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

Q 41 :

Let the function, $f (x) = {\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ a^{2} + b x, & x \geq 1 \end{matrix}$ be differentiable for all $x \in R$ , where a > 1, $b \in R$ . If the area of the region enclosed by y = f(x) and the line y = –20 is $α + β \sqrt{3}, α, β \in Z$ then the value of $α + β$ is __________. [2025]

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Given, f(x) is continuous and differentiable at x = 1.

$f (x) = {\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ a^{2} + b x, & x \geq 1 \end{matrix}$

$\Rightarrow f' (x) = {\begin{matrix} - 6 a x, & x < 1 \\ b, & x \geq 1 \end{matrix}$

L.H.L. at $x = 1 \Rightarrow - 3 a - 2$

R.H.L. at $x = 1 \Rightarrow a^{2} + b$

$∴$ L.H.L. = R.H.L. { $∵$ f(x) is continuous}

$\Rightarrow - 3 a - 2 = a^{2} + b$ ... (i)

L.H.D. at $x = 1 \Rightarrow - 6 a$

R.H.D. at $x = 1 \Rightarrow b$

$∴$ L.H.D. = R.H.D. [ $∵$ f(x) is differentiable]

$\Rightarrow - 6 a = b$ ... (ii)

From (i) and (ii), we get

$- 3 a - 2 = a^{2} - 6 a$

$\Rightarrow a^{2} - 3 a + 2 = 0 \Rightarrow (a - 1) (a - 2) = 0$

$\Rightarrow a = 1 o r a = 2 \Rightarrow a = 2$ ( $∵$ a > 1)

From (ii), b = –12

Now, $f (x) = {\begin{matrix} - 6 x^{2} - 2, & x < 1 \\ 4 - 12 x, & x \geq 1 \end{matrix}$

$∴$ Area of region $= \int_{- \sqrt{3}}^{1} (- 6 x^{2} - 2 + 20) d x + \int_{1}^{2} (4 - 12 x + 20) d x$

$= {(\frac{- 6 x^{3}}{3} - 2 x + 20 x)}_{- \sqrt{3}}^{1} + {(4 x - \frac{12 x^{2}}{2} + 20 x)}_{1}^{2}$

$\Rightarrow 16 + 12 \sqrt{3} + 6 = 22 + 12 \sqrt{3}$

$\Rightarrow 22 + 12 \sqrt{3} = α + β \sqrt{3}$

So, $α = 22 and β = 12$

$∴ α + β = 22 + 12 = 34$

Q 42 :

If the area of the larger portion bounded between the curves $x^{2} + y^{2} = 25$ and $y = | x - 1 |$ is $\frac{1}{4} (b π + c), b, c \in N$ , then b + c is equal to __________. [2025]

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Given, $x^{2} + y^{2} = 25$

$\Rightarrow x^{2} + {(x - 1)}^{2} = 25$ [ $∵$ y = |x –1|]

$\Rightarrow x = 4, - 3$

$∴$ Required area $= 25 π - (\int_{- 3}^{4} (\sqrt{25 - x^{2}} - | x - 1 |) d x)$

$= 25 π - {[\frac{1}{2} x \sqrt{25 - x^{2}} + \frac{25}{2} \sin^{- 1} \frac{x}{5}]}_{- 3}^{4} + \int_{- 3}^{1} (1 - x) d x + \int_{1}^{4} (x - 1) d x$

$= 25 π + \frac{25}{2} - (2 \times 3 + \frac{25}{2} \sin^{- 1} (\frac{4}{5}) + \frac{3}{2} \times 4 + \frac{25}{2} \sin^{- 1} (\frac{3}{5}))$

$= 25 π + \frac{25}{2} - 12 - \frac{25 π}{4}$

$= \frac{75 π}{4} + \frac{1}{2} = \frac{1}{4} (75 π + 2)$

$∴ \frac{1}{4} (b π + c) = \frac{1}{4} (75 π + 2)$

$\Rightarrow b = 75, c = 2$

$\Rightarrow b + c = 77$ .

Q 43 :

The area bounded by the curves $y =$ $| x - 1 |$ $+$ $| x - 2 |$ and $y = 3$ is equal to [2023]

6
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4

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(4)

$Required area is shaded in the given figure, which is a trapezium.$

$∴$ $Area = \frac{1}{2} \times Sum of parallel sides \times Height$

$= \frac{1}{2} \times (A B + C D) \times A M = \frac{1}{2} (1 + 3) \times 2 = 4 sq. units$

Q 44 :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is [2023]

24
20
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21

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(2)

$We have,$

$x^{2} \leq y$

$8 - x^{2} \geq y$

$y \leq 7$

$Converting the given inequations into equations, we get x^{2} = y and y = 8 - x^{2}$

$Solving these equations to find their point of intersection$

i.e., $x^{2} = 8 - x^{2}$

$\Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

$∴$ $y = 4$

$Required area = \int_{- 2}^{2} (8 - x^{2} - x^{2}) d x - \int_{- 1}^{1} (8 - x^{2} - 7) d x$

$= \int_{- 2}^{2} (8 - 2 x^{2}) d x - \int_{- 1}^{1} (1 - x^{2}) d x$ $= {[8 x - \frac{2}{3} x^{3}]}_{- 2}^{2} - {[x - \frac{x^{3}}{3}]}_{- 1}^{1}$

$= [(16 - \frac{16}{3}) - (- 16 + \frac{16}{3})] - [\frac{2}{3} + \frac{2}{3}]$

$= 16 - \frac{16}{3} + 16 - \frac{16}{3} - \frac{4}{3}$ $= 20 sq. units$

Q 45 :

Area of the region ${(x, y) : x^{2} + {(y - 2)}^{2} \leq 4, x^{2} \geq 2 y}$ is [2023]

$2 π - \frac{16}{3}$
$π - \frac{8}{3}$
$π + \frac{8}{3}$
$2 π + \frac{16}{3}$

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(1)

$Given that x^{2} + {(y - 2)}^{2} \leq 2^{2} and x^{2} \geq 2 y$

$Convert the given inequalities into equations:$

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

$Solving circle and parabola simultaneously:$

$2 y + y^{2} - 4 y + 4 = 4,$ $y^{2} - 2 y = 0,$ $y = 0, 2$

$Put y = 2 in x^{2} = 2 y \Rightarrow x = \pm 2 \Rightarrow (2, 2) and (- 2, 2)$

$A_{1} = 2 \times 2 - \frac{1}{4} \cdot π \cdot 2^{2} = 4 - π$

$Required area = 2 [\int_{0}^{2} \frac{x^{2}}{2} d x - (4 - π)]$

$= 2 [\frac{x^{3}}{6} |_{0}^{2} - 4 + π] = 2 [\frac{4}{3} + π - 4] = 2 [π - \frac{8}{3}] = 2 π - \frac{16}{3} sq. units$

Q 46 :

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is [2023]

$\frac{27}{4}$
$\frac{31}{4}$
$\frac{23}{4}$
$\frac{19}{4}$

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(1)

$Given, y = x^{3}$

$Equation of tangent to the curve y = x^{3} at point (- 1, - 1) is$

$y + 1 = 3 (x + 1) \Rightarrow y = 3 x + 2$

$Point of intersection with the curve is (2, 8)$

$So, area = \int_{- 1}^{2} [(3 x + 2) - x^{3}] d x$

$= {[3 \times \frac{x^{2}}{2} + 2 x - \frac{x^{4}}{4}]}_{- 1}^{2}$

$= \frac{27}{4} sq. units$

Q 47 :

The area of the region enclosed by the curve $f (x) = \max {\sin x, \cos x}$ , $- π \leq x \leq π$ and the $x$ -axis is [2023]

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

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(4)

We have $f (x) = \max {\sin x, \cos x}, - π \leq x \leq π$

Now, the area of the region enclosed by the given curve and the $x$ –axis is

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

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

Q 48 :

The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

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

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(4)

$Required area = 2 (\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y)$

$= 2 ({[\frac{y^{3 / 2}}{3 / 2}]}_{1}^{2} - {[\frac{{(4 - y)}^{3 / 2}}{3 / 2}]}_{2}^{4})$

$= \frac{4}{3} [(2 \sqrt{2} - 1) - (0 - 2 \sqrt{2})] = \frac{4}{3} (4 \sqrt{2} - 1)$

Q 49 :

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is [2023]

$16 \log_{e} 2 - \frac{14}{3}$
$8 \log_{e} 2 - \frac{13}{3}$
$8 \log_{e} 2 + \frac{7}{6}$
$16 \log_{e} 2 + \frac{7}{3}$

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(1)

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

$= {(\frac{x^{3}}{3} - x)}_{1}^{2} + 8 {(\log_{e} x)}_{2}^{8} - {(x)}_{2}^{8}$

$= \frac{4}{3} + 8 (2 \log_{e} 2) - 6 = 16 \log_{e} 2 - \frac{14}{3}$

Q 50 :

The area enclosed by the curves $y^{2} + 4 x = 4 and y - 2 x = 2$ is [2023]

$\frac{23}{3}$
$\frac{22}{3}$
$9$
$\frac{25}{3}$

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(3)

Given curves are $y^{2} + 4 x = 4$ and $y - 2 x = 2$

The points of intersection of the given curves are (0, 2) and (- 3, - 4)

$Required area = \int_{- 4}^{2} [(\frac{4 - y^{2}}{4}) - (\frac{y - 2}{2})] d y$

$= \int_{- 4}^{2} \frac{- y^{2} - 2 y + 8}{4} d y = \frac{1}{4} {[- \frac{y^{3}}{3} - y^{2} + 8 y]}_{- 4}^{2}$

$= \frac{1}{4} [(- \frac{8}{3} - 4 + 16) - (\frac{64}{3} - 16 - 32)]$

$= \frac{1}{4} [(\frac{28}{3}) - (- \frac{80}{3})] = \frac{1}{4} (\frac{108}{3}) = \frac{108}{12} = 9 sq. units .$

Topic Question Set

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Q 42 :

If the area of the larger portion bounded between the curves $x^{2} + y^{2} = 25$ and $y = | x - 1 |$ is $\frac{1}{4} (b π + c), b, c \in N$ , then b + c is equal to __________. [2025]

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Q 43 :

The area bounded by the curves $y =$ $| x - 1 |$ $+$ $| x - 2 |$ and $y = 3$ is equal to [2023]

Q 44 :

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is [2023]

Q 45 :

Area of the region ${(x, y) : x^{2} + {(y - 2)}^{2} \leq 4, x^{2} \geq 2 y}$ is [2023]

Q 46 :

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is [2023]

Q 47 :

The area of the region enclosed by the curve $f (x) = \max {\sin x, \cos x}$ , $- π \leq x \leq π$ and the $x$ -axis is [2023]

Q 48 :

The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

Q 49 :

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is [2023]

Q 50 :

The area enclosed by the curves $y^{2} + 4 x = 4 and y - 2 x = 2$ is [2023]

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