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

Q 21 :
The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

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(5)

Given, equation of parabola

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

and line $x - 2 y + 4 = 0$ ...(ii)

From (ii), we have

$x - 2 (y - 2) = 0 \Rightarrow y - 2 = \frac{x}{2}$

Putting in (i), we get ${(\frac{x}{2})}^{2} = (x - 1)$

$\Rightarrow x^{2} - 4 x + 4 = 0$

$\Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x = 2$

From (ii), we have, $y = 3$

$∴$ Intersection point of (i) and (ii) is (2,3).

$∴$ $Required area = \int_{0}^{3} ({(y - 2)}^{2} + 1) d y - Area of triangle$

$= \int_{0}^{3} (y^{2} - 4 y + 5) d y - \frac{1}{2} \times 2 \times 1$

$= {[\frac{y^{3}}{3} - 2 y^{2} + 5 y]}_{0}^{3} - 1 = (9 - 18 + 15) - 1 = 6 - 1 = 5 sq. units$

Q 22 :
The area of the region ${(x, y) : | x - y | \leq y \leq 4 \sqrt{x}}$ is [2025]

512
$\frac{2048}{3}$
$\frac{1024}{3}$
$\frac{512}{3}$

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3

4

(3)

We have, $| x - y |$ $\leq y \leq 4 \sqrt{x}$

Now, $y =$ $| x - y |$ $\Rightarrow y^{2} = {(x - y)}^{2}$

$\Rightarrow y = \frac{x}{2} and x = 0$

Required Area $= \int_{0}^{64} (4 \sqrt{x} - \frac{x}{2}) d x$

$= {[\frac{4 x^{3 / 2}}{3 / 2} - \frac{x^{2}}{4}]}_{0}^{64} = \frac{8}{3} \cdot 8^{3} - \frac{64^{2}}{4} = 64^{2} (\frac{1}{12}) = \frac{1024}{3}$ .

Q 23 :
Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

$\frac{1}{2}$
$\sqrt{2}$
$\sqrt{5}$
2

1

2

3

4

(1)

We have, $y = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$\Rightarrow \frac{d y}{d x} = - 2 + e^{- x} \cdot e^{x} f (x) + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$= - 2 + y + y + 2 x - 1$

$\Rightarrow \frac{d y}{d x} - 2 y = 2 x - 3$

This is a linear differential equation, so we have

$y e^{- 2 x} = \int (2 x - 3) \cdot e^{- 2 x} d x + c$ [ $∵ I . F . = e^{\int - 2 d x} = e^{- 2 x}$ ]

$\Rightarrow y e^{- 2 x} = - \frac{(2 x - 3)}{2} e^{- 2 x} + \int e^{- 2 x} \cdot d x + c$

$\Rightarrow y e^{- 2 x} = \frac{- (2 x - 3)}{2} e^{- 2 x} - \frac{e^{- 2 x}}{2} + c$

$∵ f (0) = 1 ∴ c = 1 - \frac{3}{2} + \frac{1}{2} = 0$

$\Rightarrow y = - \frac{(2 x - 3)}{2} - \frac{1}{2} \Rightarrow x + y = 1$

So, area $= \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$ .

Q 24 :
A line passing through the point A(–2, 0), touches the parabola $P : y^{2} = x - 2$ at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

$\frac{7}{3}$
2
3
$\frac{8}{3}$

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3

4

(4)

Given parabola $P : y^{2} = x - 2$ and line is passing through A(–2, 0)

Tangent is y = m(x + 2)

$\Rightarrow (m {(x + 2))}^{2} = x - 2$

$\Rightarrow m^{2} x^{2} + (4 m^{2} - 1) x + (4 m^{2} + 2) = 0$

As D = 0

$\Rightarrow {(4 m^{2} - 1)}^{2} = 4 m^{2} (4 m^{2} + 2) \Rightarrow m = \frac{1}{4}$

So, tangent is $y = \frac{1}{4} (x + 2)$

Point of tangency is (6, 2)

$∴$ Area of region $= \int_{0}^{2} (y^{2} + 2) - (4 y - 2) d y$

$= \frac{8}{3} - 8 + 8 = \frac{8}{3}$ sq. units.

Q 25 :
If the area of the region bounded by the curves $y = 4 - \frac{x^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals [2025]

240
220
210
250

1

2

3

4

(4)

Given, $y = 4 - \frac{x^{2}}{4}$ ... (i)

and $y = \frac{x - 4}{2}$ ... (ii)

From (i) & (ii), we get

x = –6, 4 and y = –5, 0

Thus, point of intersection of (i) & (ii) are (–6, 5) and (4, 0).

Required Area $= \int_{- 6}^{4} {(4 - \frac{x^{2}}{4}) - (\frac{x - 4}{2})} d x$

$\Rightarrow α = \int_{- 6}^{4} {- \frac{x^{2}}{4} - \frac{x}{2} + 6} d x$

$α =$ $- \frac{x^{3}}{12} - \frac{x^{2}}{4} + 6 x |_{- 6}^{4}$

$= - (\frac{64}{12} + \frac{216}{12}) - (\frac{16}{4} - \frac{36}{4}) + 6 (4 + 6)$

$= \frac{125}{3}$

$∴ 6 α = 250$ .

Q 26 :
If the area of the region ${(x, y) : 1 + x^{2} \leq y \leq \min {x + 7, 11 - 3 x}}$ is A, then 3A is equal to [2025]

47
49
50
46

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2

3

4

(3)

$A = \int_{- 2}^{1} (x + 7 - x^{2} - 1) d x + \int_{1}^{2} (11 - 3 x - x^{2} - 1) d x$

$= {[\frac{x^{2}}{2} + 6 x - \frac{x^{3}}{3}]}_{- 2}^{1} + {[10 x - \frac{3 x^{2}}{2} - \frac{x^{3}}{3}]}_{1}^{2}$

$= \frac{50}{3} \Rightarrow 3 A = 50$ .

Q 27 :
The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

$3 π + 8$
$3 π - 8$
$6 π - 16$
$6 π - 8$

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(3)

We have, ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ , a circle with centre $(2 \sqrt{3}, 0)$ and radius $\sqrt{12}$ and a parabola $y^{2} = 2 \sqrt{3} x$ .

$∴$ Required Area = $\frac{π (12)}{2} - 2 \int_{0}^{2 \sqrt{3}} {(2 \sqrt{3})}^{\frac{1}{2}} x^{1 / 2} d x$

$= 6 π - 2 {(2 \sqrt{3})}^{1 / 2} {[\frac{x^{3 / 2}}{3} \cdot 2]}_{0}^{2 \sqrt{3}}$

$= 6 π - \frac{4}{3} {(2 \sqrt{3})}^{1 / 2} {(2 \sqrt{3})}^{3 / 2}$

$= 6 π - \frac{4}{3} {(2 \sqrt{3})}^{2} = 6 π - 16$ .

Q 28 :
Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

$e^{4} + 1$
$e^{2} - 1$
$e^{4} - 1$
$e^{2} + 1$

1

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3

4

(2)

We have, f(x + y) = f(x)f(y)

Let $f (x) = e^{λ x}$

Differentiating w.r.t. x, we get

$f' (x) = λ \cdot e^{λ x}$

Put x = 0, we get

$f' (0) = λ$ [ $∵ f' (0) = 4 a$ ]

$\Rightarrow λ = 4 a$

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

Now, $f'' (x) - 3 a f' (x) - f (x) = 0$

$\Rightarrow \frac{d}{d x} (λ e^{λ x}) - 3 a (λ e^{λ x}) - e^{λ x} = 0$

$\Rightarrow λ^{2} - 3 a λ - 1 = 0$

$\Rightarrow 16 a^{2} - 12 a^{2} - 1 = 0 \Rightarrow 4 a^{2} = 1$

$\Rightarrow a = \frac{1}{2}$ [ $∵ a > 0$ ]

$∴ f (x) = e^{2 x}$

Now, $f (x) = f (a x) = e^{x}$

$∴$ Area of region $\int_{0}^{2} e^{x} d x = {[e^{x}]}_{0}^{2} = e^{2} - 1$ .

Q 29 :
The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

5
4/3
8/3
8

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2

3

4

(3)

The given curves are $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$

Points of intersection are (2, 0) and (0, 4).

$∴$ Required Area $= \int_{0}^{2} (\sqrt{16 - 8 x} - (x^{2} - 4 x + 4)) d x$

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

$= 2 \sqrt{2} {[\frac{2 {(2 - x)}^{3 / 2}}{- 3}]}_{0}^{2} - {[\frac{{(x - 2)}^{3}}{3}]}_{0}^{2}$

$= \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$ sq. units.

Q 30 :
If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

6
8
7
5

1

2

3

4

(4)

Required area $= a + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$

$= a + {[a x + e^{x} + e^{- x}]}_{0}^{1}$

$\Rightarrow 2 a + e + e^{- 1} - 2 = e + 8 + \frac{1}{e}$

$\Rightarrow 2 a - 2 = 8$

$\Rightarrow 2 a = 10 \Rightarrow a = 5$ .

Topic Question Set

Q 21 :
The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

0%

Q 22 :
The area of the region ${(x, y) : | x - y | \leq y \leq 4 \sqrt{x}}$ is [2025]

Q 23 :
Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Q 24 :
A line passing through the point A(–2, 0), touches the parabola $P : y^{2} = x - 2$ at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

Q 25 :
If the area of the region bounded by the curves $y = 4 - \frac{x^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals [2025]

Q 26 :
If the area of the region ${(x, y) : 1 + x^{2} \leq y \leq \min {x + 7, 11 - 3 x}}$ is A, then 3A is equal to [2025]

Q 27 :
The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

Q 28 :
Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

Q 29 :
The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

Q 30 :
If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

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Topic Question Set

Q 21 : The area of the region enclosed by the parabola (y-2)2=x-1, the line x-2y+4=0 and the positive coordinate axes is ____________. [2024]

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Q 22 : The area of the region {(x,y):|x–y|≤y≤4x} is [2025]

Q 23 : Let f:[0,∞)→R be a differentiable function such that f(x)=1–2x+∫0xex–tf(t)dt for all x∈[0,∞). Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Q 24 : A line passing through the point A(–2, 0), touches the parabola P : y2=x–2 at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

Q 25 : If the area of the region bounded by the curves y=4–x24 and y=x–42 is equal to α, then 6α equals [2025]

Q 26 : If the area of the region {(x,y):1+x2≤y≤min {x+7,11–3x}} is A, then 3A is equal to [2025]

Q 27 : The area of the region, inside the circle (x–23)2+y2=12 and outside the parabola y2=23x is : [2025]

Q 28 : Let f : R→R be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, y∈R. If f'(0)=4a and f satisfies f''(x)=3af'(x)–f(x)=0, a > 0, then the area of the region R={(x,y) | 0≤y≤f(ax), 0≤x≤2} is : [2025]

Q 29 : The area of the region enclosed by the curves y=x2–4x+4 and y2=16–8x is : [2025]

Q 30 : If the area of the region {(x,y):–1≤x≤1, 0≤y≤a+e|x|–e–x, a>0} is e2+8e+1e, then the value of a is : [2025]

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Q 21 :
The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

Q 22 :
The area of the region ${(x, y) : | x - y | \leq y \leq 4 \sqrt{x}}$ is [2025]

Q 23 :
Let $f : [0, \infty) \to R$ be a differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ .

Then the area of the region bounded by y = f(x) and the coordinates axes is [2025]

Q 24 :
A line passing through the point A(–2, 0), touches the parabola $P : y^{2} = x - 2$ at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is : [2025]

Q 25 :
If the area of the region bounded by the curves $y = 4 - \frac{x^{2}}{4}$ and $y = \frac{x - 4}{2}$ is equal to $α$ , then $6 α$ equals [2025]

Q 26 :
If the area of the region ${(x, y) : 1 + x^{2} \leq y \leq \min {x + 7, 11 - 3 x}}$ is A, then 3A is equal to [2025]

Q 27 :
The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

Q 28 :
Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

Q 29 :
The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

Q 30 :
If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]