(0)

We have, $f\left(x\right)=\frac{x}{{(1+x^n)}^{1/n}}$

$\Rightarrow f\left(f\right(x\left)\right)=\frac{x}{{(1+2x^n)}^{1/n}}$

     $f\left(f\right(f\left(x\right)\left)\right)=\frac{x}{{(1+3x^n)}^{1/n}}$

$\Rightarrow f^n\left(x\right)=\frac{x}{{(1+n{x}^n)}^{1/n}}$

Now, $\underset{n\to \infty }{\lim }{\int }_0^1{x}^{n-2}·\left(f^n\right(x\left)\right) dx=\underset{n\to \infty }{\lim }{\int }_0^1\frac{x^{n-1}}{{(1+n{x}^n)}^{1/n}}dx$

Put $1+n{x}^n=t\Rightarrow n^2{x}^{n-1} dx=dt$

When $x=0, t=1$ and $x=1, t=1+n$

$\therefore$  $\underset{n\to \infty }{\lim }·\frac{1}{n^2}{\int }_1^{1+n}\frac{dt}{t^{1/n}}=\underset{n\to \infty }{\lim }\frac{1}{n^2}{\left[\frac{t^{1-1/n}}{1-1/n}\right]}_1^{1+n}$

$=\underset{n\to \infty }{\lim }\frac{1}{n(n-1)}[{(1+n)}^{1-1/n}-1]$

$=\underset{h\to 0}{\lim }\frac{h^2}{1-h}[{(1+\frac{{\displaystyle 1}}{{\displaystyle h}})}^{1-h}-1] (\text{Put }n=\frac{1}{h}, n\to \infty , h\to 0)$

$=0$

(14)

$\text{Let }I=\frac{2}{\pi }{\int }_{\pi /6}^{5\pi /6}\left(8\right[\cos ecx]-5[\cot x\left]\right)dx$

$=\frac{2}{\pi }\left[8{\int }_{\pi /6}^{5\pi /6}\right[\cos ecx]dx-5{\int }_{\pi /6}^{5\pi /6}[\cot x\left]dx\right]$

$=\frac{2}{\pi }\left[8\right\{{\int }_{\pi /6}^{5\pi /6}1·dx\}-5\{{\int }_{\pi /6}^{\pi /4}1·dx+{\int }_{\pi /4}^{\pi /2}0·dx+{\int }_{\pi /2}^{3\pi /4}(-1)dx+{\int }_{3\pi /4}^{5\pi /6}(-2)dx\left\}\right]$

$=\frac{2}{\pi }[8\times \frac{{\displaystyle 2\pi }}{{\displaystyle 3}}-5\{\frac{{\displaystyle \pi }}{{\displaystyle 12}}-\frac{{\displaystyle \pi }}{{\displaystyle 4}}-2\times \frac{{\displaystyle \pi }}{{\displaystyle 12}}\left\}\right]=\frac{2}{\pi }[\frac{{\displaystyle 16\pi }}{{\displaystyle 3}}+\frac{{\displaystyle 5\pi }}{{\displaystyle 3}}]=14$

(6)

We have,

${\int }_0^{2.4}\left[x^2\right] dx={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx+{\int }_2^{\sqrt{5}}4dx+{\int }_{\sqrt{5}}^{2.4}5dx$

$=\sqrt{2}-1+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})+4(\sqrt{5}-2)+5(2.4-\sqrt{5})$

$=9-\sqrt{2}-\sqrt{3}-\sqrt{5}$

$\therefore$   $\alpha +\beta +\gamma +\delta =9-1-1-1=6$

(32)

We have, $\alpha (m,n)={\int }_0^2{t}^m{(1+3t)}^{n}dt$

Now, $11\alpha (10,6)+18\alpha (11,5)$

$=11{\int }_0^2{t}^{10}{(1+3t)}^{6}dt+18{\int }_0^2{t}^{11}{(1+3t)}^{5}dt$

$=11{[{(1+3t)}^6·\frac{{\displaystyle t^{11}}}{{\displaystyle 11}}]}_0^2-11{\int }_0^{2}6{(1+3t)}^5·$$3\frac{t^{11}}{11}dt+18{\int }_0^2{t}^{11}{(1+3t)}^{5}dt$

$={\left[t^{11}{(1+3t)}^6\right]}_0^2$$=2^{11}{\left(7\right)}^6=2^5{\left(14\right)}^6$

$=32{\left(14\right)}^6=p{\left(14\right)}^6$                                                    $\text{[∵ Given]}$

$\Rightarrow p=32$

(575)

Let $I={\int }_{-0.15}^{0.15}$$|100x^2-1|$$dx$

$=2{\int }_0^{0.15}$$|100x^2-1|$$dx$

$(\because {\int }_{-a}^{a}f(x) dx=\{\begin{array}{cc}2{\int }_0^{a}f\left(x\right)dx,& \text{if }f(-x)=f\left(x\right)\text{ i.e. }f\left(x\right)\text{ is even}\\ 0,& \text{if }f(-x)=-f\left(x\right)\text{ i.e. }f\left(x\right)\text{ is odd}\end{array})$

$=2\{{\int }_0^{0.1}-(100x^2-1)dx+{\int }_{0.1}^{0.15}(100x^2-1\left)dx\right\}$

$=2\{{[x-\frac{{\displaystyle 100x^3}}{{\displaystyle 3}}]}_0^{0.1}+{[\frac{{\displaystyle 100x^3}}{{\displaystyle 3}}-x]}_{0.1}^{0.15}\}$$=\frac{575}{3000}$

$\therefore$$k=575$

(18)

Given, $S_k\left(x\right)=C_{k}x+k{\int }_0^x{S}_{k-1}\left(t\right) dt \text{...(i)}$

Put $k$ = 2 and $x$ = 3 in (i),

$S_2\left(3\right)=C_2\left(3\right)+2{\int }_0^3{S}_1\left(t\right) dt \text{...(ii)}$

Put $k$ = 1 in (i),

$S_1\left(x\right)=C_{1}x+{\int }_0^x{S}_0\left(t\right)dt=C_{1}x+{\int }_0^{x}t dt [\because S_0(x)=x]$

$=C_{1}x+\frac{x^2}{2}$

Put $S_1\left(t\right)=C_{1}t+\frac{t^2}{2}$ in (ii),

$S_2\left(3\right)=C_2\left(3\right)+2{\int }_0^3(C_{1}t+\frac{{\displaystyle t^2}}{{\displaystyle 2}})dt$

$=3C_2+2{[C_1\frac{{\displaystyle t^2}}{{\displaystyle 2}}+\frac{{\displaystyle t^3}}{{\displaystyle 6}}]}_0^3=3C_2+2\left[C_1\right(\frac{{\displaystyle 9}}{{\displaystyle 2}})+(\frac{{\displaystyle 27}}{{\displaystyle 6}}\left)\right]$

$=3C_2+9C_1+9$
 

Given, $C_k=1-{\int }_0^1{S}_{k-1}\left(x\right) dx$                         $\text{...(iii)}$

Put $k$ = 1 in (iii),

$C_1=1-{\int }_0^1{S}_0\left(x\right)dx=1-{\int }_0^{1}xdx=1-{\left(\frac{{\displaystyle x^2}}{{\displaystyle 2}}\right)}_0^1=\frac{1}{2}$

Put $k$ = 2 in (iii),

$C_2=1-{\int }_0^1{S}_1\left(x\right)dx=1-{\int }_0^1(C_{1}x+\frac{{\displaystyle x^2}}{{\displaystyle 2}})dx$

$=1-{[C_1\frac{{\displaystyle x^2}}{{\displaystyle 2}}+\frac{{\displaystyle x^3}}{{\displaystyle 6}}]}_0^1 =1-[\frac{{\displaystyle 1}}{{\displaystyle 2}}\times \frac{{\displaystyle 1}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}] =1-[\frac{{\displaystyle 1}}{{\displaystyle 4}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}]=1-\frac{5}{12}=\frac{7}{12}$

Put $k$ = 3 in (iii), $C_3=1-{\int }_0^1{S}_2\left(x\right)dx$

Put $k$ = 2 in (i), $S_2\left(x\right)=C_{2}x+2{\int }_0^x{S}_1\left(t\right) dt=C_{2}x+C_1{x}^2+\frac{x^3}{3}$

So, $C_3=1-{\int }_0^1(C_{2}x+C_1{x}^2+\frac{{\displaystyle x^3}}{{\displaystyle 3}})dx$

$=1-{[C_2\frac{{\displaystyle x^2}}{{\displaystyle 2}}+C_1\frac{{\displaystyle x^3}}{{\displaystyle 3}}+\frac{{\displaystyle x^4}}{{\displaystyle 12}}]}_0^1$

$=1-\frac{{\displaystyle 7}}{{\displaystyle 12}}\times \frac{{\displaystyle 1}}{{\displaystyle 2}}-\frac{{\displaystyle 1}}{{\displaystyle 2}}\times \frac{{\displaystyle 1}}{{\displaystyle 3}}-\frac{{\displaystyle 1}}{{\displaystyle 12}}=1-\frac{{\displaystyle 7}}{{\displaystyle 24}}-\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle 1}}{{\displaystyle 12}}$

$=\frac{24-7-4-2}{24}=\frac{11}{24}$

$S_3\left(3\right)+6C_3=3C_2+9C_1+9+6C_3$

$=3\times \frac{7}{12}+9\times \frac{1}{2}+9+6\times \frac{11}{24}=\frac{7}{4}+\frac{9}{2}+9+\frac{11}{4}$

$=\frac{7+18+36+11}{4}=\frac{72}{4}=18$


 

(41)

Let $f_n={\int }_0^{\pi /2}\left(\sum _{k=1}^n{\sin }^{k-1}x\right)\left(\sum _{k=1}^n\right(2k-1\left){\sin }^{k-1}x\right)\cos x dx, x\in \mathrm{\mathbb{N}}$

$f_n\left(x\right)={\int }_0^{\pi /2}(1+\sin x+{\sin }^{2}x+\dots +{\sin }^{n-1}x)(1+3\sin x+5{\sin }^{2}x+\dots +(2n-1\left){\sin }^{n-1}x\right)·\cos x dx$

Multiply and divide by $\sqrt{\sin x}$, we get:

${\int }_0^{\pi /2}[{\left(\sin x\right)}^{1/2}+{\left(\sin x\right)}^{3/2}+{\left(\sin x\right)}^{5/2}+\dots +{\left(\sin x\right)}^{\frac{2n-1}{2}}](1+3\sin x+5{\sin }^{2}x+\dots +(2n-1\left){\sin }^{n-1}x\right)\frac{\cos x}{\sqrt{\sin x}}dx$

Put ${\left(\sin x\right)}^{1/2}+{\left(\sin x\right)}^{3/2}+{\left(\sin x\right)}^{5/2}+\dots +{\left(\sin x\right)}^{n-\frac{{\displaystyle 1}}{{\displaystyle 2}}}=t$

$\Rightarrow \frac{1}{2}\frac{(1+3\sin x+5{\sin }^{2}x+\dots +(2n-1\left){\sin }^{n-1}x\right)}{\sqrt{\sin x}}\cos x dx=dt$

$\Rightarrow f_n=2{\int }_0^{n}t dt\Rightarrow f_n=n^2$

$\therefore$   $f_{21}-f_{20}={\left(21\right)}^2-{\left(20\right)}^2=441-400=41$

(63)

${\int }_0^1(x^{21}+x^{14}+x^7){(2x^{14}+3x^7+6)}^{1/7}dx=\frac{1}{l}{\left(11\right)}^{m/n}, l,m,n\in \mathrm{\mathbb{N}}$

L.H.S.$={\int }_0^{1}x(x^{20}+x^{13}+x^6){(2x^{14}+3x^7+6)}^{1/7}dx$

$={\int }_0^1(x^{20}+x^{13}+x^6){(2x^{21}+3x^{14}+6x^7)}^{1/7}dx$

Let $2x^{21}+3x^{14}+6x^7=t^7\Rightarrow 42(x^{20}+x^{13}+x^6)dx=7t^{6}dt$

$\Rightarrow (x^{20}+x^{13}+x^6)dx=\frac{t^6}{6}dt,$ When $x$ = 0, $t$ = 0 and when $x$ = 1, 

$t={11}^{1/7}$$\Rightarrow \text{L.H.S.}={\int }_0^{{11}^{1/7}}(\frac{{\displaystyle t^6}}{{\displaystyle 6}}· t)dt$

$=\frac{1}{6}{\int }_0^{{11}^{1/7}}{t}^{7}dt$$=\frac{1}{6}{\left[\frac{{\displaystyle t^8}}{{\displaystyle 8}}\right]}_0^{{\left(11\right)}^{1/7}}=\frac{{11}^{8/7}}{48}$

Comparing with R.H.S., we get $l=48, m=8, n=7$

$\therefore$  $l+m+n=48+8+7=63$

(2)

$\text{Use }{\int }_0^{a}f\left(x\right) dx={\int }_0^{a}f(a-x) dx\text{ to solve.}$
 

(22)

$\text{Let }I=12{\int }_0^3$$|x^2-3x+2|$$dx$

$=12{\int }_0^3$$\left|\right(x-2\left)\right(x-1\left)\right|$$dx$

$=12\times \left[{\int }_0^1\right(x^2-3x+2)dx-{\int }_1^2(x^2-3x+2)dx+{\int }_2^3(x^2-3x+2\left)dx\right]$

$=12\times [{(\frac{{\displaystyle x^3}}{{\displaystyle 3}}-\frac{{\displaystyle 3x^2}}{{\displaystyle 2}}+2x)}_0^1+{(\frac{{\displaystyle x^3}}{{\displaystyle 3}}-\frac{{\displaystyle 3x^2}}{{\displaystyle 2}}+2x)}_2^3-{(\frac{{\displaystyle x^3}}{{\displaystyle 3}}-\frac{{\displaystyle 3x^2}}{{\displaystyle 2}}+2x)}_1^2]=22$