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

Q 41 :
If the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

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(176)

Let $I = 525 \int_{0}^{\frac{π}{2}} \sin 2 x \cdot \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$

Put $\cos x = t^{2}$

$- \sin x d x = 2 t d t$

$I = - 525 \int_{1}^{0} 2 \sin x \cdot t^{2} \cdot t^{11} {(1 + t^{5})}^{\frac{1}{2}} \cdot \frac{2 t d t}{\sin x}$

$= 525 \times 4 \int_{0}^{1} t^{3} \cdot t^{11} {(1 + t^{5})}^{\frac{1}{2}} d t$

$I = 2100 \int_{0}^{1} t^{14} \sqrt{(1 + t^{5})} d t$

Let $1 + t^{5} = k^{2} \Rightarrow 5 t^{4} d t = 2 k d k \Rightarrow t^{4} d t = \frac{2 k d k}{5}$

$∴ I = 2100 \int_{1}^{\sqrt{2}} {(k^{2} - 1)}^{2} \cdot k \cdot \frac{2 k d k}{5}$

$= \frac{2100 \times 2}{5} \int_{1}^{\sqrt{2}} k^{2} {(k^{2} - 1)}^{2} d k$

$= 420 \times 2 \int_{1}^{\sqrt{2}} (k^{6} - 2 k^{4} + k^{2}) d k$

$= 840 {[\frac{k^{7}}{7} - \frac{2 k^{5}}{5} + \frac{k^{3}}{3}]}_{1}^{\sqrt{2}}$

$= 840 [\frac{8 \sqrt{2}}{7} - \frac{8 \sqrt{2}}{5} + \frac{2 \sqrt{2}}{3} - (\frac{1}{7} - \frac{2}{5} + \frac{1}{3})]$

$= 840 [\frac{22 \sqrt{2} - 8}{105}] = 8 [22 \sqrt{2} - 8] = 176 \sqrt{2} - 64$

So, $n = 176$

Q 42 :
$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

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(15)

$Let I = \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

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

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

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

$= 2 π \cdot \frac{π}{4} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} - π^{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

$= \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} = \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{1 - 2 \sin^{2} x + \cos^{2} x} d x$

$= \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$

$Put \cos 2 x = t ∴ I = \frac{- π^{2}}{2} \int_{1}^{- 1} \frac{- d t}{2 (1 + t^{2})}$

$= \frac{- π^{2}}{2} \int_{0}^{1} \frac{d t}{1 + t^{2}} = \frac{- π^{2}}{2} {(\tan^{- 1} t)}_{0}^{1} = \frac{- π^{2}}{2} (\frac{π}{4}) = \frac{- π^{3}}{8}$

$∴$ $| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} |$ $= \frac{120}{π^{3}} \times \frac{π^{3}}{8} = 15$

Q 43 :
If $(a, b)$ be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

66
88
72
80

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(3)

Let the vertices of the triangle are $A (3, 1), B (1, 2)$ and $C (2, 3)$ .

$Slope of B C = \frac{3 - 2}{2 - 1} = 1$

Since, AD is perpendicular to BC.

Slope of AD = $- 1$

Equation of line $A D$ is

$y - 1 = - 1 (x - 3)$

Since, $(a, b)$ lies on AD.

$∴$ $b - 1 = - 1 (a - 3) \Rightarrow a + b = 4$

Now, $I_{1} = \int_{a}^{b} x \sin [(4 - x) x] d x$

$I_{1} = \int_{a}^{b} (a + b - x) \sin [(4 - (a + b - x)) (a + b - x)] d x$

$I_{1} = 4 \int_{a}^{b} \sin [(4 - x) x] d x - \int_{a}^{b} x \sin [(4 - x) x] d x$

$I_{1} = 4 I_{2} - I_{1}$ $[∵ \int_{a}^{b} \sin [(4 - x) x] d x = I_{2}]$

$\Rightarrow 2 I_{1} = 4 I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = 2$ $∴$ $\frac{36 I_{1}}{I_{2}} = 36 \times 2 = 72$

Q 44 :
Let $S = (- 1, \infty)$ and $f : S \to R$ be defined as $f (x) = \int_{- 1}^{x} {(e^{t} - 1)}^{11} {(2 t - 1)}^{5} {(t - 2)}^{7} {(t - 3)}^{12} {(2 t - 10)}^{61} d t .$ Let $p =$ Sum of squares of the values of $x$ , where $f (x)$ attains local maxima on $S,$ and $q =$ Sum of the values of $x,$ where $f (x)$ attains local minima on $S$ . Then, the value of $p^{2} + 2 q$ is ____________ . [2024]

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(27)

Given,

$f (x) = \int_{- 1}^{x} {(e^{t} - 1)}^{11} {(2 t - 1)}^{5} {(t - 2)}^{7} {(t - 3)}^{12} {(2 t - 10)}^{61} d t$

$f' (x) = {(e^{x} - 1)}^{11} {(2 x - 1)}^{15} {(x - 2)}^{7} {(x - 3)}^{12} {(2 x - 10)}^{61}$

$So, x = 0, x = \frac{1}{2}, x = 2, x = 3, x = 5$

$∴ p = 0 + 4 = 4$

$q = \frac{1}{2} + 5 = \frac{11}{2}$

$∴ p^{2} + 2 q = 16 + 11 = 27$

Q 45 :
$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

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

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4

(3)

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

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

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

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

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

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

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

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

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

$∴ 4 \int_{0}^{1} f (x) d x - 3 \log_{e} \sqrt{3} = 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2})$ .

Q 46 :
Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x$ is equal to : [2025]

21
18
24
27

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(3)

We have, $x^{2} = 2 y$ and y – 2x – 6 = 0

Substitute y = 2x + 6 in $x^{2} = 2 y$ , we get

$x^{2} = (4 x + 12)$

$\Rightarrow x^{2} - 4 x - 12 = 0 \Rightarrow (x - 6) (x + 2) = 0$

$\Rightarrow x = 6 or - 2 \Rightarrow y = 18 or 2$

$∴$ Intersection points are (6, 18) and (–2, 2)

Since, (a, b) is a point in second quadrant

$∴$ (a, b) = (–2, 2)

Now, $I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{x}} d x$ ... (i)

$∴ I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{- x}} d x = \int_{- 2}^{2} \frac{9 x^{2} 5^{x}}{5^{x} + 1} d x$ ... (ii)

On adding (i) and (ii), we get

$2 I = \int_{- 2}^{2} \frac{9 x^{2} (5^{x} + 1)}{(5^{x} + 1)} d x = \int_{- 2}^{2} 9 x^{2} d x$

$= {[\frac{9 x^{3}}{3}]}_{- 2}^{2} = 3 [8 - (- 8)] = 48$

$\Rightarrow I = \frac{48}{2} = 24$ .

Q 47 :
Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

8
11
9
10

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(4)

We have, $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$

$∴$ f(x) is define when $\log_{4} \log_{6} (3 + 4 x - x^{2}) > 0$

$\Rightarrow \log_{6} (3 + 4 x - x^{2}) > 1$

$\Rightarrow 3 + 4 x - x^{2} > 6$

$\Rightarrow - x^{2} + 4 x - 3 > 0$

$\Rightarrow x^{2} - 4 x + 3 < 0$

$\Rightarrow (x - 1) (x - 3) < 0$

$\Rightarrow x \in (1, 3)$

$∴$ a = 1 and b = 3

Now, $\int_{0}^{2} [x^{2}] d x = \int_{0}^{1} [0] d x + \int_{1}^{\sqrt{2}} [1] d x + \int_{\sqrt{2}}^{\sqrt{3}} [2] d x + \int_{\sqrt{3}}^{\sqrt{4}} [3] d x$

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

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

$= 5 - \sqrt{2} - \sqrt{3}$

$\Rightarrow$ p = 5, q = 2, r = 3

$∴$ p + q + r = 5 + 2 + 3 = 10.

Q 48 :
The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

$π^{2}$
$4 π^{2}$
$\frac{3 π^{2}}{2}$
$2 π^{2}$

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(4)

Let $I = \int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ ... (i)

$\Rightarrow I = \int_{0}^{π} \frac{8 (π - x) d x}{4 \cos^{2} (π - x) + \sin^{2} (π - x)}$

$\Rightarrow I = \int_{0}^{π} \frac{(8 π - 8 x) d x}{4 \cos^{2} x + \sin^{2} x} d x$ ... (ii)

Adding (i) and (ii), we get

$2 I = 8 π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + \sin^{2} x}$

$\Rightarrow 2 I = 8 π \times 2 \int_{0}^{π} \frac{\sec^{2} x}{4 + \tan^{2 x}} d x$

Put $\tan x = t \Rightarrow \sec^{2} xdx = dt$

$∴ I = 8 π \int_{0}^{\infty} \frac{d t}{4 + t^{2}} = 8 π \times \frac{1}{2} {[\tan^{- 1} \frac{t}{2}]}_{0}^{\infty}$

$= 4 π \times \frac{π}{2} = 2 π^{2}$ .

Q 49 :
The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

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

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(4)

Consider, $I = \int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}}$

$= \int_{- 1}^{1} \frac{e^{x} + e^{x} \sqrt{| x | - x} + e^{- x} \sqrt{| x - x |}}{e^{x} + e^{- x}} d x$

$= \int_{- 1}^{1} [\frac{e^{x}}{e^{x} + e^{- x}} + \frac{\sqrt{| x | - x} (e^{x} + e^{- x})}{e^{x} + e^{- x}}] d x$

$= \int_{- 1}^{1} \frac{e^{x}}{e^{x} + e^{- x}} d x + \int_{- 1}^{1} \sqrt{| x | - x} d x = I_{1} + I_{2}$

Now, $I_{1} = \int_{- 1}^{1} \frac{e^{x}}{e^{x} + e^{- x}} d x$ ... (i)

$\Rightarrow I_{1} = \int_{- 1}^{1} \frac{e^{- x}}{e^{- x} + e^{x}} d x$ ... (ii)

On adding (i) and (ii), we get

$\Rightarrow 2 I_{1} = \int_{- 1}^{1} \frac{e^{x} + e^{- x}}{e^{x} + e^{- x}} d x = \int_{- 1}^{1} 1 d x = 2 \Rightarrow I_{1} = 1$

and $I_{2} = \int_{- 1}^{1} \sqrt{| x | - x} d x = \int_{- 1}^{0} \sqrt{| x | - x} d x + \int_{0}^{1} \sqrt{| x | - x} d x$

$\Rightarrow I_{2} = \int_{- 1}^{0} \sqrt{- 2 x} d x + \int_{0}^{1} \sqrt{x - x} d x = \int_{- 1}^{0} \sqrt{- 2 x} d x$

Put $- 2 x = u \Rightarrow d x = - \frac{1}{2} d u$

$∴ I_{2} = \int_{2}^{0} \sqrt{u} (- \frac{1}{2} d u) = \frac{1}{2} \int_{0}^{2} \sqrt{u} d u = \frac{1}{2} {[\frac{u^{3 / 2}}{\frac{3}{2}}]}_{0}^{2} = \frac{2 \sqrt{2}}{3}$

$∴ I = I_{1} + I_{2} = 1 + \frac{2 \sqrt{2}}{3}$ .

Q 50 :
Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

0
11
10
1

1

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3

4

(2)

We have, $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ ... (i)

$\Rightarrow f (\frac{1}{x}) + 2 f (x) = \frac{1}{x^{2}} + 5$ ... (ii)

On solving (i) and (ii), we get

$f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}$

So, $α = \int_{1}^{2} f (x) d x = \int_{1}^{2} (\frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}) d x$

$= {(- \frac{2}{3 x} - \frac{x^{3}}{9} + \frac{5 x}{3})}_{1}^{2} = \frac{6}{3} - \frac{7}{9} = \frac{11}{9}$

Also, we have $2 g (x) - 3 g (\frac{1}{2}) = x$

$\Rightarrow 2 g (\frac{1}{2}) - 3 g (\frac{1}{2}) = \frac{1}{2} \Rightarrow g (\frac{1}{2}) = - \frac{1}{2}$

so, $g (x) = \frac{x}{2} - \frac{3}{4}$

$β = \int_{1}^{2} g (x) d x = \int_{1}^{2} (\frac{x}{2} - \frac{3}{4}) d x$

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

$∴ 9 α + β = 9 (\frac{11}{9}) + 0 = 11$ .

Topic Question Set

Q 41 :
If the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

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Q 42 :
$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

0%

Q 43 :
If $(a, b)$ be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

0%

Q 45 :
$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

Q 46 :
Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x$ is equal to : [2025]

Q 47 :
Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

Q 48 :
The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

Q 49 :
The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

Q 50 :
Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

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Topic Question Set

Q 41 : If the integral 525∫0π2sin2xcos112x(1+cos52x)12dx is equal to (n2-64), then n is equal to ______ . [2024]

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Q 42 : |120π3∫0πx2sinxcosxsin4x+cos4x dx| is equal to ________ . [2024]

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Q 43 : If (a,b) be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and I1=∫abxsin(4x-x2) dx, I2=∫absin(4x-x2) dx, then 36I1I2 is equal to : [2024]

Q 44 : Let S=(-1,∞) and f:S→R be defined as f(x)=∫-1x(et-1)11(2t-1)5(t-2)7(t-3)12(2t-10)61 dt. Let p=Sum of squares of the values of x, where f(x) attains local maxima on S, and q= Sum of the values of x, where f(x) attains local minima on S. Then, the value of p2+2q is ____________ . [2024]

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Q 45 : 4∫01(13+x2+1+x2)dx–3 loge (3) is equal to : [2025]

Q 46 : Let (a, b) be the point of intersection of the curve x2=2y and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral I=∫ab9x21+5xdx is equal to : [2025]

Q 47 : Let the domain of the function f(x)=log2 log4 log6 (3+4x–x2) be (a, b). If ∫0b–a[x2]dx=p–q–r, p, q, r∈ℕ, gcd (p, q, r) = 1, where [·] is the greatest integer function, then p + q + r is equal to [2025]

Q 48 : The integral ∫0π8 xdx4cos2x+sin2x is equal to [2025]

Q 49 : The value of ∫–11(1+|x|–x)ex+(|x|–x)e–xex+e–xdx is equal to [2025]

Q 50 : Let f(x)+2f(1x)=x2+5 and 2g(x)–3g(12)=x, x > 0. If α=∫12f(x)dx, and β=∫12g(x)dx, then the value of 9α+β is : [2025]

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Q 41 :
If the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

Q 42 :
$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

Q 43 :
If $(a, b)$ be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

Q 45 :
$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

Q 46 :
Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x$ is equal to : [2025]

Q 47 :
Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

Q 48 :
The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

Q 49 :
The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

Q 50 :
Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]