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]

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

1

2

3

4

(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

1

2

3

4

(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]

0%

(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]

0%

(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]

0%

(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 $α = 12$ .

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]

0%

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]

0%

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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Topic Question Set

Q 61 : The integral 80∫0π4(sinθ+cosθ9+16sin2θ)dθ is equal to : [2025]

Q 62 : 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 63 : Let [·] denote the greatest integer function. If ∫0e3[1ex–1]dx=α–loge2, then α3 is equal to __________. [2025]

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Q 64 : If limt→0(∫01(3x+5)tdt)1t=α5e(85)23, then α is equal to _________. [2025]

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Q 65 : If 24∫0π4(sin|4x–π12|+[2sinx])dx=2π+α, where [·] denotes the greatest integer function, then α is equal to _________. [2025]

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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]

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]

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]