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

Q 21 :
$The value of \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})} is:$ [2024]

$\frac{13 (2 \sqrt{3} - 3) π}{8}$
$\frac{π}{8 (2 \sqrt{3} + 3)}$
$\frac{(2 \sqrt{3} + 3) π}{24}$
$\frac{13 π}{8 (4 \sqrt{3} + 3)}$

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(4)

Let $L = \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})}$

$= \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{n^{4} (1 + \frac{k^{2}}{n^{2}}) (1 + \frac{3 k^{2}}{n^{2}})}$

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

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

$= \frac{- 1}{2} \int_{0}^{1} \frac{1}{x^{2} + 1} d x + \frac{3}{2} \int_{0}^{1} \frac{1}{3 x^{2} + 1} d x$

$= \frac{- 1}{2} \int_{0}^{1} \frac{1}{x^{2} + 1} d x + \frac{1}{2} \int_{0}^{1} \frac{1}{x^{2} + {(\frac{1}{\sqrt{3}})}^{2}} d x$

$= \frac{- 1}{2} {[\tan^{- 1} x]}_{0}^{1} + \frac{\sqrt{3}}{2} {[\tan^{- 1} \sqrt{3} x]}_{0}^{1}$

$= \frac{- 1}{2} [\frac{π}{4} - 0] + \frac{\sqrt{3}}{2} [\frac{π}{3} - 0] = \frac{- π}{8} + \frac{π}{2 \sqrt{3}}$

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

$= \frac{π (4 - \sqrt{3}) (4 + \sqrt{3})}{8 \sqrt{3} (4 + \sqrt{3})} = \frac{13 π}{8 (4 \sqrt{3} + 3)}$

Q 22 :
Let $f : R \to R$ be defined as $f (x) = a e^{2 x} + b e^{x} + c x .$ If $f (0) = - 1, f' (\log_{e} 2) = 21 and \int_{0}^{\log_{e} 4} (f (x) - c x) d x = \frac{39}{2},$ then the value of $| a + b + c |$ equals [2024]

12
16
8
10

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(3)

Given, $f (x) = a e^{2 x} + b e^{x} + c x$

$\Rightarrow f (0) = a (1) + b (1) + c (0)$

$\Rightarrow a + b = - 1$ ...(i)

$Also, f' (x) = 2 a e^{2 x} + b e^{x} + c$

$\Rightarrow f' (\log_{e} 2) = 8 a + 2 b + c = 21$ ...(ii)

$Now, \int_{0}^{\log_{e} 4} (f (x) - c x) d x$

$= \int_{0}^{\log_{e} 4} (a e^{2 x} + b e^{x}) d x = \frac{a}{2} (16 - 1) + b (4 - 1)$

$= \frac{15 a}{2} + 3 b = \frac{39}{2} = \frac{9 a}{2} + 3 (a + b) = \frac{39}{2}$

$\Rightarrow \frac{9 a}{2} = \frac{39}{2} + 3$ $(Since, a + b = 1)$

$\Rightarrow \frac{9 a}{2} = \frac{45}{2} \Rightarrow a = 5$

$From (i) and (ii), we get b = - 6 and c = - 7$

Q 23 :
Let $f : R \to R$ be a function defined by $f (x) = \frac{x}{{(1 + x^{4})}^{1 / 4}},$ and $g (x) = f (f (f (f (x)))) .$ Then $18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x$ is equal to [2024]

36
42
39
33

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(3)

We have, $f (x) = \frac{x}{{(1 + x^{4})}^{1 / 4}} and g (x) = f (f (f (x)))$

$∴ f (f (f (f (x)))) = \frac{x}{{(1 + 4 x^{4})}^{1 / 4}} = g (x)$

$Now, \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x = \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} \cdot \frac{x}{{(1 + 4 x^{4})}^{1 / 4}} d x$

$= \int_{0}^{\sqrt{2 \sqrt{5}}} \frac{x^{3}}{{(1 + 4 x^{4})}^{1 / 4}} d x$

$Put 1 + 4 x^{4} = t^{4} \Rightarrow 16 x^{3} d x = 4 t^{3} d t$

$= \frac{1}{4} \int_{1}^{3} t^{2} d t = \frac{1}{12} (27 - 1) = \frac{26}{12} = \frac{13}{6}$

$∴ 18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x = 18 \times \frac{13}{6} = 39$

Q 24 :
Let $a$ and $b$ be real constants such that the function $f$ defined by $f (x)$ $= {\begin{matrix} x^{2} + 3 x + a, & x \leq 1 \\ b x + 2, & x > 1 \end{matrix}$ be differentiable on $R .$ Then, the value of $\int_{- 2}^{2} f (x) d x$ equals [2024]

$\frac{19}{6}$
$21$
$17$
$\frac{15}{6}$

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(3)

We have, $f (x)$ $= {\begin{matrix} x^{2} + 3 x + a, & x \leq 1 \\ b x + 2, & x > 1 \end{matrix}$

Since, $f (x)$ be differentiable on R.

So, $f (x)$ be continuous at $x = 1 .$

$∴ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1)$

$\Rightarrow \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} f (1 + h) = f (1)$

$\Rightarrow a + 4 = b + 2 \Rightarrow b = a + 2$ ...(i)

Also, $L f' (1) = R f' (1)$

$\Rightarrow 5 = b$

$∴ a = 3 and b = 5 (Using (i))$

Now, $\int_{- 2}^{2} f (x) d x = \int_{- 2}^{1} (x^{2} + 3 x + 3) d x + \int_{1}^{2} (5 x + 2) d x$

$= {(\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 3 x)}_{- 2}^{1} + {(\frac{5 x^{2}}{2} + 2 x)}_{1}^{2}$

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

$= 3 - \frac{9}{2} + 9 + \frac{15}{2} + 2 = 17$

Q 25 :
Let $y = f (x)$ be a thrice differentiable function in $(- 5, 5) .$ Let the tangents to the curve $y = f (x)$ at $(1, f (1))$ and $(3, f (3))$ make angles $π / 6$ and $π / 4,$ respectively with the positive $x$ -axis. If $27 \int_{1}^{3} ((f' {(t))}^{2} + 1) f'' (t) d t = α + β \sqrt{3}$ where $α, β$ are integers, then the value of $α + β$ equals [2024]

- 14
36
- 16
26

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(4)

We have, $f' (1) = \tan \frac{π}{6} = \frac{1}{\sqrt{3}}$

$f' (3) = \tan \frac{π}{4} = 1$

Now, let $I = \int_{1}^{3} ((f' {(t))}^{2} + 1) f'' (t) d t$

where $f' (t) = u$

$\Rightarrow f'' (t) d t = d u$

$∴ I = \int_{1 / \sqrt{3}}^{1} (u^{2} + 1) d u$

$= {(\frac{u^{3}}{3} + u)}_{1 / \sqrt{3}}^{1} = \frac{4}{3} - \frac{10}{9 \sqrt{3}} = \frac{4}{3} - \frac{10 \sqrt{3}}{27}$

$\Rightarrow 27 I = 36 - 10 \sqrt{3}$

$∴ α = 36 and β = - 10$

$∴ α + β = 36 - 10 = 26$

Q 26 :
$Let a be the sum of all coefficients in the expansion of {(1 - 2 x + 2 x^{2})}^{2023} {(3 - 4 x^{2} + 2 x^{3})}^{2024} and$ $b = \lim_{x \to 0} (\frac{\int_{0}^{x} \frac{\log (1 + t)}{t^{2024} + 1} d t}{x^{2}}) .$ $If the equations c x^{2} + d x + e = 0 and 2 b x^{2} + a x + 4 = 0 have a common root, where c, d, e \in R, then d : c : e equals$ [2024]

2 : 1 : 4
4 : 1 : 4
1 : 1 : 4
1 : 2 : 4

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(3)

$a = 1,$ $b = \lim_{x \to 0} (\frac{\int_{0}^{x} \frac{\log (1 + t)}{t^{2024} + 1}}{x^{2}})$

Using L-Hopital's Rule

$b = \lim_{x \to 0} \frac{\log (1 + x)}{2 x (x^{2024} + 1)} = \frac{1}{2} .$

Then, $x^{2} + x + 4 = 0$ (non-real) ...(i)

and $c x^{2} + d x + e = 0$ ...(ii)

Both equation (i) and (ii) have a common root

So, $\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$

So, $d : c : e$ is equal to $1 : 1 : 4$

Q 27 :
Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t and g (x) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t .$ Then, the value of $9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}))$ is equal to [2024]

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(1)

We have, $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t$

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

$= 2 [\int_{0}^{x} t e^{- t^{2}} d t - \int_{0}^{x} t^{2} e^{- t^{2}} d t] = [1 - e^{- x^{2}} - 2 \int_{0}^{x} t^{2} e^{- t^{2}} d t]$

Let $t^{2} = p \Rightarrow 2 t \cdot d t = d p$

$\Rightarrow d t = \frac{d p}{2 \sqrt{p}}$

$∴ f (x) = 1 - e^{- x^{2}} - 2 \int_{0}^{x^{2}} \frac{p \cdot e^{- p}}{2 \sqrt{p}} d p$

$= 1 - e^{- x^{2}} - \int_{0}^{x^{2}} \sqrt{p} e^{- p} d p = 1 - e^{- x^{2}} - g (x)$

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

Now, $f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}) = 1 - e^{- \log_{e} 9} = 1 - \frac{1}{9} = \frac{8}{9}$

$∴ 9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9})) = 8$

Q 28 :
If $\int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{1 + \sin x \cos x} d x = \frac{1}{a} \log_{e} (\frac{a}{3}) + \frac{π}{b \sqrt{3}}, where a, b \in N,$ then $a + b$ is equal to ______ . [2024]

0%

(8)

Let $I = \int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{1 + \sin x \cos x} d x$

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

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

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

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

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

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

$= \frac{1}{2} \ln 2 - \frac{1}{2} \ln 3 + \frac{1}{2} \int_{0}^{1} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$

$= \frac{1}{2} \ln \frac{2}{3} + \frac{1}{2} \cdot \frac{2}{\sqrt{3}} {[\tan^{- 1} \frac{2 x + 1}{\sqrt{3}}]}_{0}^{1} = \frac{1}{2} \ln \frac{2}{3} + \frac{1}{\sqrt{3}} (\frac{π}{3} - \frac{π}{6})$

$= \frac{1}{2} \ln \frac{2}{3} + \frac{1}{\sqrt{3}} \cdot \frac{π}{6}$ $∴ a = 2, b = 6$

$∴ a + b = 8$

Q 29 :
If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - \cos^{2} t \sin^{2} x}, 0 < t < π,$ then the value of $\int_{0}^{π / 2} \frac{π^{2} d t}{f (t)}$ equals ______ . [2024]

0%

(1)

$f (t) = \int_{0}^{π} \frac{2 x}{1 - \cos^{2} t \sin^{2} x} d x$ ...(i)

$\Rightarrow f (t) = 2 \int_{0}^{π} \frac{(π - x)}{1 - \cos^{2} t \sin^{2} (π - x)} d x$

$\Rightarrow f (t) = 2 \int_{0}^{π} \frac{(π - x)}{1 - \cos^{2} t \sin^{2} x} d x$ ...(ii)

Adding (i) and (ii), we get

$2 f (t) = 2 \int_{0}^{π} \frac{π}{1 - \cos^{2} t \sin^{2} x} d x$

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

On dividing the numerator and denominator by $\cos^{2} x,$ we get

$f (t) = 2 π \int_{0}^{π / 2} \frac{\sec^{2} x d x}{\sec^{2} x - \cos^{2} t \tan^{2} x}$

Put $\tan x = v \Rightarrow \sec^{2} x d x = d v$

$When x = 0, t = 0$

$When x = \frac{π}{2}, t = \infty$

$∴ f (t) = 2 π \int_{0}^{\infty} \frac{d v}{v^{2} + 1 - \cos^{2} t v^{2}}$

$= 2 π \int_{0}^{\infty} \frac{d v}{1 + {(v \sin t)}^{2}} = 2 π {[\frac{\tan^{- 1} (v \sin t)}{\sin t}]}_{0}^{\infty}$

$= 2 π \frac{π}{2 \sin t} = \frac{π^{2}}{\sin t}$

$∴ \int_{0}^{π / 2} \frac{π^{2} d t}{f (t)} = \int_{0}^{π / 2} \sin t d t = {[- \cos t]}_{0}^{π / 2} = 1$

Q 30 :
Let $r_{k} = \frac{\int_{0}^{1} {(1 - x^{7})}^{k} d x}{\int_{0}^{1} {(1 - x^{7})}^{k + 1} d x}, k \in N .$ Then the value of $\sum_{k = 1}^{10} \frac{1}{7 (r_{k} - 1)}$ is equal to _______ . [2024]

0%

(65)

Let $I_{k} = \int_{0}^{1} 1 {(1 - x^{7})}^{k} d x$

$= {[{(1 - x^{7})}^{k} x]}_{0}^{1} + 7 k \int_{0}^{1} {(1 - x^{7})}^{k - 1} x^{7} d x$

$\Rightarrow - 7 k \int_{0}^{1} {(1 - x^{7})}^{k - 1} ((1 - x^{7}) - 1) d x = - 7 k I_{k} + 7 k I_{k - 1}$

$\Rightarrow I_{k} (7 k + 1) = 7 k I_{k - 1} \Rightarrow I_{k + 1} (7 k + 8) = 7 (k + 1) I_{k}$

$\Rightarrow \frac{I_{k}}{I_{k + 1}} = \frac{7 k + 8}{7 k + 7} \Rightarrow r_{k} = \frac{7 k + 8}{7 k + 7}$

$\Rightarrow r_{k} - 1 = \frac{1}{7 (k + 1)} \Rightarrow 7 (r_{k} - 1) = \frac{1}{k + 1}$

$∴ \sum_{k = 1}^{10} \frac{1}{7 (r_{k} - 1)} = \sum_{k = 1}^{10} (k + 1) = \frac{11 \times 12}{2} - 1 = 65$

Topic Question Set

Q 21 :
$The value of \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})} is:$ [2024]

Q 22 :
Let $f : R \to R$ be defined as $f (x) = a e^{2 x} + b e^{x} + c x .$ If $f (0) = - 1, f' (\log_{e} 2) = 21 and \int_{0}^{\log_{e} 4} (f (x) - c x) d x = \frac{39}{2},$ then the value of $| a + b + c |$ equals [2024]

Q 23 :
Let $f : R \to R$ be a function defined by $f (x) = \frac{x}{{(1 + x^{4})}^{1 / 4}},$ and $g (x) = f (f (f (f (x)))) .$ Then $18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x$ is equal to [2024]

Q 24 :
Let $a$ and $b$ be real constants such that the function $f$ defined by $f (x)$ $= {\begin{matrix} x^{2} + 3 x + a, & x \leq 1 \\ b x + 2, & x > 1 \end{matrix}$ be differentiable on $R .$ Then, the value of $\int_{- 2}^{2} f (x) d x$ equals [2024]

Q 27 :
Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t and g (x) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t .$ Then, the value of $9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}))$ is equal to [2024]

Q 28 :
If $\int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{1 + \sin x \cos x} d x = \frac{1}{a} \log_{e} (\frac{a}{3}) + \frac{π}{b \sqrt{3}}, where a, b \in N,$ then $a + b$ is equal to ______ . [2024]

0%

Q 29 :
If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - \cos^{2} t \sin^{2} x}, 0 < t < π,$ then the value of $\int_{0}^{π / 2} \frac{π^{2} d t}{f (t)}$ equals ______ . [2024]

0%

Q 30 :
Let $r_{k} = \frac{\int_{0}^{1} {(1 - x^{7})}^{k} d x}{\int_{0}^{1} {(1 - x^{7})}^{k + 1} d x}, k \in N .$ Then the value of $\sum_{k = 1}^{10} \frac{1}{7 (r_{k} - 1)}$ is equal to _______ . [2024]

0%

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Topic Question Set

Q 21 : The value of limn→∞∑k=1nn3(n2+k2)(n2+3k2) is: [2024]

Q 22 : Let f:R→R be defined as f(x)=ae2x+bex+cx. If f(0)=-1, f'(loge2)=21 and∫0loge4(f(x)-cx) dx=392, then the value of |a+b+c| equals [2024]

Q 23 : Let f:R→R be a function defined by f(x)=x(1+x4)1/4, and g(x)=f(f(f(f(x)))). Then 18∫025x2g(x) dx is equal to [2024]

Q 24 : Let a and b be real constants such that the function f defined by f(x)={x2+3x+a,x≤1bx+2,x>1 be differentiable on R. Then, the value of ∫-22f(x) dx equals [2024]

Q 25 : Let y=f(x) be a thrice differentiable function in (-5,5). Let the tangents to the curve y=f(x) at (1,f(1)) and (3,f(3)) make angles π/6 and π/4, respectively with the positive x-axis. If 27∫13((f'(t))2+1)f''(t) dt=α+β3 where α,β are integers, then the value of α+β equals [2024]

Q 26 : Let a be the sum of all coefficients in the expansion of (1-2x+2x2)2023(3-4x2+2x3)2024 and b=limx→0(∫0xlog(1+t)t2024+1 dtx2). If the equations cx2+dx+e=0 and 2bx2+ax+4=0 have a common root, where c,d,e∈R, then d:c:e equals [2024]

Q 27 : Let f,g:(0,∞)→R be two functions defined by f(x)=∫-xx(|t|-t2)e-t2 dt and g(x)=∫0x2t1/2e-t dt. Then, the value of 9(f(loge9)+g(loge9)) is equal to [2024]

Q 28 : If ∫0π4sin2x1+sinxcosx dx=1aloge(a3)+πb3, where a,b∈N, then a+b is equal to ______ . [2024]

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Q 29 : If f(t)=∫0π2x dx1-cos2tsin2x,0<t<π, then the value of ∫0π/2π2 dtf(t) equals ______ . [2024]

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Q 30 : Let rk=∫01(1-x7)k dx∫01(1-x7)k+1 dx, k∈N. Then the value of ∑k=11017(rk-1) is equal to _______ . [2024]

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Q 21 :
$The value of \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})} is:$ [2024]

Q 22 :
Let $f : R \to R$ be defined as $f (x) = a e^{2 x} + b e^{x} + c x .$ If $f (0) = - 1, f' (\log_{e} 2) = 21 and \int_{0}^{\log_{e} 4} (f (x) - c x) d x = \frac{39}{2},$ then the value of $| a + b + c |$ equals [2024]

Q 23 :
Let $f : R \to R$ be a function defined by $f (x) = \frac{x}{{(1 + x^{4})}^{1 / 4}},$ and $g (x) = f (f (f (f (x)))) .$ Then $18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g (x) d x$ is equal to [2024]

Q 24 :
Let $a$ and $b$ be real constants such that the function $f$ defined by $f (x)$ $= {\begin{matrix} x^{2} + 3 x + a, & x \leq 1 \\ b x + 2, & x > 1 \end{matrix}$ be differentiable on $R .$ Then, the value of $\int_{- 2}^{2} f (x) d x$ equals [2024]

Q 27 :
Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t and g (x) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t .$ Then, the value of $9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}))$ is equal to [2024]

Q 28 :
If $\int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{1 + \sin x \cos x} d x = \frac{1}{a} \log_{e} (\frac{a}{3}) + \frac{π}{b \sqrt{3}}, where a, b \in N,$ then $a + b$ is equal to ______ . [2024]

Q 29 :
If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - \cos^{2} t \sin^{2} x}, 0 < t < π,$ then the value of $\int_{0}^{π / 2} \frac{π^{2} d t}{f (t)}$ equals ______ . [2024]

Q 30 :
Let $r_{k} = \frac{\int_{0}^{1} {(1 - x^{7})}^{k} d x}{\int_{0}^{1} {(1 - x^{7})}^{k + 1} d x}, k \in N .$ Then the value of $\sum_{k = 1}^{10} \frac{1}{7 (r_{k} - 1)}$ is equal to _______ . [2024]