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]

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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,

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

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

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

$∴ p = 0 + 4 = 4$

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

$∴ p^{2} + 2 q = 16 + 11 = 27$

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]

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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 $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

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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 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]