\(12\sqrt { 3 } \)

2 sq.units

(i) not exist sq.units.
(ii) \(\frac { 32 }{ 15 } \)

Given x = t², y = t - \(\frac { { t }^{ 3 } }{ 3 } \) 
Point of intersection of the curve with x-axis is obtained by putting y = 0.
∴ y = 0
⇒ \(t-\frac { { t }^{ 3 } }{ 3 } =0\)
⇒ t = 0 or ± \(\sqrt3\)
∴ The limit is from t = 0 to t = \(\sqrt { 3 } \)
∴ Volume = \(\pi \int _{ 0 }^{ \sqrt { 3 } }{ { y }^{ 2 }dx } =\pi \int _{ 0 }^{ \sqrt { 3 } }{ { \left( t-\frac { { t }^{ 3 } }{ 3 } \right) }^{ 2 }}(2t \ dt)\)
[∵ x = t² ⇒ dx = 2t dt]
\(=2\pi \int _{ 0 }^{ \sqrt { 3 } }{ { \left( \frac { 3t-{ t }^{ 3 } }{ 3 } \right) }^{ 2 }t \ dt } \)
\(=\frac { 2\pi }{ 9 } \int _{ 0 }^{ \sqrt { 3 } }{ (9{ t }^{ 2 }-6{ t }^{ 4 }+{ t }^{ 6 })t \ dt } \)
\(=\frac { 2\pi }{ 9 } \int _{ 0 }^{ \sqrt { 3 } }{ (9{ t }^{ 3 }-6{ t }^{ 5 }+{ t }^{ 7 })dt } \)
\(=\frac { 2\pi }{ 9 } { \left[ \frac { { 9 }t^{ 4 } }{ 4 } -\frac { 6{ t }^{ 6 } }{ 6 } +\frac { { t }^{ 8 } }{ 8 } \right] }_{ 0 }^{ \sqrt { 3 } }\)
\(=\frac { 2\pi }{ 9 } \left[ \frac { 81 }{ 4 } -27+\frac { 81 }{ 8 } -0 \right] \)
Volume \(=\frac { 2\pi }{ 9 } \left( \frac { 27 }{ 8 } \right) =\frac { 3\pi }{ 4 } \) cubic units

\(Let\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 4x{ sin }^{ 2 }x+5{ cos }^{ 2 }x } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 4{ tan }^{ 2 }x+5 } } dx\)
(Dividing both numerator and denominator by cos² x).
Let u = tan x 
Then du = sec² x dx
When x = 0, u = tan 0 = 0
When x = \(\frac{\pi}{2}\), u = tan\(\frac{\pi}{2}\) = \(\infty\)
\(\therefore I=\int _{ 0 }^{ \infty }{ \frac { du }{ 4{ u }^{ 2 }+5 } } \) (This is an improper integral)
\(\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ \frac { du }{ \left[ { u }^{ 2 }+\left( \frac { \sqrt { 5 } }{ 2 } \right) \right] } =\frac { 1 }{ 4 } \times \frac { 2 }{ \sqrt { 5 } } { \left[ { tan }^{ -1 }\left( \frac { u }{ \frac { \sqrt { 5 } }{ 2 } } \right) \right] }_{ 0 }^{ \infty } } =\frac { 1 }{ 2\sqrt { 5 } } \left( { tan }^{ -1 }\infty -{ tan }^{ -1 }0 \right) =\frac { 1 }{ 2\sqrt { 5 } } \left( \frac { \pi }{ 2 } \right) =\frac { \pi }{ 4\sqrt { 5 } } \)

Let I =\(\int ^{\pi}_{0} \frac{x}{1+sin x}\)
= \(\int ^{\pi}_{0} x \frac{x}{1+sin x}\)dx
Let f (x) = \(\frac{1}{1+ sin x}\) 
Then f(π-x) = \(\frac{1}{1+ sin (\pi - x)}\) = \(\frac{1}{1+ sin x}\) = f(x)
∴  \(\int ^{\pi}_{0} \frac{x}{1+sin x}\) dx = \(\frac{\pi}{2}\)  \(\int ^{\pi}_{0} \frac{1}{1+sin x}\) dx,      \((\because \int _{ 0 }^{ a }{ xf(x)dx=\frac { a }{ 2 } \int _{ 0 }^{ a }{ f(x)dx } if } f(a-x)=f(x))\)
= \(\pi\) \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+sin x}\) dx,       (∴  \(\int ^{\pi}_0\) g(sin x)dx = 2  \(\int ^{ \frac{\pi}{2}}_0\) g(sin x)dx)
= \(\pi\)  \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+sin (\frac{\pi}{2} -x)}\) dx  (∴ \(\int ^{a}_{0} \) f(x) dx =  \(\int ^{a}_{0} \) f(a-x)dx)
= \(\pi\)  \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+cos x}\) dx = \(\pi\) \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{2cos^2 \frac{x}{2}}\) dx = \(\frac{\pi}{2}\) \(\int ^{ \frac{\pi}{2}}_0 sec ^2 \frac{x}{2}\) dx
= \(\pi\)  \([tan \frac{x}{2}]^\frac{\pi}{2}_0\) =  \(\pi\) \([tan \frac{\pi}{4} - tan 0]\) = \(\pi\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos\theta }{ (1+sin\theta )(2+sin\theta ) } } d\theta \)
Put u = 1 + sin\(\theta\)
Then, du = cos\(\theta\) d\(\theta\)
When \(\theta\) = 0, u = 1 
When \(\theta =\frac{\pi}{2}, u=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ u(1+u) } } =\int _{ 1 }^{ 2 }{ \frac { (1+u)-u }{ u(1+u) } du } =\int _{ 1 }^{ 2 }{ \left( \frac { 1 }{ u } -\frac { 1 }{ 1+u } \right) du=[logu-log(1+u)]_{ 1 }^{ 2 } } \)
\(=(log2-log3)-(log1-log2)=2log2-log3=log\frac { 4 }{ 3 } .\)

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

To find \(\int _{ 1 }^{ 1.5 }{ xdx } \) in  {1.1, 1.2, 1.3, 1.4, 1.5}.
Here a = 1,
b = 1.5, n = 5, f(x) = x
\(\therefore h=\Delta x=\frac { b-a }{ n } =\frac { 1.5-1 }{ 5 } =\frac { 0.5 }{ 5 } =0.1\)
The partition of the interval is given by 
x₀ = 1
x₁ = x₀ + h = 1 + 0.1 = 1.1
x₂ = x₁ + h = 1.1 + 0.1 = 1.2
x₃ = x₂ + h = 1.2 + 0.1 = 1.3
x₄ = x₃ + h = 1.3 + 0.1 = 1.4
x₅ = x₄ + h = 1.4 + 0.1 = 1.5
The left end rule for Riemann sum with equal width \(\Delta\)x is
\(\int _{ a }^{ b }{ xdx } =[f({ x }_{ 0 })+f({ x }_{ 1 })+...+f({ x }_{ (n-1 })]\Delta x\)
S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)](0.1)
= (1 + 1.1 + 1.2 + 1.3 + 1.4) (0.1)
= 6 \(\times\) 0.1
S = 0.6

Here a = 0, b = 0.5, n = 5, f(x) = x²
So, the width of each subinterval is \(h=\Delta x=\frac { b-a }{ n } =\frac { 0.5-0 }{ 5 } =0.1\)
The partition of the interval is given by the points
x₀ = 0,
x₁ = x₀ + h = 0 + 0.1 = 0.1
x₂ = x₁ + h = 0.1+ 0.1 = 0.2
x₃ = x₂ + h = 0.2 + 0.1 = 0.3
x₄ = x₃ + h = 0.3+ 0.1 = 0.4
x₅ = x₄ + h = 0.4 + 0.1 = 0.5
(i) The left-end rule for Riemann sum with equal width \(\Delta\)x is
S = [f(x₀) + f(x₁) +....+f(x_n-1)]\(\Delta x\)
\(\therefore\) S = [f(0) + f(0.1) + f(0.2) + f(0.3) + f(0.4)](0.1)
= [0.00 + 0.01+ 0.04 + 0.09 + 0.16](0.1) = 0.03
\(\therefore \int _{ 0 }^{ 0.5 }{ x^{ 2 } }\) dx is approximately 0.03.
(ii) The right-end rule for Riemann sum with equal width \(\Delta\)x is
S = [f(x₁)+f(x₂)+...+f(x_n)]\(\Delta\)x
\(\therefore\) S = [f(0.1) + f(0.2) + f(0.3) + f(0.4) + f(0.5)](0.1)
= [0.01 + 0.04 + 0.09 + 0.16 + 0.25](0.1) = 0.055
\(\therefore \int _{ 0 }^{ 0.5 }{ x^{ 2 } }\) dx is approximately 0.055.
(iii) The mid-point rule for Riemann sum with equal width \(\Delta\)x is
\(S=\left[ f\left( \frac { { x }_{ 0 }+{ x }_{ 1 } }{ 2 } \right) +f\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \right) +..+f\left( \frac { { x }_{ n-1 }+{ x }_{ n } }{ 2 } \right) \right] \Delta x\)
\(\therefore\) S = [f[f(0.05) + f(0.15) + f(0.25) + f(0.35) + f(0.45)](0.1)
= [0.0025 + 0.0225 + 0.0625 + 0.1225 + 0.2025](0.1)
= 0.04125
\(\therefore \int _{ 0 }^{ 0.5 }{ { x }^{ 2 } } \) dx is approximately 0.04125

\(Let\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+5{ cos }^{ 2 }x } } \)
Put U = tan x \(\Rightarrow\) du = sec 2x dx Dividing the numerator and denominator by cos²x we get,
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \frac { 1 }{ { cos }^{ 2 }x } }{ \frac { 1 }{ { cos }^{ 2 }x } +5 } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ { sec }^{ 2 }x+5 } dx } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 1+{ tan }^{ 2 }+x+5 } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x\quad dx }{ { tan }^{ 2 }x+6 } } \)

\(=\int _{ 0 }^{ \infty }{ \frac { du }{ { u }^{ 2 }+6 } } \)
\(=\int _{ 0 }^{ \infty }{ \frac { du }{ { u }^{ 2 }+{ (\sqrt { 6 } ) }^{ 2 } } } \)
\(I=\frac { \pi }{ 2\sqrt { 6 } } \left[ \because \int _{ 0 }^{ \infty }{ \frac { dx }{ { a }^{ 2 }+{ x }^{ 2 } } =\frac { \pi }{ 2a } ,Here\quad a=\sqrt { 6 } } \right] \)

Taking u = 1 + x − 2x³ and v = e^-2x, and applying the Bernoulli’s formula, we get
I = \(\\ \int _{ 0 }^{ 1 }{ { e }^{ -2x }(1+x-{ 2x }^{ 3 })dx } \)
\(={ \left[ (1+x-{ 2x }^{ 3 })\left( \frac { { e }^{ -2x } }{ -2 } \right) -(1-6{ x }^{ 2 })\left( \frac { { e }^{ -2x } }{ -4 } \right) +(-12x)\left( \frac { { e }^{ -2x } }{ -8 } \right) -(-12)\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 }\)
\(={ \left[ \frac { { e }^{ -2x } }{ 16 } (16{ x }^{ 3 }+24{ x }^{ 2 }+16x) \right] }_{ 0 }^{ 1 }\)
\(\\ =\frac { 7 }{ 2{ e }^{ 2 } } \)

Let \(I=\int _{ 0 }^{ \pi }{ xlog\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) \left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx } \quad ...(1)\)
\(I=\int _{ 0 }^{ 2\pi }{ (2\pi -x)log\left( \frac { 3+cos(2\pi -x) }{ 3-cos(2\pi -x) } \right) } dx\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ (a-x)dx } } \right] \)
\(=\int _{ 0 }^{ 2\pi }{ 2\pi log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \)
\(-\int _{ 0 }^{ 2\pi }{ xlog\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \)
\(\left[ \because cos(2\pi -x)=cos\quad x \right] \)
\(I=2\pi \int _{ 0 }^{ 2\pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx-I } [using(1)]\)
\(\\ 2I=2\pi \int _{ 0 }^{ 2\pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \quad \quad ...(2)\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=2\int _{ 0 }^{ \frac { a }{ 2 } }{ f(x)dx\quad if(a-x)=f(x) } } \right] \)
I = 2 pi integral limit 0 to p_i
\(\left[ log3+\frac { x }{ 3-x } \right] dx\)
\(\Rightarrow I=2\pi \int _{ 0 }^{ \pi }{ log\left( \frac { 3+cos(\pi -x) }{ 3-cos(\pi -x) } \right) dx } \)
\(\Rightarrow I=2\pi \int _{ 0 }^{ \pi }{ log\left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx...(3) } \)
\([\because cos(\pi -x)=cosx]\)
Adding (2) and (3) we get,
\(2I=\int _{ 0 }^{ \pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) \left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx } \)

\(\Rightarrow 2\pi \int _{ 0 }^{ 2\pi }{ 0dx=0 } \)
\(\Rightarrow 2I=0\Rightarrow I=0\)
\(\therefore \int _{ 0 }^{ 2\pi }{ xlog\left( \frac { 3+cos\ x }{ 3-cos\ x } \right) dx=0 } \)

Let \(f(x)={ sin }^{ 2 }x\)
\(f(-x)={ (sin(-x)) }^{ 2 }={ sin }^{ 2 }x=f(x)\)
\(\therefore \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } =2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } \)
\(cos\ 2x=1-2{ sin }^{ 2 }x\)
\(2{ sin }^{ 2 }x=1-cos2\)
\({ sin }^{ 2 }x=\frac { 1-cos\quad 2x }{ 2 } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\(=\frac { 2 }{ 2 } { \left[ x-\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { \pi }{ 4 } -\frac { sin\frac { \pi }{ 4 } }{ 2 } -0+\frac { sin0 }{ 2 } \)
\(\\ =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } =\frac { \pi -2 }{ 4 } \)

Let \(f(x)=xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(f(-x)=-x\quad cos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(=-x\quad cos\left( \frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } \right) \)
\(=-x\quad cos\left( \frac { 1-{ e }^{ x } }{ 1+{ e }^{ x } } \right) \)
\(=-x\quad cos\left( -\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \right) \)
\(=-xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\([\because cos(-\theta )=cos\theta ]\)
= -f(x)
\(\therefore\) f(x) is an odd function
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

We write \(\int _{ 0 }^{ 2a }{ f(x)dx\int _{ 0 }^{ a }{ f(x)dx } } +\int _{ 0 }^{ 2a }{ f(x)dx } \) ....(1)
Consider \(\int _{ 0 }^{ 2a }{ f(x)dx } \)
Substituting x = a + u, we have dx = du ; when x = a, u = 0 and when x = 2a,u = a.
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a+u)du } =\int _{ 0 }^{ a }{ d(u)du } } \), since f(x) = f(a+x)
\(\\ \\ \\ =\int _{ 0 }^{ a }{ f(x)dx } \) ..(2)
Substituting (2) in (1), we get
\(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)

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

\(\frac { 2 }{ a } \)

\(\frac { 1 }{ 2 } \)

\(\frac { 28 }{ 3 } \)

2

\(\frac { 1 }{ 2 } \)

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) -\(\frac { \pi }{ 4 } \)

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

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