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

We use the formula
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { b-a }{ n } \sum _{ r=1 }^{ n }{ f } \left( a+(b-a)\frac { r }{ n } \right) \)
Here f(x) = 2x²+3, a = 1 and b = 4
So, we get
\(f\left( a+(b-a)\frac { r }{ n } \right) =f\left( 1+(4-1)\frac { r }{ n } \right) =f\left( 1+\frac { 3r }{ n } \right) =2{ \left( 1+\frac { 3r }{ n } \right) }^{ 2 }+3=5+\frac { { 18r }^{ 2 } }{ { n }^{ 2 } } +\frac { 12r }{ n } \)Hence, we get
\(\int _{ 1 }^{ 4 }{ ({ 2x }^{ 2 }+3) } dx=\underset { n\rightarrow \infty }{ lim } \frac { 3 }{ n } \sum _{ r=1 }^{ n }{ \left( 5+\frac { { 18r }^{ 2 } }{ { n }^{ 2 } } +\frac { 12r }{ n } \right) } =\underset { n\rightarrow \infty }{ lim } \left[ \frac { 15 }{ n } \sum _{ r=1 }^{ n }{ 1+\frac { 54 }{ { n }^{ 3 } } \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\frac { 36 }{ { n }^{ 2 } } \sum _{ r=1 }^{ n }{ r } } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ \frac { 15 }{ n } n+\frac { 54 }{ { n }^{ 3 } } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 })+\frac { 36 }{ { n }^{ 2 } } (1+2+..+n) \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 15+\frac { 54 }{ { n }^{ 3 } } \frac { n(n+1)(2n+1) }{ 6 } +\frac { 36 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } \right] \)
\(=\underset { n\rightarrow \infty }{ lim } \left[ 15+9\left( 1+\frac { 1 }{ n } \right) \left( 2+\frac { 1 }{ n } \right) +18\left( 1+\frac { 1 }{ n } \right) \right] \)
= 15+9(1+ 0)(2 + 0) +18(1+ 0) = 51.

I =\(\int ^{1}_{0} (tan ^{-1} x + tan ^{-1}(1-x))\) dx
=\(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-x)) dx\)
= \(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-(1-x)) dx\),  Since\(\int ^{a}_{0} f (x) dx = \int ^{a}_{0} f(a-x) dx \)
=\(\int ^{1}_{0} tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1} x dx\)
= 2\(\int ^{1}_{0} tan^{-1} x dx\)
= \([2 \int udv]^1_0\) ,  Where u = tan^-1 x and dν = dx
= 2\([uν - \int udv]^1_0,\) applying integration by parts
= 2\((x tan^{-1} x - \int x \frac{dx}{1+x^2})^{1}_{0}\)
= 2\((x tan^{-1} x - \frac {1}{2} log (1+x^2))^{1}_{0}\)
= \(\frac{\pi}{2} \)- log 2

The region is sketched. The upper boundary of the region is y = sin x for \(\\ \\ \frac { \pi }{ 4 } \le x\le \frac { 5\pi }{ 4 } \) and the lower boundary of the region is y = cos x for \(\frac { \pi }{ 4 } \le x\le \frac { 5\pi }{ 4 } \). So the required area A is given by

\(A=\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ ({ y }_{ U }-{ y }_{ L })dx= } \int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ (sinx-cosx)dx={ [-cosx=sinx] }_{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } } } \)
\(=\left( -sin\frac { 5\pi }{ 4 } -cos\frac { 5\pi }{ 4 } \right) -\left( -sin\frac { \pi }{ 4 } -cos\frac { \pi }{ 4 } \right) \)
\(=\left( -\left( -\frac { 1 }{ \sqrt { 2 } } \right) -\left( -\frac { 1 }{ \sqrt { 2 } } \right) -\left( -\left( \frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } \right) \right) \right) \)
\(=\frac { 2 }{ \sqrt { 2 } } +\frac { 2 }{ \sqrt { 2 } } =2\sqrt { 2 } \)

Given equation of the parabola is (y-1) = (x-2)²
\(\Rightarrow\) y = (x - 2)² + 1
It vertex is (2, 1) which is the minimum point P. Let Q(x, y) be a point on the parabola given slope of PQ = 2
\(\Rightarrow \frac { y-1 }{ x-2 } =2\ \left[ \because slope=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\(\Rightarrow\) y-1 = 2(x-2) \(\Rightarrow\) y-1 = 2x-4 \(\Rightarrow\) y = 2x-4+1
\(\Rightarrow\) y = 2x + 3
From (1) and (2), (x-2)²+1 = 2x-3
\(\Rightarrow\) x²-4x + 4 + 1 = 2x - 3 \(\Rightarrow\) x²- 6x + 8 = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4

\(\therefore\) Required area \(=\int _{ 2 }^{ 4 }{ ({ y }_{ 1 }-{ y }_{ 2 }) } dx\)
\(=\int _{ 2 }^{ 4 }{ (2x-3)-{ (x-2) }^{ 2 }-1dx } \)
\(=\int _{ 2 }^{ 4 }{ (2x-3-{ x }^{ 2 }+4x-4-1)dx } \)
\(=\int _{ 2 }^{ 4 }{ (-{ x }^{ 2 }+6x-8)dx } \)
\({ \left[ \frac { -{ x }^{ 3 } }{ 3 } +3{ x }^{ 2 }-8x \right] }_{ 2 }^{ 4 }=\left( \frac { -64 }{ 3 } +48-32 \right) -\left( \frac { -8 }{ 3 } +12-16 \right) \)
\(=\left( \frac { -64 }{ 3 } +16 \right) -\left( -\frac { 8 }{ 3 } -4 \right) \)
\(=\left( \frac { -64+48 }{ 3 } \right) -\left( \frac { -8-12 }{ 3 } \right) \)
\(=\frac { 16 }{ 3 } \) sq.units