Discussion Forum

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 sides 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 Δ PQR , then the value of n is........

Answer:

Explanation:

We have

                    y= xⁿ  n>1

  $\because$  P(0,0), Q(1,1) and R(2,0) are vertices of  Δ PQR.

$\because$   Area of shaded region= 30% of area of   Δ PQR

   $\Rightarrow$    $\int_{0}^{1} (x-x^{n})dx=\frac{30}{100}\times\frac{1}{2}\times2\times1$

$\Rightarrow$        $[\frac{x^{2}}{2}-\frac{x^{n+1}}{n+1}]_0^1=\frac{3}{10}$

$\Rightarrow$       $[\frac{1^{}}{2}-\frac{1^{}}{n+1}]=\frac{3}{10}$

$\Rightarrow$     $\frac{1^{}}{n+1}=\frac{1}{2}-\frac{3}{10}=\frac{2}{10}=\frac{1}{5}$

$\Rightarrow$     n+1=5 $\Rightarrow$  n=4