A farmer $F_{1}$ has a land in the shape of a triangle with vertices at P(0, 0), Q(1, 1) and R(2, 0). From this land, a neighbouring farmer $F_{2}$ takes away the region which lies between the side PQ and a curve of the form $y = x^{n} (n > 1).$ If the area of the region taken away by the farmer $F_{2}$ is exactly 30% of the area of ΔPQR, then the value of n is ______.

A farmer F₁ has a land in the shape of a triangle with vertices at P(0, 0), Q(1, 1) and R(2, 0). From this land, a neighbouring farmer F₂ takes away the region which lies between the side PQ and a curve of the form y = xⁿ (n > 1). If the area of the region taken away by the farmer F₂ is exactly 30% of the area of Δ PQR, then the value of n is ________. $P(0,0),\ Q(1,1),\ R(2,0)$

$\text{region between side } PQ \text{ and } y=x^{n},\ n>1$

To find: n such that taken area = 30% of area(triangle PQR)

$PQ:\ y=x \quad (\text{through } (0,0)\text{ and }(1,1))$

Area(triangle PQR)=1/2× 2× 1=1

$0\le x\le 1,\ n>1 \;\Rightarrow\; x^{n}\le x$

$\text{Plan: } \text{Area}(\triangle)=1;\ \int_0^1(x-x^{n})dx=30\% \Rightarrow n$

$\text{limits: } 0 \text{ to } 1 \ (\text{they meet at } x=0,1)$

$A_{\text{taken}}=\int_0^1(x-x^{n})\,dx$

$=\left[\dfrac{x^{2}}{2}-\dfrac{x^{n+1}}{n+1}\right]_0^1=\dfrac{1}{2}-\dfrac{1}{n+1}$

$\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{3}{10}$

$\dfrac{1}{n+1}=\dfrac{1}{2}-\dfrac{3}{10}=\dfrac{1}{5} \;\Rightarrow\; n+1=5 \;\Rightarrow\; n=4$

$n=4:\ \dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}=30\%\ \checkmark$

$n=4$