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

Q 21 :

The integral $80 \int_{0}^{\frac{π}{4}} (\frac{\sin θ + \cos θ}{9 + 16 \sin 2 θ}) d θ$ is equal to : [2025]

$4 \log_{e} 3$
$2 \log_{e} 3$
$3 \log_{e} 4$
$6 \log_{e} 4$

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(1)

Let $I = 80 \int_{0}^{\frac{π}{4}} \frac{(\sin θ + \cos θ) d θ}{(9 + 16 \sin 2 θ)}$

$= 80 \int_{0}^{\frac{π}{4}} \frac{(\sin θ + \cos θ) d θ}{9 + 16 (1 - 1 + 2 \sin θ \cos θ)}$

$= 80 \int_{0}^{\frac{π}{4}} \frac{(\sin θ + \cos θ) d θ}{25 - 16 {(\sin θ - \cos θ)}^{2}}$

Put $\sin θ - \cos θ = t \Rightarrow (\cos θ + \sin θ) d θ = d t$

When $θ = 0, t = - 1 and θ = π / 4, t = 0$ .

Q 22 :

Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ . If the range of f is $[α, β]$ , then $4 (α + β)$ equals : [2025]

253
154
157
125

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(3)

We have, $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$

$\Rightarrow f' (x) = x^{3} - 9 x^{2} + 20 x = x (x - 4) (x - 5)$

Where $1 \leq x \leq 5$

On integration, we get

$f (x) = \frac{x^{4}}{4} - \frac{9 x^{3}}{3} + \frac{20 x^{2}}{2}$

$= \frac{x^{4}}{4} - 3 x^{3} + 10 x^{2}$ , where $1 \leq x \leq 5$

$∴ f (1) = \frac{1}{4} - 3 + 10 = \frac{29}{4} = α$

$f (4) = \frac{{(4)}^{4}}{4} - 3 {(4)}^{3} + 10 {(4)}^{2} = 32 = β$

$f (5) = \frac{{(5)}^{4}}{4} - 3 {(5)}^{3} + 10 {(5)}^{2} = \frac{125}{4}$

$∴ 4 (α + β) = 4 (\frac{29}{4} + 32) = 157$

Q 23 :

Let $[\cdot]$ denote the greatest integer function. If $\int_{0}^{e^{3}} [\frac{1}{e^{x - 1}}] d x = α - \log_{e} 2$ , then $α^{3}$ is equal to __________. [2025]

(8)

Let $f (x) = \frac{1}{e^{x - 1}} = e^{1 - x}$

$f (0) = e^{1} = 2.71; f (e^{3}) = e^{1 - e^{3}} \in (0, 1)$

Let $f (x) = 2 \Rightarrow \frac{1}{e^{x - 1}} = 2 \Rightarrow x = 1 - \log_{e} 2$

and $f (x) = 1 \Rightarrow x = 1$

$∴ I = \int_{0}^{1 - \log_{e} 2} 2 d x + \int_{1 - \log_{e} 2}^{1} 1 d x + \int_{1}^{e^{3}} 0 d x$

$= 2 (1 - \log_{e} 2 - 0) + (1 - 1 + \log_{e} 2) + 0 = 2 - \log_{e} 2$

$∴$ $α = 2$

Hence, $α^{3} = 8$ .

Q 24 :

If $\lim_{t \to 0} (\int_{0}^{1} {{(3 x + 5)}^{t} d t)}^{\frac{1}{t}} = \frac{α}{5 e} {(\frac{8}{5})}^{\frac{2}{3}}$ , then $α$ is equal to _________. [2025]

(64)

Let $L = \lim_{t \to 0} (\int_{0}^{1} {{(3 x + 5)}^{t} d x)}^{\frac{1}{t}}$

$= e^{\lim_{t \to 0} \frac{1}{t} (\int_{0}^{1} {(3 x + 5)}^{t} d x - 1)}$

$= e^{\lim_{t \to 0} \frac{1}{t} [{(\frac{{(3 x + 5)}^{t + 1}}{3 (t + 1)})}_{0}^{1} - 1]}$ [ $∵ 1^{\infty} form$ ]

$= e^{\lim_{t \to 0} [\frac{8^{t + 1} - 5^{t + 1} - 3 t - 3}{3 t (t + 1)}]}$

$= e^{\lim_{t \to 0} {[\frac{8 \cdot 8^{t} - 5 \cdot 5^{t} - 3 t - (8 - 5)}{3 t}] \cdot \lim_{t \to 0} (\frac{1}{t + 1})}}$

$= e^{\lim_{t \to 0} \frac{1}{3} [\frac{8 (8^{t} - 1)}{t} - \frac{5 (5^{t} - 1)}{t} - \frac{3 t}{t}]}$

$= e^{\frac{(8 \ln (8) - 5 \ln (5) - 3)}{3}} = e^{[\ln {(8)}^{8 / 3} - \ln {(5)}^{5 / 3} - \ln e]}$

$= \frac{{(8)}^{8 / 3}}{{(5)}^{5 / 3} \cdot e} \Rightarrow {(\frac{8}{5})}^{\frac{2}{3}} \cdot \frac{8^{2}}{5} \cdot \frac{1}{e} = \frac{α}{5 e} {(\frac{8}{5})}^{\frac{2}{3}}$ [Given]

On comparing, we get

$∴ α = 8^{2} = 64$ .

Q 25 :

If $24 \int_{0}^{\frac{π}{4}} (\sin | 4 x - \frac{π}{12} | + [2 \sin x]) d x = 2 π + α$ , where $[\cdot]$ denotes the greatest integer function, then $α$ is equal to _________. [2025]

(12)

Let $I = 24 \int_{0}^{\frac{π}{4}} (\sin | 4 x - \frac{π}{12} | + [2 \sin x]) d x$

$= 24 [\int_{0}^{π / 48} [- \sin (4 x - \frac{π}{12})] d x + \int_{π / 48}^{π / 4} \sin (4 x - \frac{π}{12}) d x + \int_{0}^{π / 6} [0] d x + \int_{π / 6}^{π / 4} [1] d x]$

$= 24 [\frac{(1 - \cos \frac{π}{12}) - (- 1 - \cos \frac{π}{12}) + \frac{π}{4} - \frac{π}{6}}{4}]$

$= 24 (\frac{1}{2} + \frac{π}{12}) = 12 + 2 π$

$\Rightarrow I = 12 + 2 π = 2 π + α$ [Given]

On comparing, we get $α = 12$ .

Q 26 :

$Let 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0 .$ $Then 18 \int_{1}^{2} f (x) d x is equal to$ [2023]

$5 \log_{e} 2 + 3$
$10 \log_{e} 2 + 6$
$5 \log_{e} 2 - 3$
$10 \log_{e} 2 - 6$

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(4)

$We have, 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3 ...(i)$

$Replace x by \frac{1}{x}, we get 5 f (\frac{1}{x}) + 4 f (x) = x + 3 ...(ii)$

$Solving (i) and (ii), we get 25 f (x) - 16 f (x) = \frac{5}{x} + 15 - 4 x - 12$

$\Rightarrow 9 f (x) = \frac{5}{x} - 4 x + 3$ $\Rightarrow f (x) = \frac{1}{9} {\frac{5}{x} - 4 x + 3}$

$∴$ $18 \int_{1}^{2} f (x) d x = 18 \times \frac{1}{9} \int_{1}^{2} (\frac{5}{x} - 4 x + 3) d x$

$= 2 {[5 \log x - \frac{4 x^{2}}{2} + 3 x]}_{1}^{2}$

$= 2 [5 \log 2 - 6 + 3]$ $= 10 \log 2 - 6$

Q 27 :

$Let f (x) be a function satisfying f (x) + f (π - x) = π^{2}, \forall x \in ℝ .$ $Then \int_{0}^{π} f (x) \sin x d x is equal to$ [2023]

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

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(1)

$Let I = \int_{0}^{π} f (x) \sin x d x ...(i)$

$\Rightarrow I = \int_{0}^{π} f (π - x) \sin (π - x) d x$

$\Rightarrow I = \int_{0}^{π} f (π - x) \sin x d x ...(ii)$

$Adding (i) and (ii), we get$

$2 I = \int_{0}^{π} [f (x) + f (π - x)] \sin x d x$

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

$= \frac{π^{2}}{2} {[- \cos x]}_{0}^{π}$ $= \frac{π^{2}}{2} [2] = π^{2}$

Q 28 :

$If I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x and I (0) = 1,$ $then I (\frac{π}{3}) is equal to$ [2023]

$- e^{\frac{3}{4}}$
$e^{\frac{3}{4}}$
$\frac{1}{2} e^{\frac{3}{4}}$
$- \frac{1}{2} e^{\frac{3}{4}}$

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(3)

$I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x$

$= \int e^{g (x)} [f (x) g' (x) + f' (x)] d x$ $[where g (x) = \sin^{2} x and f (x) = \cos x]$

$= e^{g (x)} f (x) + C = e^{\sin^{2} x} \cdot \cos x + C$

$As I (0) = 1,$

$∴$ $e^{\sin^{2} 0} \cos (0) + C = 1$ $\Rightarrow C = 0$ $∴$ $I (x) = e^{\sin^{2} x} \cos x$

$Hence, I (\frac{π}{3}) = e^{\sin^{2} \frac{π}{3}} \cos \frac{π}{3} = \frac{1}{2} e^{\frac{3}{4}}$

Q 29 :

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0 .$ Then $f (\frac{π^{2}}{4})$ is equal to [2023]

$π (1 - \frac{π^{3}}{16})$
$- π^{2} (1 + \frac{π^{3}}{16})$
$- π (1 + \frac{π^{3}}{16})$
$π^{2} (1 - \frac{π^{2}}{16})$

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(1)

$Given, \int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$

Using Leibnitz Rule, we get

$[f (t^{2}) + {(t^{2})}^{2}] \cdot 2 t = 4 t^{2} \Rightarrow f (t^{2}) + t^{4} = 2 t \Rightarrow f (t^{2}) = 2 t - t^{4}$

$Put t = \frac{π}{2}$

$f ({(\frac{π}{2})}^{2}) = 2 \frac{π}{2} - {(\frac{π}{2})}^{4} = π - \frac{π^{4}}{16} = π (1 - \frac{π^{3}}{16})$

Q 30 :

The value of the integral $\int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\log_{e} (e^{x} + \sqrt{1 + e^{2 x}})) d x$ is equal to [2023]

$\log_{e} (\frac{2 (2 + \sqrt{5})}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$
$\log_{e} (\frac{\sqrt{2} {(3 - \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}}) + \frac{\sqrt{5}}{2}$
$\log_{e} (\frac{{(2 + \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}}) + \frac{\sqrt{5}}{2}$
$\log_{e} (\frac{\sqrt{2} {(2 + \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$

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(4)

$Let I = \int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\ln (e^{x} + \sqrt{1 + e^{2 x}})) d x$

$Put e^{x} = t \Rightarrow e^{x} d x = d t$

$When x = - \log_{e} 2, t = \frac{1}{2}; when x = \log_{e} 2, t = 2$

$∴$ $I = \int_{1 / 2}^{2} \ln (t + \sqrt{1 + t^{2}}) d t$

Applying integration by parts, we get

$I = {[t \ln (t + \sqrt{1 + t^{2}})]}_{1 / 2}^{2} - \int_{1 / 2}^{2} \frac{t}{t + \sqrt{1 + t^{2}}} (1 + \frac{2 t}{2 \sqrt{1 + t^{2}}}) d t$

$= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2}) - \int_{1 / 2}^{2} \frac{t}{\sqrt{1 + t^{2}}} d t$

$= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2}) - {[\sqrt{1 + t^{2}}]}_{1 / 2}^{2}$

$= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2}) - \frac{\sqrt{5}}{2}$ $= \ln (\frac{{(2 + \sqrt{5})}^{2}}{{(\frac{\sqrt{5} + 1}{2})}^{\frac{1}{2}}}) - \frac{\sqrt{5}}{2}$

Q 31 :

If $f : ℝ \to ℝ$ be a continuous function satisfying $\int_{0}^{π / 2} f (\sin 2 x) \sin x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0,$ then the value of $α$ is [2023]

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

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(1)

$I = \int_{0}^{π / 4} f (\sin 2 x) \sin x d x + \int_{π / 4}^{π / 2} f (\sin 2 x) \sin x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0$

$Apply \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x in the first part and put x - \frac{π}{4} = t in the second part, we get$

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

$= \int_{0}^{π / 4} f (\cos 2 x) [2 \sin \frac{π}{4} \cdot \cos x + α \cos x] d x = 0$

$(α + \sqrt{2}) \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0 ∴ α = - \sqrt{2}$

$[∵ In (0, \frac{π}{4}), f (\cos (2 x)) and \cos x are not zero.]$

Q 32 :

Let the function $f : [0, 2] \to ℝ$ be defined as $f (x) =$ ${\begin{matrix} e^{\min {x^{2}, x - [x]}}, & x \in [0, 1) \\ e^{[x - \log_{e} x]}, & x \in [1, 2) \end{matrix}$ where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is [2023]

$(e - 1) (e^{2} + \frac{1}{2})$
$2 e - \frac{1}{2}$
$1 + \frac{3 e}{2}$
$2 e - 1$

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(2)

$f (x) =$ ${\begin{matrix} e^{\min {x^{2}, x - [x]}}, & x \in [0, 1) \\ e^{[x - \log_{e} x]}, & x \in [1, 2) \end{matrix}$

$For x \in [0, 1);$ $\min {x^{2}, {x}} = x^{2}$

and for $x \in$ [1,2); $[x - \log_{e} x] = 1$

$∴$ $f (x) =$ ${\begin{matrix} e^{x^{2}}, & x \in [0, 1) \\ e, & x \in [1, 2) \end{matrix}$

$Now, \int_{0}^{2} x f (x) = \int_{0}^{1} x e^{x^{2}} d x + \int_{1}^{2} x e d x$

$= \frac{1}{2} (e - 1) + \frac{1}{2} (4 - 1) e = 2 e - \frac{1}{2}$

Q 33 :

$\int_{0}^{\infty} \frac{6}{e^{3 x} + 6 e^{2 x} + 11 e^{x} + 6} d x =$ [2023]

$\log_{e} (\frac{64}{27})$
$\log_{e} (\frac{512}{81})$
$\log_{e} (\frac{256}{81})$
$\log_{e} (\frac{32}{27})$

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(4)

Put $e^{x} = t \Rightarrow e^{x} d x = d t$ and on solving, we get $\log_{e} (\frac{32}{27})$

Q 34 :

$The value of \frac{e^{- \frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{- x} \tan^{50} x d x}{\int_{0}^{\frac{π}{4}} e^{- x} (\tan^{49} x + \tan^{51} x) d x} is:$ [2023]

51
49
25
50

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(4)

$\frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$

First, simplify, $\int_{0}^{π / 4} e^{- x} \tan^{50} x d x$

$= {[- e^{- x} {(\tan x)}^{50}]}_{0}^{π / 4} + \int_{0}^{π / 4} e^{- x} (50) \tan^{49} x \cdot \sec^{2} x d x$

$= - e^{- π / 4} + 0 + 50 \int_{0}^{π / 4} e^{- x} {(\tan x)}^{49} (1 + \tan^{2} x) d x$

$= - e^{- π / 4} + 50 \int_{0}^{π / 4} e^{- x} [{(\tan x)}^{51} + {(\tan x)}^{49}] d x$

Now, $\frac{- e^{- π / 4} + \int_{0}^{π / 4} e^{- x} {(\tan x)}^{50} d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x}$

$= \frac{50 \int_{0}^{π / 4} e^{- x} (\tan^{51} x + \tan^{49} x) d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x} = 50$

$∴$ $\frac{e^{- π / 4} + \int_{0}^{π / 4} e^{- x} \tan^{50} x d x}{\int_{0}^{π / 4} e^{- x} (\tan^{49} x + \tan^{51} x) d x} = 50$

Q 35 :

$If \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x = \frac{1}{α} \log_{e} (\frac{α + 1}{β}),$ $α, β > 0, then α^{4} - β^{4} is equal to$ [2023]

- 21
0
21
19

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(3)

Let $I = \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x$

$I = \int_{0}^{1} \frac{1}{(5 + 2 x (1 - x))} \cdot \frac{e^{4 x}}{(e^{4 x} + e^{2})} d x$ ...(i)

We know that, $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (b + a - x) d x$

$∴$ $I = \int_{0}^{1} \frac{1}{5 + 2 (1 - x) \cdot x} \cdot \frac{e^{4 (1 - x)}}{e^{4 (1 - x)} + e^{2}} d x$ ...(ii)

Adding (i) and (ii), we get

$\Rightarrow 2 I = \int_{0}^{1} \frac{1}{5 + 2 (1 - x) \cdot x} [\frac{e^{4 x} e^{4 (1 - x)} + e^{4 x} e^{2} + e^{4 x} e^{4 (1 - x)} + e^{4 (1 - x)} \cdot e^{2}}{(e^{4 (1 - x)} + e^{2}) (e^{4 x} + e^{2})}] d x$

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

$= \int_{0}^{1} \frac{d x}{(\frac{11}{4} - {(x - \frac{1}{2})}^{2})}$

$\Rightarrow I = \frac{1}{\sqrt{11}} \ln (\frac{\sqrt{11} + 1}{\sqrt{10}})$

$∴$ $α = \sqrt{11} and β = \sqrt{10} ∴ α^{4} - β^{4} = 121 - 100 = 21$

Q 36 :

$\lim_{n \to \infty} [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}] is equal to$ [2023]

$\log_{e} 2$
$\log_{e} (\frac{2}{3})$
0
$\log_{e} (\frac{3}{2})$

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(1)

$\lim_{n \to \infty} [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}]$

$= \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{r + n} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}} = \int_{0}^{1} \frac{d x}{1 + x} = \ln [1 + x] |_{0}^{1} = \ln 2$

Q 37 :

$The value of the integral \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - \cos 2 x} d x is$ [2023]

$\frac{π^{2}}{12 \sqrt{3}}$
$\frac{π^{2}}{6}$
$\frac{π^{2}}{3 \sqrt{3}}$
$\frac{π^{2}}{6 \sqrt{3}}$

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(4)

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

Replace $x$ by $- x$ , we get

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

Adding (i) and (ii), we get

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

$\Rightarrow I = \frac{π}{4} \cdot 2 \int_{0}^{\frac{π}{4}} \frac{(1 + \tan^{2} x) d x}{2 (1 + \tan^{2} x) - (1 - \tan^{2} x)}$

$= \frac{π}{2} \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x d x}{2 (1 + \tan^{2} x) - (1 - \tan^{2} x)}$

Let $\tan x = t \Rightarrow \sec^{2} x d x = d t$

at $x = 0, t = 0$ and $x = \frac{π}{4}, t = 1$

$I = \frac{π}{2} \int_{0}^{1} \frac{d t}{3 t^{2} + 1} \Rightarrow I = \frac{π}{6} \int_{0}^{1} \frac{d t}{t^{2} + {(\frac{1}{\sqrt{3}})}^{2}} = \frac{π}{6} \sqrt{3} \cdot {[\tan^{- 1} \sqrt{3} t]}_{0}^{1}$

$\Rightarrow I = \frac{π}{2 \sqrt{3}} \tan^{- 1} \sqrt{3} I = \frac{π^{2}}{6 \sqrt{3}}$

Q 38 :

$\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4 x^{2}}} d x is equal to$ [2023]

$2 π$
$\frac{π}{6}$
$\frac{π}{3}$
$\frac{π}{2}$

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(1)

Q 39 :

$The minimum value of the function f (x) = \int_{0}^{2} e^{| x - t |} d t is$ [2023]

2
2(e - 1)
2e - 1
e(e - 1)

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(2)

Case 1: When $x < 0$

$f (x) = \int_{0}^{2} e^{t - x} d t = e^{- x} {(e^{t})}_{0}^{2} = e^{- x} (e^{2} - 1)$

Case 2: When $0 < x < 2$

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

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

Case 3: When $x \geq 2$

$f (x) = \int_{0}^{2} e^{(x - t)} d t = e^{x} {(- e^{- t})}_{0}^{2} = - e^{x} (e^{- 2} - 1)$

Hence,

$f (x) =$ ${\begin{matrix} e^{- x} (e^{2} - 1), & x < 0 \\ e^{x} + e^{- x} e^{2} - 2, & 0 \leq x < 2 \\ (1 - e^{- 2}) e^{x}, & x \geq 2 \end{matrix}$

$f (x)$ is decreasing for $x < 0$ and increasing for $x \geq 2$ .

$f (x)$ is also continuous.

For $0 \leq x \leq 2$ , $f (x)$ is minimum at $x = 1$ .

(Using A.M.–G.M. inequality)

$Hence, the minimum value f (x) = f (1) = 2 (e - 1)$

Q 40 :

$The integral 16 \int_{1}^{2} \frac{d x}{x^{3} {(x^{2} + 2)}^{2}} is equal to$ [2023]

$\frac{11}{6} - \log_{e} 4$
$\frac{11}{12} + \log_{e} 4$
$\frac{11}{6} + \log_{e} 4$
$\frac{11}{12} - \log_{e} 4$

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(1)

Let $I = 16 \int_{1}^{2} \frac{d x}{x^{3} {(x^{2} + 2)}^{2}} = 16 \int_{1}^{2} \frac{1}{x^{7} {(1 + \frac{2}{x^{2}})}^{2}} d x$

Let $1 + \frac{2}{x^{2}} = p \Rightarrow - \frac{4}{x^{3}} d x = d p \Rightarrow \frac{d x}{x^{3}} = \frac{- 1}{4} d p$

$I = - 4 \int_{3}^{3 / 2} \frac{d p}{\frac{4}{{(p - 1)}^{2}} \cdot p^{2}}$

$= - 4 \int_{3}^{3 / 2} \frac{{(p - 1)}^{2} d p}{4 p^{2}} = - \int_{3}^{3 / 2} \frac{p^{2} + 1 - 2 p}{p^{2}} d p$

$= - \int_{3}^{3 / 2} (1 + \frac{1}{p^{2}} - \frac{2}{p}) d p = - {[p - \frac{1}{p} - 2 \log p]}_{3}^{3 / 2}$

$= - [{\frac{5}{6} - 2 \log_{e} (\frac{3}{2})} - {\frac{8}{3} - 2 \log_{e} 3}]$

$= - [2 \log_{e} 2 - \frac{11}{6}] = \frac{11}{6} - \log_{e} 2^{2} = \frac{11}{6} - \log_{e} 4$

Q 41 :

Let $[x]$ denote the greatest integer $\leq x$ . Consider the function $f (x) = \max {x^{2}, 1 + [x]}$ . Then the value of the integral $\int_{0}^{2} f (x) d x$ is [2023]

$\frac{5 + 4 \sqrt{2}}{3}$
$\frac{8 + 4 \sqrt{2}}{3}$
$\frac{1 + 5 \sqrt{2}}{3}$
$\frac{4 + 5 \sqrt{2}}{3}$

1

2

3

4

(1)

Given $f (x) = \max {x^{2}, 1 + [x]}$

So, $\int_{0}^{2} f (x) d x = \int_{0}^{1} 1 d x + \int_{1}^{\sqrt{2}} 2 d x + \int_{\sqrt{2}}^{2} x^{2} d x$

$= 1 + 2 (\sqrt{2} - 1) + \frac{8 - 2 \sqrt{2}}{3} = \frac{5 + 4 \sqrt{2}}{3}$

Q 42 :

$Let f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x, x \in ℝ$ be a function which satisfies $f (x) = x + \int_{0}^{π / 2} \sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to [2023]

$- π (π + 2)$
$- π (π - 2)$
$- 2 π (π + 2)$
$- 2 π (π - 2)$

1

2

3

4

(3)

$f (x) = x + \int_{0}^{π / 2} (\sin x \cos y + \cos x \sin y) f (y) d y$

$= x + \int_{0}^{π / 2} ((\cos y f (y) d y) \sin x + (\sin y f (y) d y) \cos x)$ ...(i)

On comparing with $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x$

$\Rightarrow \frac{a}{π^{2} - 4} = \int_{0}^{π / 2} \cos y f (y) d y$ ...(ii)

$\Rightarrow \frac{b}{π^{2} - 4} = \int_{0}^{π / 2} \sin y f (y) d y$ ...(iii)

Adding (ii) and (iii), we get

$\frac{a + b}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) f (y) d y$ ...(iv)

$\frac{a + b}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) f (\frac{π}{2} - y) d y$ ...(v)

Adding (iv) and (v), we get

$\frac{2 (a + b)}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) [\frac{π}{2} + \frac{(a + b)}{π^{2} - 4} (\sin y + \cos y)] d y$

$= π + \frac{a + b}{π^{2} - 4} (\frac{π}{2} + 1)$ $\Rightarrow (a + b) = - 2 π (π + 2)$

Q 43 :

$The value of the integral \int_{1 / 2}^{2} \frac{\tan^{- 1} x}{x} d x is equal to$ [2023]

$\frac{1}{2} \log_{e} 2$
$π \log_{e} 2$
$\frac{π}{4} \log_{e} 2$
$\frac{π}{2} \log_{e} 2$

1

2

3

4

(4)

We have the integral as follows

$I = \int_{1 / 2}^{2} \frac{\tan^{- 1} x}{x} d x \dots (1)$

By substituting $x = \frac{1}{t}$ and $d x = - \frac{1}{t^{2}} d t$ in equation (1),

$I = - \int_{2}^{1 / 2} \frac{\tan^{- 1} \frac{1}{t}}{\frac{1}{t}} \cdot \frac{1}{t^{2}} \cdot d t$

$I = - \int_{2}^{1 / 2} \frac{\tan^{- 1} \frac{1}{t}}{t} d t = \int_{1 / 2}^{2} \frac{\cot^{- 1} t}{t} d t = \int_{1 / 2}^{2} \frac{\cot^{- 1} x}{x} d x \dots (2)$

On adding equation (1) and (2), we have

$2 I = \int_{1 / 2}^{2} \frac{\tan^{- 1} x + \cot^{- 1} x}{x} d x;$ $2 I = \frac{π}{2} \int_{1 / 2}^{2} \frac{d x}{x} = \frac{π}{2} {[\log x]}_{1 / 2}^{2}$

$2 I = \frac{π}{2} [\log_{e} 2 - \log_{e} \frac{1}{2}] = π \log_{e} 2$

$\Rightarrow$ $2 I = {πlog}_{e} 2$ $∴$ $I = \frac{π}{2} \log_{e} 2$

Hence, option (4) is the correct answer.

Q 44 :

$The value of the integral \int_{1}^{2} (\frac{t^{4} + 1}{t^{6} + 1}) d t is$ [2023]

$\tan^{- 1} 2 - \frac{1}{3} \tan^{- 1} 8 + \frac{π}{3}$
$\tan^{- 1} 2 + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$
$\tan^{- 1} \frac{1}{2} + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$
$\tan^{- 1} \frac{1}{2} - \frac{1}{3} \tan^{- 1} 8 + \frac{π}{3}$

1

2

3

4

(2)

Let $I = \int_{1}^{2} [\frac{t^{4} + 1}{t^{6} + 1}] d t$

$= \int_{1}^{2} \frac{1}{t^{2} + 1} d t + \frac{1}{3} \int_{1}^{2} \frac{3 t^{2}}{{(t^{3})}^{2} + 1} d t$

$= {[\tan^{- 1} t]}_{1}^{2} + {[\frac{1}{3} (\tan^{- 1} t^{3})]}_{1}^{2}$

$= \tan^{- 1} 2 + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$

Q 45 :

If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\frac{3 (e - 1)}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ is [2023]

$e^{8} - 1$
$e^{7} - 1$
$e^{9} - e$
$e^{8} - e$

1

2

3

4

(4)

Let $I = \frac{3 (e - 1)}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$

Between 1 and 2, $[x]$ = 1, so $e^{[x]} = e$ .

$∴$ $I = 3 (e - 1) \int_{1}^{2} x^{2} \cdot e^{[x^{3}]} d x$

Let $x^{3} = t \Rightarrow x^{2} d x = \frac{d t}{3}$

$I = \frac{3 (e - 1)}{3} \int_{1}^{8} e^{[t]} d t = (e - 1) \int_{1}^{8} e^{[t]} d t$

$= (e - 1) [e + e^{2} + e^{3} + e^{4} + e^{5} + e^{6} + e^{7}]$ $= (e - 1) [\frac{e (e^{7} - 1)}{e - 1}]$

$\Rightarrow e (e^{7} - 1) = e^{8} - e$

Q 46 :

$\lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$ is equal to [2023]

$\frac{19}{3}$
12
0
19

1

2

3

4

(4)

We have, $\lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$

$= \lim_{n \to \infty} \frac{3}{n} {{(2 + \frac{0}{n})}^{2} + {(2 + \frac{1}{n})}^{2} + \dots + (2 + {(\frac{n - 1}{n}))}^{2}}$

$= \lim_{n \to \infty} \frac{3}{n} \sum_{r = 0}^{n - 1} {(2 + \frac{r}{n})}^{2} = 3 \int_{0}^{1} {(2 + x)}^{2} d x$

Let $2 + x = t d x = d t$ $\Rightarrow 3 \int_{2}^{3} t^{2} d t = 3 \times {[\frac{t^{3}}{2}]}_{2}^{3} = \frac{3}{3} [27 - 8] = 19$

Q 47 :

Let $α \in (0, 1)$ and $β = \log_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1) .$ Then the integral $\int_{0}^{α} \frac{t^{50}}{1 - t} d t$ is equal to [2023]

$P_{50} (α) - β$
$- (β + P_{50} (α))$
$β + P_{50} (α)$
$β - P_{50} (α)$

1

2

3

4

(2)

We have, $P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1)$ ...(i)

Let $I = \int_{0}^{α} \frac{t^{50}}{1 - t} d t = \int_{0}^{α} \frac{t^{50} + 1 - 1}{1 - t} d t = \int_{0}^{α} (\frac{- (1 - t^{50})}{1 - t} + \frac{1}{1 - t}) d t$

$= \int_{0}^{α} - (1 + t + t^{2} + \dots + t^{49}) d t + \int_{0}^{α} \frac{1}{1 - t} d t$

$= - {[t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + \dots + \frac{t^{50}}{50}]}_{0}^{α} - {[\log (1 - t)]}_{0}^{α}$

$= - \log (1 - α) - [α + \frac{α^{2}}{2} + \frac{α^{3}}{3} + \dots + \frac{α^{50}}{50}]$

$= - β - P_{50} (α)$ $[∵ β = \log_{e} (1 - α) and from (i)]$

$= - (β + P_{50} (α))$

Q 48 :

The value of $\int_{π / 3}^{π / 2} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to [2023]

$\frac{7}{2} - \sqrt{3} - \log_{e} \sqrt{3}$
$\frac{10}{3} - \sqrt{3} - \log_{e} \sqrt{3}$
$\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$
$- 2 + 3 \sqrt{3} + \log_{e} \sqrt{3}$

1

2

3

4

(3)

Let $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$

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

$= \int_{π / 3}^{π / 2} \frac{2}{\sin^{3} x} d x - \int_{π / 3}^{π / 2} \frac{2 \cos x}{\sin^{3} x} d x + \frac{3}{2} \int_{π / 3}^{π / 2} \frac{1}{\cos^{2} \frac{x}{2}} d x$

$= 2 \int_{π / 3}^{π / 2} \sqrt{c o s e c^{2} x} c o s e c^{2} x d x - 2 \int_{π / 3}^{π / 2} \cot x \cdot c o s e c^{2} x d x + \frac{3}{2} {[2 \tan \frac{x}{2}]}_{π / 3}^{π / 2}$

$= 2 \int_{π / 3}^{π / 2} \sqrt{1 + \cot^{2} x} c o s e c^{2} d x - 2 \int_{π / 3}^{π / 2} \cot x \cdot c o s e c^{2} d x + 3 [\tan \frac{π}{4} - \tan \frac{π}{6}]$

Let $\cot x = t \Rightarrow c o s e c^{2} x d x = - d t$ and

For $x = \frac{π}{2}; t = 0$ and for $x = \frac{π}{3}; t = \frac{1}{\sqrt{3}}$

$∴$ $I = - 2 \int_{1 / \sqrt{3}}^{0} \sqrt{1 + t^{2}} d t + 2 \int_{1 / \sqrt{3}}^{0} t d t + 3 [1 - \frac{1}{\sqrt{3}}]$

$= - 2 {[\frac{t}{2} \sqrt{1 + t^{2}} + \frac{1}{2} \log_{e} (t + \sqrt{1 + t^{2}})]}_{1 / \sqrt{3}}^{0} + {[t^{2}]}_{1 / \sqrt{3}}^{0} + (3 - \sqrt{3})$

$= \frac{2}{3} + \log \sqrt{3} - \frac{1}{3} + 3 - \sqrt{3} = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$

Q 49 :

If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t, x > 0,$ then $ϕ' (\frac{π}{4})$ is equal to [2023]

$\frac{4}{6 + \sqrt{π}}$
$\frac{8}{6 + \sqrt{π}}$
$\frac{8}{\sqrt{π}}$
$\frac{4}{6 - \sqrt{π}}$

1

2

3

4

(2)

By Leibnitz rule,

$ϕ' (x) = \frac{1}{\sqrt{x}} [(4 \sqrt{2} \sin x - 3 ϕ' (x)) \cdot 1 - 0] - \frac{1}{2} x^{- 3 / 2} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t$

Put $x = \frac{π}{4}$ , we get $ϕ' (\frac{π}{4}) = \frac{2}{\sqrt{π}} [4 - 3 ϕ' (\frac{π}{4})] + 0$

$\Rightarrow (1 + \frac{6}{\sqrt{π}}) ϕ' (\frac{π}{4}) = \frac{8}{\sqrt{π}}$ $= \frac{8}{\sqrt{π}} \times (\frac{\sqrt{π}}{6 + \sqrt{π}}) = \frac{8}{6 + \sqrt{π}}$

Q 50 :

Let $α > 0$ . If $\int_{0}^{α} \frac{x}{\sqrt{x + α} - \sqrt{x}} d x = \frac{16 + 20 \sqrt{2}}{15},$ then $α$ is equal to [2023]

$2$
$2 \sqrt{2}$
$4$
$\sqrt{2}$

1

2

3

4

(1)

After rationalising, $I = \int_{0}^{α} \frac{x}{α} (\sqrt{x + α} + \sqrt{x}) d x$

$= \int_{0}^{α} \frac{1}{α} [{(x + α)}^{3 / 2} - α {(x + α)}^{1 / 2} + x^{3 / 2}] d x$

$= \frac{1}{α} {[\frac{2}{5} {(x + α)}^{5 / 2} - α \frac{2}{3} {(x + α)}^{3 / 2} + \frac{2}{5} x^{5 / 2}]}_{0}^{α}$

$= \frac{1}{α} (\frac{2}{5} {(2 α)}^{5 / 2} - \frac{2 α}{3} {(2 α)}^{3 / 2} + \frac{2}{5} α^{5 / 2} - \frac{2}{5} α^{5 / 2} + \frac{2}{3} α^{5 / 2})$

$= \frac{1}{α} (\frac{2^{7 / 2} α^{5 / 2}}{5} - \frac{2^{5 / 2} α^{5 / 2}}{3} + \frac{2}{3} α^{5 / 2})$ $= α^{3 / 2} (\frac{2^{7 / 2}}{5} - \frac{2^{5 / 2}}{3} + \frac{2}{3})$

$= \frac{α^{3 / 2}}{15} (24 \sqrt{2} - 20 \sqrt{2} + 10) = \frac{α^{3 / 2}}{15} (4 \sqrt{2} + 10)$

Now, $\frac{α^{3 / 2}}{15} (4 \sqrt{2} + 10) = \frac{16 + 20 \sqrt{2}}{15} = \frac{2 \sqrt{2} (4 \sqrt{2} + 10)}{15}$

$\Rightarrow α^{3 / 2} = 2 \sqrt{2} = 2^{3 / 2} \Rightarrow α = 2$

Topic Question Set

Q 21 :

The integral $80 \int_{0}^{\frac{π}{4}} (\frac{\sin θ + \cos θ}{9 + 16 \sin 2 θ}) d θ$ is equal to : [2025]

Q 22 :

Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ . If the range of f is $[α, β]$ , then $4 (α + β)$ equals : [2025]

Q 23 :

Let $[\cdot]$ denote the greatest integer function. If $\int_{0}^{e^{3}} [\frac{1}{e^{x - 1}}] d x = α - \log_{e} 2$ , then $α^{3}$ is equal to __________. [2025]

Q 24 :

If $\lim_{t \to 0} (\int_{0}^{1} {{(3 x + 5)}^{t} d t)}^{\frac{1}{t}} = \frac{α}{5 e} {(\frac{8}{5})}^{\frac{2}{3}}$ , then $α$ is equal to _________. [2025]

Q 25 :

If $24 \int_{0}^{\frac{π}{4}} (\sin | 4 x - \frac{π}{12} | + [2 \sin x]) d x = 2 π + α$ , where $[\cdot]$ denotes the greatest integer function, then $α$ is equal to _________. [2025]

Q 26 :

$Let 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0 .$ $Then 18 \int_{1}^{2} f (x) d x is equal to$ [2023]

Q 27 :

$Let f (x) be a function satisfying f (x) + f (π - x) = π^{2}, \forall x \in ℝ .$ $Then \int_{0}^{π} f (x) \sin x d x is equal to$ [2023]

Q 28 :

$If I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x and I (0) = 1,$ $then I (\frac{π}{3}) is equal to$ [2023]

Q 29 :

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0 .$ Then $f (\frac{π^{2}}{4})$ is equal to [2023]

Q 30 :

The value of the integral $\int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\log_{e} (e^{x} + \sqrt{1 + e^{2 x}})) d x$ is equal to [2023]

Q 31 :

If $f : ℝ \to ℝ$ be a continuous function satisfying $\int_{0}^{π / 2} f (\sin 2 x) \sin x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0,$ then the value of $α$ is [2023]

Q 33 :

$\int_{0}^{\infty} \frac{6}{e^{3 x} + 6 e^{2 x} + 11 e^{x} + 6} d x =$ [2023]

Q 34 :

$The value of \frac{e^{- \frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{- x} \tan^{50} x d x}{\int_{0}^{\frac{π}{4}} e^{- x} (\tan^{49} x + \tan^{51} x) d x} is:$ [2023]

Q 35 :

$If \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x = \frac{1}{α} \log_{e} (\frac{α + 1}{β}),$ $α, β > 0, then α^{4} - β^{4} is equal to$ [2023]

Q 36 :

$\lim_{n \to \infty} [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}] is equal to$ [2023]

Q 37 :

$The value of the integral \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - \cos 2 x} d x is$ [2023]

Q 38 :

$\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4 x^{2}}} d x is equal to$ [2023]

(1)

Q 39 :

$The minimum value of the function f (x) = \int_{0}^{2} e^{| x - t |} d t is$ [2023]

Q 40 :

$The integral 16 \int_{1}^{2} \frac{d x}{x^{3} {(x^{2} + 2)}^{2}} is equal to$ [2023]

Q 41 :

Let $[x]$ denote the greatest integer $\leq x$ . Consider the function $f (x) = \max {x^{2}, 1 + [x]}$ . Then the value of the integral $\int_{0}^{2} f (x) d x$ is [2023]

Q 42 :

$Let f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x, x \in ℝ$ be a function which satisfies $f (x) = x + \int_{0}^{π / 2} \sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to [2023]

Q 43 :

$The value of the integral \int_{1 / 2}^{2} \frac{\tan^{- 1} x}{x} d x is equal to$ [2023]

Q 44 :

$The value of the integral \int_{1}^{2} (\frac{t^{4} + 1}{t^{6} + 1}) d t is$ [2023]

Q 45 :

If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\frac{3 (e - 1)}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ is [2023]

Q 46 :

$\lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$ is equal to [2023]

Q 47 :

Let $α \in (0, 1)$ and $β = \log_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1) .$ Then the integral $\int_{0}^{α} \frac{t^{50}}{1 - t} d t$ is equal to [2023]

Q 48 :

The value of $\int_{π / 3}^{π / 2} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to [2023]

Q 49 :

If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t, x > 0,$ then $ϕ' (\frac{π}{4})$ is equal to [2023]

Q 50 :

Let $α > 0$ . If $\int_{0}^{α} \frac{x}{\sqrt{x + α} - \sqrt{x}} d x = \frac{16 + 20 \sqrt{2}}{15},$ then $α$ is equal to [2023]

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Topic Question Set

Q 21 : The integral 80∫0π4(sinθ+cosθ9+16sin2θ)dθ is equal to : [2025]

Q 22 : Let f(x)=∫0xt(t2–9t+20)dt, 1≤x≤5. If the range of f is [α,β], then 4(α+β) equals : [2025]

Q 23 : Let [·] denote the greatest integer function. If ∫0e3[1ex–1]dx=α–loge2, then α3 is equal to __________. [2025]

Q 24 : If limt→0(∫01(3x+5)tdt)1t=α5e(85)23, then α is equal to _________. [2025]

Q 25 : If 24∫0π4(sin|4x–π12|+[2sinx])dx=2π+α, where [·] denotes the greatest integer function, then α is equal to _________. [2025]

Q 26 : Let 5f(x)+4f(1x)=1x+3, x>0. Then 18∫12f(x) dx is equal to [2023]

Q 27 : Let f(x) be a function satisfying f(x)+f(π-x)=π2,∀x∈ℝ. Then ∫0πf(x)sinx dx is equal to [2023]

Q 28 : If I(x)=∫esin2x(cosxsin2x-sinx) dx and I(0)=1, then I(π3) is equal to [2023]

Q 29 : Let f be a continuous function satisfying ∫0t2(f(x)+x2) dx=43t3,∀t>0. Then f(π24) is equal to [2023]

Q 30 : The value of the integral ∫-loge2loge2ex(loge(ex+1+e2x))dx is equal to [2023]

Q 31 : If f:ℝ→ℝ be a continuous function satisfying ∫0π/2f(sin2x)sinx dx+α∫0π/4f(cos2x)cosx dx=0, then the value of α is [2023]

Q 32 : Let the function f:[0,2]→ℝ be defined as f(x)={emin{x2, x-[x]},x∈[0,1)e[x-logex],x∈[1,2) where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫02xf(x)dx is [2023]

Q 33 : ∫0∞6e3x+6e2x+11ex+6dx= [2023]

Q 34 : The value of e-π4+∫0π4e-xtan50x dx∫0π4e-x(tan49x+tan51x) dx is: [2023]

Q 35 : If ∫011(5+2x-2x2)(1+e(2-4x))dx=1αloge(α+1β), α,β>0, then α4-β4 is equal to [2023]

Q 36 : limn→∞[11+n+12+n+13+n+…+12n] is equal to [2023]

Q 37 : The value of the integral ∫-π4π4x+π42-cos2xdx is [2023]

Q 38 : ∫324334489-4x2dx is equal to [2023]

(1)

Q 39 : The minimum value of the function f(x)=∫02e|x-t| dt is [2023]

Q 40 : The integral 16∫12dxx3(x2+2)2 is equal to [2023]

Q 41 : Let [x] denote the greatest integer ≤x. Consider the function f(x)=max{x2, 1+[x]}. Then the value of the integral ∫02f(x) dx is [2023]

Q 42 : Let f(x)=x+aπ2-4sinx+bπ2-4cosx, x∈ℝ be a function which satisfies f(x)=x+∫0π/2sin(x+y)f(y) dy. Then (a+b) is equal to [2023]

Q 43 : The value of the integral ∫1/22tan-1xxdx is equal to [2023]

Q 44 : The value of the integral ∫12(t4+1t6+1)dt is [2023]

Q 45 : If [t] denotes the greatest integer ≤t, then the value of 3(e-1)e∫12x2e[x]+[x3] dx is [2023]

Q 46 : limn→∞3n{4+(2+1n)2+(2+2n)2+⋯+(3-1n)2} is equal to [2023]

Q 47 : Let α∈(0,1) and β=loge(1-α). Let Pn(x)=x+x22+x33+…+xnn,x∈(0,1). Then the integral ∫0αt501-tdt is equal to [2023]

Q 48 : The value of ∫π/3π/2(2+3sinx)sinx(1+cosx)dx is equal to [2023]

Q 49 : If ϕ(x)=1x∫π/4x(42sint-3ϕ'(t))dt,x>0, then ϕ'(π4) is equal to [2023]

Q 50 : Let α>0. If ∫0αxx+α-xdx=16+20215, then α is equal to [2023]

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Q 21 :

The integral $80 \int_{0}^{\frac{π}{4}} (\frac{\sin θ + \cos θ}{9 + 16 \sin 2 θ}) d θ$ is equal to : [2025]

Q 22 :

Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ . If the range of f is $[α, β]$ , then $4 (α + β)$ equals : [2025]

Q 23 :

Let $[\cdot]$ denote the greatest integer function. If $\int_{0}^{e^{3}} [\frac{1}{e^{x - 1}}] d x = α - \log_{e} 2$ , then $α^{3}$ is equal to __________. [2025]

Q 24 :

If $\lim_{t \to 0} (\int_{0}^{1} {{(3 x + 5)}^{t} d t)}^{\frac{1}{t}} = \frac{α}{5 e} {(\frac{8}{5})}^{\frac{2}{3}}$ , then $α$ is equal to _________. [2025]

Q 25 :

If $24 \int_{0}^{\frac{π}{4}} (\sin | 4 x - \frac{π}{12} | + [2 \sin x]) d x = 2 π + α$ , where $[\cdot]$ denotes the greatest integer function, then $α$ is equal to _________. [2025]

Q 26 :

$Let 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0 .$ $Then 18 \int_{1}^{2} f (x) d x is equal to$ [2023]

Q 27 :

$Let f (x) be a function satisfying f (x) + f (π - x) = π^{2}, \forall x \in ℝ .$ $Then \int_{0}^{π} f (x) \sin x d x is equal to$ [2023]

Q 28 :

$If I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x and I (0) = 1,$ $then I (\frac{π}{3}) is equal to$ [2023]

Q 29 :

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0 .$ Then $f (\frac{π^{2}}{4})$ is equal to [2023]

Q 30 :

The value of the integral $\int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\log_{e} (e^{x} + \sqrt{1 + e^{2 x}})) d x$ is equal to [2023]

Q 31 :

If $f : ℝ \to ℝ$ be a continuous function satisfying $\int_{0}^{π / 2} f (\sin 2 x) \sin x d x + α \int_{0}^{π / 4} f (\cos 2 x) \cos x d x = 0,$ then the value of $α$ is [2023]

Q 33 :

$\int_{0}^{\infty} \frac{6}{e^{3 x} + 6 e^{2 x} + 11 e^{x} + 6} d x =$ [2023]

Q 34 :

$The value of \frac{e^{- \frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{- x} \tan^{50} x d x}{\int_{0}^{\frac{π}{4}} e^{- x} (\tan^{49} x + \tan^{51} x) d x} is:$ [2023]

Q 35 :

$If \int_{0}^{1} \frac{1}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})} d x = \frac{1}{α} \log_{e} (\frac{α + 1}{β}),$ $α, β > 0, then α^{4} - β^{4} is equal to$ [2023]

Q 36 :

$\lim_{n \to \infty} [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}] is equal to$ [2023]

Q 37 :

$The value of the integral \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - \cos 2 x} d x is$ [2023]

Q 38 :

$\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4 x^{2}}} d x is equal to$ [2023]

Q 39 :

$The minimum value of the function f (x) = \int_{0}^{2} e^{| x - t |} d t is$ [2023]

Q 40 :

$The integral 16 \int_{1}^{2} \frac{d x}{x^{3} {(x^{2} + 2)}^{2}} is equal to$ [2023]

Q 41 :

Let $[x]$ denote the greatest integer $\leq x$ . Consider the function $f (x) = \max {x^{2}, 1 + [x]}$ . Then the value of the integral $\int_{0}^{2} f (x) d x$ is [2023]

Q 42 :

$Let f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x, x \in ℝ$ be a function which satisfies $f (x) = x + \int_{0}^{π / 2} \sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to [2023]

Q 43 :

$The value of the integral \int_{1 / 2}^{2} \frac{\tan^{- 1} x}{x} d x is equal to$ [2023]

Q 44 :

$The value of the integral \int_{1}^{2} (\frac{t^{4} + 1}{t^{6} + 1}) d t is$ [2023]

Q 45 :

If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\frac{3 (e - 1)}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ is [2023]

Q 46 :

$\lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$ is equal to [2023]

Q 47 :

Let $α \in (0, 1)$ and $β = \log_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1) .$ Then the integral $\int_{0}^{α} \frac{t^{50}}{1 - t} d t$ is equal to [2023]

Q 48 :

The value of $\int_{π / 3}^{π / 2} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to [2023]

Q 49 :

If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t, x > 0,$ then $ϕ' (\frac{π}{4})$ is equal to [2023]

Q 50 :

Let $α > 0$ . If $\int_{0}^{α} \frac{x}{\sqrt{x + α} - \sqrt{x}} d x = \frac{16 + 20 \sqrt{2}}{15},$ then $α$ is equal to [2023]