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

Q 61 :

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]

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(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 62 :

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]

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(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 63 :

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]

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(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 64 :

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]

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(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 65 :

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]

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(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 66 :

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]

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(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 67 :

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]

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(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 68 :

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]

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(12)

Q 69 :

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)

[IMAGE 77]

$Required area = area of ellipse - shaded area$

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

Q 70 :

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)

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