u = x; v = e^-2x
u' = 1; \({ v }_{ 1 }=\frac { { e }^{ -2x } }{ -2 } \)
\({ v }_{ 2 }=\frac { { e }^{ -2x } }{ 4 } \)
Bernoulli's formula
\(\int { uv\ dx={ uv }^{ (1) }-{ uv }^{ (2) } } \)
\(\therefore \int _{ 0 }^{ 1 }{ { xe }^{ -2x } } dx={ \left[ x\left( \frac { { e }^{ -2x } }{ -2 } \right) -1\left( \frac { { e }^{ -2x } }{ 4 } \right) \right] }_{ 0 }^{ 1 }\)
= \(-{ \left[ \frac { x }{ 2 } { e }^{ -2x }+\frac { { e }^{ -2x } }{ 4 } \right] }_{ 0 }^{ 1 }\)
= \({ -e }^{ -2x }{ \left[ \frac { x }{ 2 } +\frac { 1 }{ 4 } \right] }_{ 0 }^{ 1 }\)
= \({ -e }^{ -2 }\left( \frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) +{ e }^{ 0 }\left( 0+\frac { 1 }{ 4 } \right) \)
= \(-{ e }^{ -2 }\left( \frac { 3 }{ 4 } \right) +\frac { 1 }{ 4 } =\frac { 1 }{ 4 } (1-{ 3e }^{ -2 })\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sin \ x }{ 9+{ cos }^{ 2 } } dx } \)

Put t = cos x ⇒ - sin x dx = dt ⇒ sin x dx = - dt
\(\therefore I=\int _{ 1 }^{ 0 }{ -\frac { dt }{ { 3 }^{ 2 }+t^{ 2 } } } =\int _{ 0 }^{ 1 }{ \frac { dt }{ { 3 }^{ 2 }+t^{ 2 } } } =\frac { 1 }{ 3 } { \left[ { tan }^{ -1 }\left( \frac { t }{ 3 } \right) \right] }_{ 0 }^{ 1 }\)
= \(\frac { 1 }{ 3 } \left[ { tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) -{ tan }^{ -1 }(0) \right] \)
= \(\frac { 1 }{ 3 } { tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) \) [∵ tan^-1(0) = 0]

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot \ x } }{ \sqrt { cot \ x } +\sqrt { tan \ x } } dx } \) ....(1)
By the property, \(\int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x) } dx\)
∴ I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot(\frac { \pi }{ 2 } -x) } }{ \sqrt { cot(\frac { \pi }{ 2 } -x) } +\sqrt { tan(\frac { \pi }{ 2 } -x) } } dx } \)
= \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { tanx } }{ \sqrt { tanx } +\sqrt { cotx } } dx } \) ...(2)
(1) + (2) ⇒ 2I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot \ x } +\sqrt { tan \ x } }{ \sqrt { cot \ x } +\sqrt { tan \ x } } dx } \)
=\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ dx } ={ \left[ x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 2 } -0=\frac { \pi }{ 2 } \)
∴ I = \(\frac { \pi }{ 4 } \)

Taking u = x² and v = sin nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ 2\pi }{ { x }^{ 2 } } sin\quad nx\quad dx={ \left[ \left( { x }^{ 2 } \right) \left( -\frac { cos\quad nx }{ n } \right) -(2x)\left( -\frac { sin\quad nx }{ { n }^{ 2 } } \right) +(2)\left( \frac { cos\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ 2x }\)
\(=\left[ (4{ \pi }^{ 2 })\left( -\frac { 1 }{ n } \right) -0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] -\left[ 0-0+(2)\left( \frac { 1 }{ { n }^{ 3 } } \right) \right] \) since cos 2n\(\pi\) = 1 and sin 2n \(\pi\) = 0
\(=-\frac { 4{ \pi }^{ 2 } }{ n } +\frac { 2 }{ { n }^{ 3 } } -\frac { 2 }{ { n }^{ 3 } } =\frac { { 4\pi }^{ 2 } }{ n } \)

\(\int _{ 0 }^{ 1 }{ \frac { 2x+7 }{ { 5x }^{ 2 }+9 } dx } =\int _{ 0 }^{ 1 }{ \frac { 2x }{ { 5x }^{ 2 }+9 } } +7\int _{ 0 }^{ 1 }{ \frac { dx }{ (5{ x }^{ 2 })+{ 3 }^{ 2 } } =\frac { 1 }{ 5 } } log{ [{ 5x }^{ 2 }+9] }_{ 0 }^{ 1 }+\frac { 7 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }{ \left( \frac { 3 }{ \sqrt { 5 } } \right) }^{ 2 } } } \)
\(=\frac { 1 }{ 5 } [log14-log9]+\frac { 7 }{ 5 } \times \frac { \sqrt { 5 } }{ 3 } { \left[ { tan }^{ -1 }\frac { x }{ \left[ \frac { 3 }{ \sqrt { 5 } } \right] } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 5 } log\frac { 14 }{ 9 } +\frac { 7 }{ 3\sqrt { 5 } } { tan }^{ -1 }\frac { \sqrt { 5 } }{ 3 } \)

\(\int _{ 0 }^{ 3 }{ (3{ x }^{ 2 }-4x+5) } dx=\int _{ 0 }^{ 3 }{ 3{ x }^{ 2 } } dx-\int _{ 0 }^{ 3 }{ 4x } dx+\int _{ 0 }^{ 3 }{ 5 } dx\)
\(=3\int _{ 0 }^{ 3 }{ { x }^{ 2 } } dx-4\int _{ 0 }^{ 3 }{ c } dx+5\int _{ 0 }^{ 3 }{ dx } \)
\(=3{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }-4{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }+5{ \left[ x \right] }_{ 0 }^{ 3 }\)
= (27 − 0) − 2(9 − 0) + 5(3− 0)
= 27 −18 +15 = 24.

Given sides are y = 2x + 1.....(1)
y = 3x + 1...(2)
x = 4...(3)
Solving (1) & (2), x = 0, y = 1
Solving (2) & (3), x = 4, y = 13
Solving (1) & (3), x = 4, y = 9

∴ Required area \(\int _{ 0 }^{ 4 }{ (3x+1) } dx-\int _{ 0 }^{ 4 }{ (2x+1) } dx\)
\({ =\left( \frac { { 3x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }-{ \left( \frac { { 2x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }\)
\(=\left( \frac { 48 }{ 2 } +4 \right) -(16+4)=28-20\)
Area = 8 sq. units

y = 2at ⇒ \(t=\frac { y }{ 2a } \) ...(1)
x = at² ... (2)
substituting (1) in (2) we get

\(x=a\left( \frac { { y }^{ 2 } }{ 4{ a }^{ 2 } } \right) =\frac { { y }^{ 2 } }{ 4a } \)
y² = 4ax
since y² = 4ax is symmetrical about x - axis,
Required area = \(2\int _{ 0 }^{ 4a }{ ydx } =2\int _{ 0 }^{ 4a }{ \sqrt { 4a \ x \ } } dx\)
\(=4\sqrt { a } \int _{ 0 }^{ 4a }{ { \sqrt { x } dx=4\sqrt { a } \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ a }^{ 4a } } \)
\(=\frac { 8\sqrt { a } }{ 3 } [4a\sqrt { 4a } -a\sqrt { a } ]\)
\(=\frac { 8\sqrt { a } }{ 3 } [8a\sqrt { a } -a\sqrt { a } ]\)
\(=\frac { 8\sqrt { a } }{ 3 } \times 7a\sqrt { a } \)
Area = \(\frac { { 56 }a^{ 2 } }{ 3 } \) sq.units

\(=\int _{ 0 }^{ \frac { 3 }{ 5 } }{ -(5x-3)dx+\int _{ \frac { 3 }{ 5 } }^{ 1 }{ -(5x-3)dx } } \)
\(={ { \left[ 3x-\frac { { 5x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { 3 }{ 5 } }+\left[ \frac { { 5x }^{ 2 } }{ 2 } -3x \right] }_{ 3 }^{ 1 }\)
\(=\frac { 9 }{ 5 } -\frac { 5 }{ 2 } \left( \frac { 9 }{ 25 } \right) -0+\frac { 5 }{ 2 } -3-\frac { 5 }{ 2 } \left( \frac { 9 }{ 25 } \right) +\frac { 9 }{ 5 } \)
\(=\frac { 9 }{ 5 } -\frac { 9 }{ 10 } +\frac { 5 }{ 2 } -3-\frac { 9 }{ 10 } +\frac { 9 }{ 5 } \)
\(=\frac { 18-9+25-30-9+18 }{ 10 } =\frac { 13 }{ 10 } \)

I = \(\int ^\frac{\pi}{4}_{0} \frac{1}{sin x+cos x}\) dx = \(\int ^\frac{\pi}{4}_{0} \frac{1} {\sqrt 2( \frac {1}{\sqrt 2}sin x+ \frac {1}{\sqrt 2}cos x)}\) dx
= \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {( cos \frac {\pi}{4}cos x+ sin \frac {\pi}{4}sin x)}\) dx = \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {cos (\frac{\pi}{4}- x)}\) dx
= \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {cos x}\)dx since\(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\) f (a-x)dx
\(=\frac { 1 }{ \sqrt { 2 } } \int _{ 0 }^{ \frac { \pi }{ 4 } } secxdx=\frac { 1 }{ \sqrt { 2 } } { \left[ log(secx+tanx) \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
=  \(\frac{1}{\sqrt 2} [log ( \sqrt {2} + 1) - log (1+0)]\)
= \(\frac{1}{\sqrt 2} [log ( \sqrt {2} + 1)\).

\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ sin^{ 4 }x+cos^{ 4 }x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (sin^{ 2 }x+cos^{ 2 }x)-2sin^{ 2 }xcos^{ 2 }x } \)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (1-\frac { 1 }{ 2 } (2sinxcosx)^{ 2 } } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 2-sin^{ 2 }2x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 1+cos^{ 2 }2x } \)
Put u = cos 2x, Then, du = −2sin 2x dx
When x = 0 , we have u = cos 0 = 1. When x = \(\frac{\pi}{4}\) we have u = cos\(\frac{\pi}{4}\) = 0
∴ I =  \(\int^{0}_{1} \frac {-du}{1+u^2}\) = \(\int^{0}_{1} \frac {du}{1+u^2}\) = [tan^-1 u\(]^1_0\) = \(\frac{\pi}{4}\)

Limits are from 0 to 4

3x - 4y = 0
4y = 3x
\(\Rightarrow y=\frac { 3 }{ 4 } x\)
∴ Required volume = \(\pi \int _{ 0 }^{ 4 }{ { y }^{ 2 }dx } \)
= \(\pi \int _{ 0 }^{ 4 }{ \frac { 9 }{ 16 } { x }^{ 2 }dx } \)
= \(\frac { 9\pi }{ 16 } { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 4 }=\frac { 9\pi }{ 16 } \times \frac { 64 }{ 3 } \)
∴ V = \(12\pi \) Cubic units

Let I = \(\int _{ 0 }^{ 1 }{ \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } dx } \)

Put e^x = t ⇒ e^x dx = dt
∴ \(\int _{ 1 }^{ e }{ \frac { dt }{ 1+{ t }^{ 2 } } dx } { \left[ { tan }^{ -1 }(t) \right] }_{ 1 }^{ e }\)
= tan^-1(e) - tan^-1(1)
= tan^-1(e) -\(\frac { \pi }{ 4 } \)

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\((-\frac {1}{2}+1)=1\).