SATHEE: Area Question 42

Area Question 42

Question 42

In what ratio, does the $X$ -axis divide the area of the region bounded by the parabolas $y = 4 x - x^{2}$ and $y = x^{2} - x ?$

$(1994, 5$ M)

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Solution:

Given parabolas are $y = 4 x - x^{2}$

and

$y = - (x - 2)^{2} + 4$

$(x - 2)^{2} = - (y - 4)$

Therefore, it is a vertically downward parabola with vertex at $(2, 4)$ and its axis is $x = 2$

and

$y = x^{2} - x \Rightarrow y = x - \frac{1}{2} - \frac{1}{4}$

$\Rightarrow x - {\frac{1}{2}}^{2} = y + \frac{1}{4}$

This is a parabola having its vertex at $\frac{1}{2}, - \frac{1}{4}$ .

Its axis is at $x = \frac{1}{2}$ and opening upwards.

The points of intersection of given curves are

$\begin{aligned} 4 x - x^{2} = x^{2} - x \Rightarrow 2 x^{2} = 5 x & \Rightarrow x (2 - 5 x) = 0 \Rightarrow x = 0, \frac{5}{2} \end{aligned}$

Also, $y = x^{2} - x$ meets $X$ -axis at $(0, 0)$ and $(1, 0)$ .

$∴$ Area, $A_{1} = \int_{0}^{5 / 2} [(4 x - x^{2}) - (x^{2} - x)] d x$

$\begin{aligned} = \int_{0}^{5 / 2} (5 x - 2 x^{2}) d x & = \frac{5}{2} x^{2} - \frac{2}{3} x^{3}_{0}^{5 / 2} = \frac{5}{2} {\frac{5}{2}}^{2} - \frac{2}{3} \cdot {\frac{5}{2}}^{3} & = \frac{5}{2} \cdot \frac{25}{4} - \frac{2}{3} \cdot \frac{125}{8} & = \frac{125}{8} 1 - \frac{2}{3} = \frac{125}{24} sq units \end{aligned}$

This area is considering above and below $X$ -axis both. Now, for area below $X$ -axis separately, we consider

$A_{2} = - \int_{0}^{1} (x^{2} - x) d x = \frac{x^{2}}{2} - \frac{x^{3}}{3}_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} sq units$

Therefore, net area above the $X$ -axis is

$A_{1} - A_{2} = \frac{125 - 4}{24} = \frac{121}{24} sq units$

Hence, ratio of area above the $X$ -axis and area below $X$ -axis

$= \frac{121}{24} : \frac{1}{6} = 121 : 4$

Area Question 42

Question 42

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Area Question 42

Question 42

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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