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

Q 101 :

If $\int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x}{1 + 5^{\cos x}} = \frac{k π}{16},$ then $k$ is equal to ______. [2023]

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(13)

$Let$

$I = \int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x)}{1 + 5^{\cos x}} d x$ ...(i)

$I = \int_{0}^{π} \frac{5^{\cos (π - x)} (1 + \cos (π - x) \cdot \cos 3 (π - x) + \cos^{2} (π - x) + \cos^{3} (π - x) \cdot \cos 3 (π - x))}{1 + 5^{\cos (π - x)}} d x$

$\Rightarrow I = \int_{0}^{π} \frac{5^{- \cos x} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x) d x}{1 + 5^{- \cos x}}$ ...(ii)

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

$2 I = \int_{0}^{π} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x) d x$

$\Rightarrow 2 I = 2 \int_{0}^{π / 2} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x) d x$ ...(iii)

$\Rightarrow I = \int_{0}^{π / 2} (1 + \sin x (- \sin 3 x) + \sin^{2} x - \sin^{3} x \cdot \sin 3 x) d x$ ...(iv)

$Adding (iii) and (iv), we get$

$\Rightarrow 2 I = \int_{0}^{π / 2} (3 + \cos 4 x + \cos^{3} x \cdot \cos 3 x - \sin^{3} x \cdot \sin 3 x) d x$

$2 I = \int_{0}^{π / 2} 3 + \cos 4 x + (\frac{\cos 3 x + 3 \cos x}{4}) \cos 3 x - \sin 3 x (\frac{3 \sin x - \sin 3 x}{4}) d x$

$\Rightarrow 2 I = \int_{0}^{π / 2} (3 + \cos 4 x + \frac{1}{4} + \frac{3}{4} \cos 4 x) d x$

$\Rightarrow 2 I = \frac{13}{4} \times \frac{π}{2} + \frac{7}{4} {(\frac{\sin 4 x}{4})}_{0}^{π / 2} \Rightarrow I = \frac{13 π}{16}$

$Hence, k = 13$

Q 102 :

$If \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \frac{m}{n} \log_{e} (\frac{n^{2}}{e}), where m and n are coprime natural numbers, then m^{2} + n^{2} - 5 is equal to$ ______ . [2023]

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(20)

$Let I = \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \int_{1 / 3}^{1} (- \log_{e} x) d x + \int_{1}^{3} (\log_{e} x) d x$

$= - {[x \log_{e} x - x]}_{1 / 3}^{1} + {[x \log_{e} x - x]}_{1}^{3}$

$= - [- 1 - (\frac{1}{3} \log_{e} \frac{1}{3} - \frac{1}{3})] + [3 \log_{e} 3 - 3 - (- 1)]$

$= - \frac{4}{3} + \frac{8}{3} \log_{e} 3 = \frac{4}{3} (2 \log_{e} 3 - 1) = \frac{4}{3} (\log_{e} \frac{9}{e})$

$Comparing with the given condition, we get m = 4, n = 3 .$

$Now, m^{2} + n^{2} - 5 = 16 + 9 - 5 = 20$

Q 103 :

$\lim_{x \to 0} \frac{48}{x^{4}} \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t is equal to$ ______. [2023]

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(12)

Let $I = \lim_{x \to 0} \frac{48 \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t}{x^{4}}$ $(\frac{0}{0} form)$

$I = \lim_{x \to 0} \frac{48 (\frac{x^{3}}{x^{6} + 1})}{4 x^{3}} = \lim_{x \to 0} \frac{48 \cdot x^{3}}{x^{6} + 1} \times \frac{1}{4 x^{3}} = \lim_{x \to 0} \frac{12}{x^{6} + 1}$

$I = 12$

Q 104 :

The value of the integral $\int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\tan 2 x}}$ is: [2026]

$\frac{π}{3}$
$\frac{π}{18}$
$\frac{π}{12}$
$\frac{π}{6}$

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(3)

Q 105 :

Let $f$ be a twice differentiable non-negative function such that $(f {(x))}^{2} = 25 + \int_{0}^{x} ((f {(t))}^{2} + (f' {(t))}^{2}) d t .$ Then the mean of $f (\log_{e} (1)), f (\log_{e} (2)), \dots, f (\log_{e} (625))$ is equal to _________ . [2026]

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(1565)

Q 106 :

The value of $\int_{- π / 6}^{π / 6} (\frac{π + 4 x^{11}}{1 - \sin (| x | + π / 6)}) d x$ is equal to: [2026]

$8 π$
$2 π$
$6 π$
$4 π$

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(4)

$= 2 π \int_{0}^{π / 6} \frac{1}{1 - \sin (x + \frac{π}{6})} d x let x + \frac{π}{6} = t \Rightarrow d x = d t$

$= 2 π \int_{π / 6}^{π / 3} \frac{d t}{1 - \sin t} = 2 π \int_{π / 6}^{π / 3} \frac{1 + \sin t}{\cos^{2} t} d t$

$= 2 π [\int_{π / 6}^{π / 3} \sec^{2} t d t + \int_{π / 6}^{π / 3} \sec t \tan t d t]$

$= 2 π [{(\tan t)}_{π / 6}^{π / 3} + {(\sec t)}_{π / 6}^{π / 3}]$

$= 2 π [(\sqrt{3} - \frac{1}{\sqrt{3}}) + (2 - \frac{2}{\sqrt{3}})]$

$= 2 π [\sqrt{3} + 2 - \sqrt{3}] = 4 π$

Q 107 :

$6 \int_{0}^{π}$ $| (\sin 3 x + \sin 2 x + \sin x) |$ $d x$ is equal to_____. [2026]

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(17)

$6 \int_{0}^{π}$ $| 2 \sin 2 x \cos x + \sin 2 x |$ $d x$

$= 6 \int_{0}^{π}$ $| 4 \sin x \cos^{2} x + 2 \sin x \cos x |$ $d x$

$I = 12 \int_{0}^{π}$ $| \sin x (2 \cos^{2} x + \cos x) |$ $d x$

Put $\cos x = t, - \sin x d x = d t$

$I = - 12 \int_{1}^{- 1}$ $| 2 t^{2} + t |$ $d t$

$I = 12 (\int_{- 1}^{- \frac{1}{2}} (2 t^{2} + t) d t + \int_{- \frac{1}{2}}^{0} - (2 t^{2} + t) d t + \int_{0}^{1} (2 t^{2} + t) d t)$

$I = 17$

Q 108 :

Let $f : [1, \infty) \to ℝ$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is [2026]

$- 4$
$3$
$- 3$
$4$

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(2)

Q 109 :

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[x] + 4}) d x,$ where $[\cdot]$ denotes the greatest integer function, is [2026]

$\frac{7}{60} (3 π - 1)$
$\frac{1}{60} (21 π - 1)$
$\frac{7}{60} (π - 3)$
$\frac{1}{60} (π - 7)$

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(1)

Q 110 :

Let [.] denote the greatest integer function. Then $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{12 (3 + [x])}{3 + [\sin x] + [\cos x]})$ is equal to: [2026]

$13 π + 1$
$15 π + 4$
$11 π + 2$
$12 π + 5$

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