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

Q 61 :

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 62 :

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 63 :

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]

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(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 64 :

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]

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(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 65 :

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]

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(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 66 :

$Let 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0 .$ $Then 18 \int_{1}^{2} f (x) d x is equal to$ [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 67 :

$Let f (x) be a function satisfying f (x) + f (π - x) = π^{2}, \forall x \in ℝ .$ $Then \int_{0}^{π} f (x) \sin x d x is equal to$ [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 68 :

$If I (x) = \int e^{\sin^{2} x} (\cos x \sin 2 x - \sin x) d x and I (0) = 1,$ $then I (\frac{π}{3}) is equal to$ [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 69 :

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 70 :

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

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

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

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

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

Topic Question Set

Q 61 :

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 62 :

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 63 :

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]

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Q 64 :

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]

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Q 65 :

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]

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Q 66 :

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

Q 67 :

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

Q 68 :

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

Q 69 :

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 70 :

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]

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Topic Question Set

Q 61 : The integral 80∫0π4(sinθ+cosθ9+16sin2θ)dθ is equal to : [2025] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 66 : Let 5f(x)+4f(1x)=1x+3, x>0. Then 18∫12f(x) dx is equal to [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Want Us to Email you About Special Offers & Updates?

Q 61 :

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 66 :

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