The region to be revolved is sketched.
Since revolution is made about the y-axis, the volume of the solid generated is given by

\(V=\pi \int _{ 0 }^{ 2 }{ { x }^{ 2 }dy=\pi \int _{ 0 }^{ 2 }{ { e }^{ y }dy } } \)
\(=\pi { \left[ { e }^{ y } \right] }_{ 0 }^{ 2 }=\pi ({ e }^{ 2 }-1)\)

Using the substitution X = \(\sqrt u\), we get dx = \(\frac { 1 }{ 2\sqrt { u } } du\)
When x = 0, we get u = 0
When x = \(\infty\), we get u = \(\infty\)
\(\therefore \int _{ 0 }^{ \infty }{ { e }^{ -{ x }^{ 2 } }{ x }^{ 2n-1 }dx } =2\int _{ 0 }^{ \infty }{ { e }^{ -u }{ \left( \sqrt { u } \right) }^{ 2n-1 } } \frac { 1 }{ 2\sqrt { u } } du=\int _{ 0 }^{ \infty }{ { e }^{ -u }{ u }^{ n-1 }du } \) = Γ(n)

Let I = \(\int _{ 1 }^{ 2 }{ \frac { x }{ (x+1)(x+2) } dx } \)
\(I=\int _{ 1 }^{ 2 }{ \left[ \frac { -1 }{ (x+1) } +\frac { 2 }{ x+2 } \right] } dx\) (Using partial fractions)
\(={ [-log(x+1)+2log(x+2)] }_{ 1 }^{ 2 }\)
\(=log{ \left[ \frac { { (x+2) }^{ 2 } }{ x+1 } \right] }_{ 1 }^{ 2 }\)
\(=log\frac { 16 }{ 3 } -log\frac { 9 }{ 2 } \)
\(=log\frac { 32 }{ 27 } \)

Here f (x) = x³, a = 0 and b = 1. Hence, we get
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f } \left( \frac { r }{ n } \right) \Rightarrow \int _{ 0 }^{ 1 }{ x^3dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { r^3 }{ n^3 } } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 4 } } [{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }]=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 4 } } \frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 4 } { \left( 1+\frac { 1 }{ n } \right) }^{ 2 }=\frac { 1 }{ 4 } \)

Here f (x) = x, a = 0 and b = 1. Hence, we get
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f } \left( \frac { r }{ n } \right) \Rightarrow \int _{ 0 }^{ 1 }{ xdx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { r }{ n } } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 2 } } [1+2+...+n]\)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 2 } \left( 1+\frac { 1 }{ n } \right) =\frac { 1 }{ 2 } \)

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ n }x } =\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2\)
\(Let\quad { I }_{ 10 }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 10 }xdx=\frac { 9 }{ 10 } { I }_{ 8 } } \)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times { I }_{ 6 }=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times { I }_{ 4 }\)
\(\\ =\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } { I }_{ 2 }\)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } \)
\(=\frac { 315 }{ 1280 } \times \frac { \pi }{ 2 } =\frac { 63\pi }{ 256(2) } =\frac { 63\pi }{ 512 } \)

Given that I =\(\int ^\frac {\pi}{2}_{0} \)( sin²x + cos⁴x)dx =\(\int ^\frac {\pi}{2}_{0} \) sin²x dx+\(\int ^\frac {\pi}{2}_{0} \)cos⁴x dx\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } \)
\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } =\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| +c\right] \)
\(=\frac { 1 }{ 4 } \left[ log\left( \frac { 4-2 }{ 4+2 } \right) -log\left( \frac { 3-2 }{ 3+2 } \right) \right] \)
\(=\frac { 1 }{ 4 } log\left[ \left( \frac { 2 }{ 6 } \right) - log \ \frac { 1 }{ 5 } \right] \\ =\frac { 1 }{ 4 } log\left( \frac { 1 }{ 3 } \times 5 \right) \)
\(=\frac { 1 }{ 4 } log\left( \frac { 5 }{ 3 } \right) \)

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

\(\int _{ 0 }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ [2x] } dx+\int _{ \frac { 1 }{ 2 } }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 0dx+ } \int _{ \frac { 1 }{ 2 } }^{ 1 }{ 1 dx} = 0+[x]^1_{\frac{1}{2}} = 1 -\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

Making the substitution t = ax, we get dt = adx and x = 0 \(\Rightarrow\) t = 0 and x = \(\infty\) \(\Rightarrow\) t = \(\infty\)
Hence, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -ax }{ x }^{ n }dx } =\int _{ 0 }^{ \infty }{ { e }^{ -t } } { \left( \frac { t }{ a } \right) }^{ n }\frac{dt}{a}= \int^\infty_0 e^{-t }t^n dt\)
\(=\frac { 1 }{ { a }^{ n+1 } } \int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx } =\frac { n! }{ { a }^{ n+1 } } \)
Thus
\(\int _{ 0 }^{ \infty }{ { e }^{ -ax }{ x }^{ n }dx } =\frac { n! }{ { a }^{ n+1 } } \)

\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 3 }\theta { cos }^{ 5 }\theta d\theta } \)
\( \mathrm{m}=3, \mathrm{n}=5 \)
\( \int_{0}^{\pi / 2} (cos ^5 \theta - cos^7 \theta) sin \theta d\theta\)
 \(t = cos \theta\)
\(dt = -sin \theta d\theta\)
\(= \int ^0_1(t^2-t^7)(-dt)\\ = \int ^0_1(t^5-t^7)(dt)= [\frac{t^6}{6}-\frac{t^8}{8}]\)
\(\frac{1}{6}-\frac{1}{8}= \frac{8-6}{48}\)
\( =\frac{1}{24} \)
Aliter method:
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 3 }\theta { cos }^{ 5 }\theta d\theta } \)
Here m = 3, which is odd and n = 5, which is odd
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ m }\theta { cos }^{n } }x dx \)
\( \frac{n-1}{m+n} \cdot \frac{n-3}{m+n-2} \cdot \frac{n-5}{m+n-4} \cdots \frac{2}{m+3} \cdot \frac{1}{m+1} \\ \)
\( \int_{0}^{\pi / 2} \sin ^{3} \theta \cos ^{5} \theta d \theta=\frac{A}{8} \times \frac{2}{6} \times \frac{1}{A} \\ =\frac{1}{24} \)

\(Let\ I=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 2 }x{ cos }^{ 4 }xdx } \)
\({ I }_{ m,n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ m } } x{ cos }^{ n }xdx=\frac { n-1 }{ m+n } { I }_{ m,m-2 }n\ge 2\)
\(=\left( \frac { m-1 }{ n+m } \right) \left( \frac { m-3 }{ n+m-2 } \right) \left( \frac { m-5 }{ m+m-4 } \right) ...\frac { 2 }{ n+3 } .\frac { 1 }{ n+1 } \)
Here m = 2, n = 4
\(\therefore I=\frac { 3 }{ 6 } \times \frac { 1 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 32 } \)

\(\int _{ 0 }^{ 1 }{ { x }^{ m } } { (1-x) }^{ n }dx=\frac { m!\times n! }{ (m+n+1)! } \)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ (1-x) }^{ 4 }dx } =\frac { 3!\times 4! }{ (3+4+1)! } =\frac { 3!\times 4! }{ 8! } =\frac { 3\times 2\times 1\times 4\times 3\times 2\times 1 }{ 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 } =\frac { 1 }{ 280 } \)

\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }x{ cos }^{ 4 }xdx= } \frac { (3) }{ (9) } \frac { (1) }{ (7) } \frac { (4) }{ (5) } \frac { (2) }{ (3) } =\frac { (4) }{ (9) } \frac { (2) }{ (7) } \frac { (1) }{ (5) } =\frac { 8 }{ 315 } \)
Also, \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }x{ cos }^{ 4 }xdx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 5 }xdx } =\frac { (4) }{ (9) } \frac { (2) }{ (7) } \frac { (1) }{ (5) } =\frac { 8 }{ 315 } \)

\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx=\frac { (6-1) }{ (6+4) } .\frac { (6-3) }{ (6+4-2) } .\frac { (6-5) }{ (6+4-4) } .\frac { (4-1) }{ (4) } .\frac { (4-3) }{ (4-2) } .\frac { \pi }{ 2 } } \)
\(=\frac { (5) }{ (10) } \frac { (3) }{ (8) } \frac { (1) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)
Also, \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 6 }x{ cos }^{ 4 }xdx } =\frac { (3) }{ (10) } \frac { (1) }{ (8) } \frac { (5) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)

I =  \(\int ^\frac{\pi}{2} _0\)(3cos ⁴x-7sin⁵x)dx = 3\(\int ^\frac{\pi}{2} _0\)cos⁴ x dx-7\(\int ^\frac{\pi}{2} _0\)sin⁵ x dx
= 3 × \(\frac {3}{4}\) ×  \(\frac {1}{2}\) ×  \(\frac {\pi}{2}\) - 7 ×  \(\frac {4}{3}\) ×  \(\frac {2}{3}\)= \(\frac {9\pi}{16}\) - \(\frac {56}{15}\).
By applying the reduction formula III iteratively, we get the following results (stated without proof):
(i) If n is even and m is even,
\(\int ^\frac{\pi}{2}_{0}\)  sin ^m x cos ⁿ x dx = \(\frac {(n-1)}{m+n}\)  \(\frac {(n-3)}{m+n-2}\) \(\frac {(n-5)}{m+n-4}\) ......\(\frac {1}{m+2}\) \(\frac {m-1}{m}\) \(\frac {m-3}{m-2}\)\(\frac {m-5}{m-4}\) ... \(\frac {1}{2}\)\(\frac {\pi}{2}\)
(ii) If n is odd and m is any positive integer (even or odd), then
\(\int ^\frac{\pi}{2}_{0}\)  sin ^m x cosⁿx dx = \(\frac {(n-1)}{m+n}\) \(\frac {(n-3)}{m+n-2}\) \(\frac {(n-5)}{m+n-4}\) ... \(\frac {2}{m+3}\)\(\frac {1}{m+1}\)

\(\int _{ b }^{ \infty }{ \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } dx } ={ \left[ \frac { 1 }{ a } { tan }^{ -1 }\frac { x }{ a } \right] }_{ b }^{ \infty }=\frac { 1 }{ a } { tan }^{ -1 }\infty -\frac { 1 }{ a } { tan }^{ -1 }\frac { b }{ a } =\frac { 1 }{ a } \left[ \frac { \pi }{ 2 } -{ tan }^{ -1 }\frac { b }{ a } \right] \)

\(\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } } \)
\(=\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+1+4 } } =\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| \right] \)
\(={ \left[ \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { x+1 }{ 2 } \right) \right] }_{ -1 }^{ 1 }\)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }(1)-{ tan }^{ -1 }(0) \right] \)
\(\\ =\frac { 1 }{ 2 } \left[ \frac { \pi }{ 4 } \right] =\frac { \pi }{ 8 } \)

Let \(\sqrt{x}\) = u
Then x = u², and so dx = 2u du
When x = 0, u = 0 
When x = 9, u = 3
\(\therefore \int _{ 0 }^{ 9 }{ \frac { 1 }{ x+\sqrt { x } } dx=\int _{ 0 }^{ 3 }{ \frac { 1 }{ { u }^{ 2 }+u } (2u)du=2\int _{ 0 }^{ 3 }{ \frac { 1 }{ 1+u } du=2{ \left[ log|1+u \right] }_{ 0 }^{ 3 }=2[log4-0]=log16 } } } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \frac { sec\ x\ tan\ x }{ 1+{ sec }^{ 2 }x } dx } \)
Put sec x = u. Then, sec x tan x dx = du.
When x = 0, u = sec0 = 1. When x = \(\frac{\pi}{3}, u=sec\frac{\pi}{2}=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ 1+{ u }^{ 2 } } ={ [{ tan }^{ -1 }u] }_{ 1 }^{ 2 }={ tan }^{ -1 }1={ tan }^{ -1 } } (2)-\frac { \pi }{ 4 } \)