Equation of AB is \(\frac { y-1 }{ 2-1 } =\frac { x+1 }{ 3+1 } or\quad y=\frac { 1 }{ 4 } (x+5)\)
Equation of BC is \(\frac { y-5 }{ 2-5 } =\frac { x-0 }{ 3-0 } or\quad y=-x+5\)
Equation of AC is \(\frac { y-1 }{ 5-1 } =\frac { x+1 }{ 0+1 } or\quad y=4x+5\)
\(\therefore\) Area of \(\Delta\)ABC = Area DACO+ Area of OCBE − Area of DABE
\(=\int _{ -1 }^{ 0 }{ (4x+5)dx+\int _{ 0 }^{ 3 }{ (-x+5)dx-\frac { 1 }{ 4 } \int _{ -1 }^{ 3 }{ (x+5)dx } } } \)
\(\\ \\ \\ ={ \left[ \frac { { 4x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 0 }+{ \left[ -\frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ 0 }^{ 3 }-\frac { 1 }{ 4 } { \left[ \frac { { x }^{ 2 } }{ 2 } +5x \right] }_{ -1 }^{ 3 }\)
\(=0-(+2-5)+\left( -\frac { 9 }{ 2 } +15 \right) -0-\frac { 1 }{ 4 } \left[ \frac { 9 }{ 2 } +15 \right] +\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2 } -5 \right] =\frac { 15 }{ 2 } \)

First, we get the points of intersection of the parabolas. For this, we solve y² x = 4 and x² y = 4 simultaneously Eliminating y between them, we get x⁴ = 64x and so x = 0 and x = 4. Then the points of intersection are (0, 0) and (4, 4). The required region is sketched.

Viewing in the direction of y -axis, the equation of the upper boundary is y = 2\(\sqrt x\) for 0\(\le x \le\) 4 and the equation of the lower boundary is \(y =\frac {x^2}{4}\)for \(0 \leq x \leq 4\). So, the required area \(\Delta\) is
\(A=\int_{0}^{4}\left(y_{U}-Y_{L}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x=\left[2\left(\frac{2 x^{3 / 2}}{3}\right)-\frac{x^{3}}{12}\right]_{0}^{4}=\left[2\left(\frac{2 \times 8}{3}\right)-\frac{64}{12}\right]-0=\frac{16}{3}\)

The equation of the latus-rectum is x = a. It intersects the parabola at the points L(a, 2a) and L₁ (a, −2). The required area is sketched. By symmetry, the required area A is twice the area bounded by the portion of the parabola

y = 2\(\sqrt a \sqrt x\), x -axis, x = 0 and x = a.
Hence, by taking vertical strips, we get
\(A=2\int _{ 0 }^{ a }{ ydx=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx=4 } \sqrt { a } } { \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }\)
\(=4\sqrt { a } \times \frac { 2 }{ 3 } { a }^{ \frac { 3 }{ 2 } }=\frac { 8{ a }^{ 2 } }{ 3 } \)

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.

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

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

Put x = sin \(\theta\). Then, dx = cos \(\theta\) d \(\theta\)
when x = 0, sin \(\theta\) = 0 and so \(\theta\) = 0. When x = 1, sin \(\theta\) = 1 and so \(\theta =\frac{\pi}{2}\)
Hence, we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { (1-{ sin }^{ 2 }\theta ) }^{ 5 }cos\theta d\theta } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { cos }^{ 11 }\theta d\theta =\frac { 10 }{ 16 } \times \frac { 8 }{ 14 } \times \frac { 6 }{ 12 } \times \frac { 4 }{ 10 } \times \frac { 2 }{ 8 } \times \frac { 1 }{ 6 } =\frac { 1 }{ 336 } } \)
By applying the reduction formula III iteratively, we get the following results (stated without proof):
\(\int _{ 0 }^{ 1 }{ { x }^{ m }{ (1-x) }^{ n }dx } =\frac { m!\times n! }{ (m+n+1)! } \), where m and n are positive integers

Taking u = 1−x² and v = e^{−\(\lambda\)x}, and applying the Bernoulli’s formula, we get
\(I=\int _{ -1 }^{ 1 }{ { e }^{ -\lambda x } } (1-{ x }^{ 2 })dx={ \left[ (1-{ x }^{ 2 })\left( \frac { { e }^{ -\lambda x } }{ -\lambda } \right) -(-2x)\left( \frac { { e }^{ -\lambda x } }{ { \lambda }^{ 2 } } \right) +(-2)\left( \frac { { e }^{ -\lambda x } }{ -{ \lambda }^{ 3 } } \right) \right] }_{ -1 }^{ 1 }\)
\(=2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 2 } } \right) +2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 3 } } \right) +2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 2 } } \right) -2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 3 } } \right) \)
\(=\frac { 2 }{ { \lambda }^{ 2 } } ({ e }^{ \lambda }+{ e }^{ -\lambda })-\frac { 2 }{ { \lambda }^{ 3 } } ({ e }^{ \lambda }-{ e }^{ -\lambda })\)

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

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\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)