [16 \cdot \frac{π}{2} - 2 \sqrt{16} - 16 \cdot \frac{π}{4}]$

$= 8 + [8 π - 8 - 4 π] = 4 π sq. units.$

Q 46 :

The area of the region bounded by the curve $y = \sqrt{16 - x^{2}}$ and $X$ -axis is:

$8 π$ sq. units
$20 π$ sq. units
$16 π$ sq. units
$256 π$ sq. units

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(1)

Ans. $8 π$ sq. units

Explanation:

Given equation of curve is $y = \sqrt{16 - x^{2}}$ and the equation of line is $X$ -axis i.e., $y = 0 .$

$∴$ $\sqrt{16 - x^{2}} = 0$ ...(i)

$\Rightarrow$ $16 - x^{2} = 0$

$\Rightarrow$ $x^{2} = 16 \Rightarrow x = \pm 4$

So, the intersection points are (4, 0) and $(- 4, 0)$ .

Therefore, area of the curve,

$A = \int_{- 4}^{4} \sqrt{16 - x^{2}} d x$

$= \int_{- 4}^{4} \sqrt{4^{2} - x^{2}} d x$

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

$= [\frac{4}{2} \sqrt{4^{2} - 4^{2}} + 8 \sin^{- 1} (\frac{4}{4})] - [- \frac{4}{2} \sqrt{4^{2} - {(- 4)}^{2}} + 8 \sin^{- 1} (- \frac{4}{4})]$

$= [2 \cdot 0 + 8 \cdot \frac{π}{2} - 0 + 8 \cdot \frac{π}{2}] = 8 π sq. units.$

Q 47 :

The area of the region bounded by the circle $x^{2} + y^{2} = 1$ is:

$2 π$ sq. units
$π$ sq. units
$3 π$ sq. units
$4 π$ sq. units

1

2

3

4

(2)

Ans. $π$ sq. units

Explanation:

We have, $x^{2} + y^{2} = 1^{2}$ $[∵ r = \pm 1]$

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

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

Therefore, area enclosed by the circle,

$= 2 \int_{- 1}^{1} \sqrt{1^{2} - x^{2}} d x$ $[∵ \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x]$

$= 2 \cdot 2 \int_{0}^{1} \sqrt{1^{2} - x^{2}} d x$

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

$= 4 [\frac{1}{2} \cdot 0 + \frac{1}{2} \cdot \frac{π}{2} - 0 - \frac{1}{2} \cdot 0]$

$= 4 \cdot \frac{π}{4} = π sq. units.$

Q 48 :

Area lying in the first quadrant and bounded by the circle $x^{2} + y^{2} = 4$ and the lines $x = 0$ and $x = 2$ is:

$π$ sq. units
$\frac{π}{2}$ sq. units
$\frac{π}{3}$ sq. units
$\frac{π}{4}$ sq. units

1

2

3

4

(1)

Ans. $π$ sq. units

Explanation:

The area bounded by the circle and the lines, $x = 0$ and $x = 2$ , in the first quadrant is represented as shaded region.

$Area O A B = \int_{0}^{2} y d x$

$= \int_{0}^{2} \sqrt{4 - x^{2}} d x$

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

$= 2 (\frac{π}{2}) = π$

Q 49 :

The area of smaller part between the circle $x^{2} + y^{2} = 4$ and the line $x = 1$ is:

$\frac{4 π}{3} - \sqrt{3}$ sq. units
$\frac{8 π}{3} - \sqrt{3}$ sq. units
$\frac{4 π}{3} + \sqrt{3}$ sq. units
$\frac{5 π}{3} + \sqrt{3}$ sq. units

1

2

3

4

(1)

Ans. $\frac{4 π}{3} - \sqrt{3}$ sq. units

Explanation :

$Area of smaller part = 2 \int_{1}^{2} \sqrt{4 - x^{2}} d x$

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

$= 2 [2 \cdot \frac{π}{2} - (\frac{\sqrt{3}}{2} + 2 \cdot \frac{π}{6})]$

$= 2 [π - (\frac{\sqrt{3}}{2} + \frac{π}{3})] = \frac{4 π}{3} - \sqrt{3} sq. units.$

Q 50 :

The area enclosed by the circle $x^{2} + y^{2} = 2$ is equal to:

$4 π$ sq. units
$2 \sqrt{2} π$ sq. units
$4 π^{2}$ sq. units
$2 π$ sq. units

1

2

3

4

(4)

Ans. $2 π$ sq. units

Explanation :

$∵$ $Area = 4 \int_{0}^{\sqrt{2}} \sqrt{2 - x^{2}} d x$

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

$= 4 [0 + \frac{π}{2}]$

$= 2 π sq. units.$

Topic Question Set

Q 21 :

Let A be the area bounded by the curve $y = x$ $| x - 3 |$ , the $x$ -axis and the ordinates $x = - 1$ and $x = 2$ . Then 12A is equal to _____. [2023]

Q 22 :

If the area of the region bounded by the curves $y^{2} - 2 y = - x$ and $x + y = 0$ is A, then 8A is equal to _________ . [2023]

Q 23 :

If the area enclosed by the parabolas $P_{1} : 2 y = 5 x^{2}$ and $P_{2} : x^{2} - y + 6 = 0$ is equal to the area enclosed by $P_{1}$ and $y = α x, α > 0$ , then $α^{3}$ is equal to _______ . [2023]

Q 24 :

Let $α$ be the area of the larger region bounded by the curve $y^{2} = 8 x$ and the lines $y = x$ and $x = 2$ , which lies in the first quadrant. Then the value of $3 α$ is equal to _____ . [2023]

Q 25 :

Let A be the area of the region ${(x, y) : y \geq x^{2}, y \geq {(1 - x)}^{2}, y \leq 2 x (1 - x)}$ . Then 540A is equal to ___________ . [2023]

Q 26 :

Let for $x \in R,$ $f (x) = \frac{x + | x |}{2}$ and $g (x) =$ ${\begin{matrix} x, & x < 0 \\ x^{2}, & x \geq 0 \end{matrix}$ . Then area bounded by the curve $y = (f o g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to _______ . [2023]

Q 27 :

Let the area of the region ${(x, y) : | 2 x - 1 | \leq y \leq | x^{2} - x |, 0 \leq x \leq 1}$ be A. Then ${(6 A + 11)}^{2}$ is equal to __________. [2023]

Q 28 :

Let the area of the region bounded by the curve $y = \max {\sin x, \cos x},$ lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be A. Then $A + A^{2}$ is equal to ________ . [2026]

Q 29 :

The area of the region inside the ellipse $x^{2} + 4 y^{2} = 4$ and outside the region bounded by the curves $y = | x | - 1 and y = 1 - | x |$ is [2026]

Q 31 :

Let the line $x = - 1$ divide the area of the region ${(x, y) : 1 + x^{2} \leq y \leq 3 - x}$ in the ratio $m : n$ , $\gcd (m, n) = 1$ . Then $m + n$ is equal to [2026]

Q 32 :

Let $P_{1} : y = 4 x^{2} and 𝑃_{2} : 𝑦 = 𝑥^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1} and P_{2}$ is six times the area of the bounded region enclosed between the line $y = α x, α > 0$ and $P_{1}$ , then $α$ is equal to : [2026]

Q 33 :

Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y =$ $| α x - 5 |$ $-$ $| 1 - α x |$ $+ α x^{2} .$

Then $(f (0) + f (1))$ is equal to [2026]

Q 34 :

The area of the region enclosed between the circles $x^{2} + y^{2} = 4$ and $x^{2} + {(y - 2)}^{2} = 4$ is: [2026]

Q 35 :

If the area of the region ${(x, y) : 1 - 2 x \leq y \leq 4 - x^{2}, x \geq 0, y \geq 0}$ is $\frac{α}{β}, α, β \in ℕ, \gcd (α, β) = 1$ , then the value of $(α + β)$ is: [2026]

Q 36 :

The area of the region $A = {(x, y) : 4 x^{2} + y^{2} \leq 8 and y^{2} \leq 4 x}$ is: [2026]

Q 37 :

The area of the region $R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}$ is [2026]

Q 38 :

The area of smaller region bounded by the curve $9 x^{2} + 4 y^{2} - 36 x + 16 y + 16 = 0$ and the line $3 x + 2 y = 8$ is

Q 39 :

Let S be the region bounded by the curves $y = x^{3}$ and $y^{2} = x .$ The curve $y = 2$ $| x |$ divides S into two regions of areas $R_{1}$ and $R_{2} .$ If $\max {R_{1}, R_{2}} = R_{2}$ ,then $\frac{R_{2}}{R_{1}} =$

Q 40 :

The area of the region bounded by the curve $x = 2 y + 3$ and the lines $y = 1, y = - 1$ is:

Q 41 :

The area of the region bounded by the curves $y =$ $| x - 2 |$ $,$ $x = 1,$ $x = 3$ and the $X$ -axis is:

Q 42 :

The area bounded by the curve $y = x$ , $X$ -axis and ordinates $x = - 1$ to $x = 2$ is:

Q 43 :

The area of the region bounded by the line $2 y + x = 8$ , $X$ -axis and the lines $x = 2$ and $x = 4$ is:

Q 44 :

The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2, x = 3$ is:

Q 45 :

Area of the region in the first quadrant enclosed by the $X$ -axis, the line $y = x$ and the circle $x^{2} + y^{2} = 32$ is:

Q 46 :

The area of the region bounded by the curve $y = \sqrt{16 - x^{2}}$ and $X$ -axis is:

Q 47 :

The area of the region bounded by the circle $x^{2} + y^{2} = 1$ is:

Q 48 :

Area lying in the first quadrant and bounded by the circle $x^{2} + y^{2} = 4$ and the lines $x = 0$ and $x = 2$ is:

Q 49 :

The area of smaller part between the circle $x^{2} + y^{2} = 4$ and the line $x = 1$ is:

Q 50 :

The area enclosed by the circle $x^{2} + y^{2} = 2$ is equal to:

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Topic Question Set

Q 21 : Let A be the area bounded by the curve y=x|x-3|, the x-axis and the ordinates x=-1 and x=2. Then 12A is equal to _____. [2023]

Q 22 : If the area of the region bounded by the curves y2-2y=-x and x+y=0 is A, then 8A is equal to _________ . [2023]

Q 23 : If the area enclosed by the parabolas P1:2y=5x2 and P2:x2-y+6=0 is equal to the area enclosed by P1 and y=αx, α>0, then α3 is equal to _______ . [2023]

Q 24 : Let α be the area of the larger region bounded by the curve y2=8x and the lines y=x and x=2, which lies in the first quadrant. Then the value of 3α is equal to _____ . [2023]

Q 25 : Let A be the area of the region {(x,y):y≥x2,y≥(1-x)2,y≤2x(1-x)}. Then 540A is equal to ___________ . [2023]

Q 26 : Let for x∈R, f(x)=x+|x|2 and g(x)={x,x<0x2,x≥0. Then area bounded by the curve y=(fog)(x) and the lines y=0, 2y-x=15 is equal to _______ . [2023]

Q 27 : Let the area of the region {(x,y):|2x-1|≤y≤|x2-x|,0≤x≤1} be A. Then (6A+11)2 is equal to __________. [2023]

Q 28 : Let the area of the region bounded by the curve y=max{sinx, cosx}, lines x=0, x=3π2, and the x-axis be A. Then A+A2 is equal to ________ . [2026]

Q 29 : The area of the region inside the ellipse x2+4y2=4 and outside the region bounded by the curves y=|x|-1 and y=1-|x| is [2026]

Q 30 : Let A1 be the bounded area enclosed by the curves y=x2+2, x+y=8 and y-axis that lies in the first quadrant. Let A2 be the bounded area enclosed by the curves y=x2+2, y2=x, x=2, and y-axis that lies in the first quadrant. Then A1-A2 is equal to [2026]

Q 31 : Let the line x=-1 divide the area of the region {(x,y):1+x2≤y≤3-x} in the ratio m:n, gcd(m,n)=1. Then m+n is equal to [2026]

Q 32 : Let P1:y=4x2 and 𝑃2:𝑦=𝑥2+27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y=αx, α>0 and P1, then α is equal to : [2026]

Q 33 : Let f(α) denote the area of the region in the first quadrant bounded by x=0, x=1, y2=x and y=|αx-5|-|1-αx|+αx2. Then (f(0)+f(1)) is equal to [2026]

Q 34 : The area of the region enclosed between the circles x2+y2=4 and x2+(y-2)2=4 is: [2026]

Q 35 : If the area of the region {(x,y):1-2x≤y≤4-x2, x≥0, y≥0} is αβ, α,β∈ℕ, gcd(α,β)=1, then the value of (α+β) is: [2026]

Q 36 : The area of the region A={(x,y):4x2+y2≤8 and y2≤4x} is: [2026]

Q 37 : The area of the region R={(x,y):xy≤8, 1≤y≤x2, x≥0} is [2026]

Q 38 : The area of smaller region bounded by the curve 9x2+4y2-36x+16y+16=0 and the line 3x+2y=8 is

Q 39 : Let S be the region bounded by the curves y=x3 and y2=x. The curve y=2|x| divides S into two regions of areas R1 and R2. If max{R1,R2}=R2,then R2R1=

Q 40 : The area of the region bounded by the curve x=2y+3 and the lines y=1, y=-1 is:

Q 41 : The area of the region bounded by the curves y=|x-2|, x=1, x=3 and the X-axis is:

Q 42 : The area bounded by the curve y=x, X-axis and ordinates x=-1 to x=2 is:

Q 43 : The area of the region bounded by the line 2y+x=8, X-axis and the lines x=2 and x=4 is:

Q 44 : The area of the region bounded by the curve y=x+1 and the lines x=2, x=3 is:

Q 45 : Area of the region in the first quadrant enclosed by the X-axis, the line y=x and the circle x2+y2=32 is:

Q 46 : The area of the region bounded by the curve y=16-x2 and X-axis is:

Q 47 : The area of the region bounded by the circle x2+y2=1 is:

Q 48 : Area lying in the first quadrant and bounded by the circle x2+y2=4 and the lines x=0 and x=2 is:

Q 49 : The area of smaller part between the circle x2+y2=4 and the line x=1 is:

Q 50 : The area enclosed by the circle x2+y2=2 is equal to:

Want Us to Email you About Special Offers & Updates?

Q 21 :

Let A be the area bounded by the curve $y = x$ $| x - 3 |$ , the $x$ -axis and the ordinates $x = - 1$ and $x = 2$ . Then 12A is equal to _____. [2023]

Q 22 :

If the area of the region bounded by the curves $y^{2} - 2 y = - x$ and $x + y = 0$ is A, then 8A is equal to _________ . [2023]

Q 23 :

If the area enclosed by the parabolas $P_{1} : 2 y = 5 x^{2}$ and $P_{2} : x^{2} - y + 6 = 0$ is equal to the area enclosed by $P_{1}$ and $y = α x, α > 0$ , then $α^{3}$ is equal to _______ . [2023]

Q 24 :

Let $α$ be the area of the larger region bounded by the curve $y^{2} = 8 x$ and the lines $y = x$ and $x = 2$ , which lies in the first quadrant. Then the value of $3 α$ is equal to _____ . [2023]

Q 25 :

Let A be the area of the region ${(x, y) : y \geq x^{2}, y \geq {(1 - x)}^{2}, y \leq 2 x (1 - x)}$ . Then 540A is equal to ___________ . [2023]

Q 26 :

Let for $x \in R,$ $f (x) = \frac{x + | x |}{2}$ and $g (x) =$ ${\begin{matrix} x, & x < 0 \\ x^{2}, & x \geq 0 \end{matrix}$ . Then area bounded by the curve $y = (f o g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to _______ . [2023]

Q 27 :

Let the area of the region ${(x, y) : | 2 x - 1 | \leq y \leq | x^{2} - x |, 0 \leq x \leq 1}$ be A. Then ${(6 A + 11)}^{2}$ is equal to __________. [2023]

Q 28 :

Let the area of the region bounded by the curve $y = \max {\sin x, \cos x},$ lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be A. Then $A + A^{2}$ is equal to ________ . [2026]

Q 29 :

The area of the region inside the ellipse $x^{2} + 4 y^{2} = 4$ and outside the region bounded by the curves $y = | x | - 1 and y = 1 - | x |$ is [2026]

Q 31 :

Let the line $x = - 1$ divide the area of the region ${(x, y) : 1 + x^{2} \leq y \leq 3 - x}$ in the ratio $m : n$ , $\gcd (m, n) = 1$ . Then $m + n$ is equal to [2026]

Q 32 :

Let $P_{1} : y = 4 x^{2} and 𝑃_{2} : 𝑦 = 𝑥^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1} and P_{2}$ is six times the area of the bounded region enclosed between the line $y = α x, α > 0$ and $P_{1}$ , then $α$ is equal to : [2026]

Q 33 :

Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y =$ $| α x - 5 |$ $-$ $| 1 - α x |$ $+ α x^{2} .$

Then $(f (0) + f (1))$ is equal to [2026]

Q 34 :

The area of the region enclosed between the circles $x^{2} + y^{2} = 4$ and $x^{2} + {(y - 2)}^{2} = 4$ is: [2026]

Q 35 :

If the area of the region ${(x, y) : 1 - 2 x \leq y \leq 4 - x^{2}, x \geq 0, y \geq 0}$ is $\frac{α}{β}, α, β \in ℕ, \gcd (α, β) = 1$ , then the value of $(α + β)$ is: [2026]

Q 36 :

The area of the region $A = {(x, y) : 4 x^{2} + y^{2} \leq 8 and y^{2} \leq 4 x}$ is: [2026]

Q 37 :

The area of the region $R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}$ is [2026]

Q 38 :

The area of smaller region bounded by the curve $9 x^{2} + 4 y^{2} - 36 x + 16 y + 16 = 0$ and the line $3 x + 2 y = 8$ is

Q 39 :

Let S be the region bounded by the curves $y = x^{3}$ and $y^{2} = x .$ The curve $y = 2$ $| x |$ divides S into two regions of areas $R_{1}$ and $R_{2} .$ If $\max {R_{1}, R_{2}} = R_{2}$ ,then $\frac{R_{2}}{R_{1}} =$

Q 40 :

The area of the region bounded by the curve $x = 2 y + 3$ and the lines $y = 1, y = - 1$ is:

Q 41 :

The area of the region bounded by the curves $y =$ $| x - 2 |$ $,$ $x = 1,$ $x = 3$ and the $X$ -axis is:

Q 42 :

The area bounded by the curve $y = x$ , $X$ -axis and ordinates $x = - 1$ to $x = 2$ is:

Q 43 :

The area of the region bounded by the line $2 y + x = 8$ , $X$ -axis and the lines $x = 2$ and $x = 4$ is:

Q 44 :

The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2, x = 3$ is:

Q 45 :

Area of the region in the first quadrant enclosed by the $X$ -axis, the line $y = x$ and the circle $x^{2} + y^{2} = 32$ is:

Q 46 :

The area of the region bounded by the curve $y = \sqrt{16 - x^{2}}$ and $X$ -axis is:

Q 47 :

The area of the region bounded by the circle $x^{2} + y^{2} = 1$ is:

Q 48 :

Area lying in the first quadrant and bounded by the circle $x^{2} + y^{2} = 4$ and the lines $x = 0$ and $x = 2$ is:

Q 49 :

The area of smaller part between the circle $x^{2} + y^{2} = 4$ and the line $x = 1$ is:

Q 50 :

The area enclosed by the circle $x^{2} + y^{2} = 2$ is equal to: