The two region under whose we have to find area is
f(x)=x
[tex]g(x)=x^{\frac{1}{n}}\\\\x \geq 0\\\\n\geq 2[/tex]
The Point of Intersection of two curves is always , x=0 and x=1.
Area of the Region
=Area under the line - Area Under the curve g(x), when n take different value, that is ≥2.
[tex]\rightarrow[- \int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{2}} \, dx]+[ -\int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{3}} \, dx]+[ -\int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{4}} \, dx]+[ -\int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{5}} \, dx]+......[/tex]
[tex]=\int\limits^1_0({x^{\frac{1}{2}}+x^{\frac{1}{3}}+x^{\frac{1}{4}}+x^{\frac{1}{5}}+.......+x^{\frac{1}{200}}}) \, dx -\int\limits^1_0 ({x}+{x}+{x}+{x}...........+200\text{times}) \, dx[/tex]
When, n=200, the first quadrant is completely occupied by the curve
[tex]g(x)=x^{\frac{1}{n}},x\geq 0\\\\2\leq n \leq 200[/tex]
[tex]=\int\limits^1_0{x^{\frac{1}{n}} \, dx=\frac{n\times x^{(\frac{1}{n}+1)}}{n+1}\left \{ {{x=1} \atop {x=0}} \right.}\\\\= \frac{n}{n+1}\\\\\int\limits^1_0{x^{n} \, dx=\frac{x^{n+1}}{n+1}\left \{ {{x=1} \atop {x=0}} \right.}\\\\=\frac{1}{n+1}[/tex]
[tex]=\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...........+\frac{200}{201}-199 \times \frac{1}{2}\\\\=1-\frac{1}{3}+1-\frac{1}{4}+1-\frac{1}{5}+...........+1-\frac{1}{201}-99.5\\\\=199-99.5+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...........+\frac{1}{201}\\\\=99.5+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...........+\frac{1}{201}-1-\frac{1}{2}\\\\=98+\frac{1}{d} \times\ln(\frac{2a+(2n-1)d}{2a-d})\\\\=98+\ln(\frac{2 \times 1+(2\times 201-1)\times 1}{2\times 1-1})\\\\=98+\ln 403\\\\=98+6({\text{approx}})\\\\=104 \text{square units}[/tex]
Sum of n terms of Harmonic Progression is
[tex]=\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+\frac{1}{a+3d}+\frac{1}{a+4d}.....+\frac{1}{a+(n-1)d}\\\\=\frac{1}{d} \times \ln(\frac{2a+(2n-1)d}{2a-d})[/tex]
