# Find the area between the curves. y = x ^\frac{15}{14} , y = 13x ^\frac{1}{14}

## Question:

Find the area between the curves.

{eq}y = x ^\frac{15}{14} , y = 13x ^\frac{1}{14} {/eq}

## Area between Curves:

The total area of a region bounded between two curves, which are located in such a way that within the points of intersection one curve completely lies above the other, will be difference of the areas of two curves bounded with {eq}x {/eq} axis and common points as extreme points.

## Answer and Explanation:

Solving the given two curves to find points at which they intersect each other:-

{eq}x^\frac{15}{14} = 13x^\frac{1}{14} \Rightarrow x=0,13 {/eq} ,

As the given curves are located in such a way that between the points of intersection one curve completely lies above the other. Hence the area contained between them will be difference of the areas of two curves bounded with {eq}x {/eq} axis and common points as corner points i.e. the area locked between curves is bounded above by {eq}13x^{\frac{1}{14}} {/eq} and below by {eq}x \cdot x^{\frac{1}{14}} {/eq} within lines {eq}x = 0, x = 13,y = 0 {/eq}:-

Area = {eq}\int_{0}^{13}(13x^{\frac{1}{14}}-x^{\frac{15}{14}})dx {/eq} , Applying {eq}\int x^n = \frac{x^{n+1}}{n+1} {/eq} :-

{eq}\Rightarrow \int_{0}^{13}(13x^{\frac{1}{14}}-x^{\frac{15}{14}})dx = \frac{13 \cdot 14}{15}(13^{\frac{15}{14}} - 0) - (\frac{14}{29}13^{\frac{29}{14}}-0) = 169 \cdot 14 \cdot13^{\frac{1}{14}}(\frac{1}{15}-\frac{1}{29}) = \frac{33124}{15 \cdot 29}.13^{\frac{1}{14}} = 13785.009 {/eq} (approx.) square units.