If a line x = -1 divides the area of region bounded by \(\{(x,y):1+x^2 \leq y \leq 3-x\}\) in the ratio \(\frac{m}{n}\) then (m + n) equal (where HCF of (m,n) = 1) :
Show Hint
To find the area between two curves, first find their points of intersection to determine the limits of integration. Then, integrate the difference between the upper function and the lower function over this interval.
Step 1: Understanding the Question:
We are given a region bounded by a parabola \(y = 1+x^2\) (upward opening) and a line \(y = 3-x\). We need to find the total area of this region and then find the two sub-areas created by the vertical line \(x=-1\). The ratio of these sub-areas will give us m and n.
Step 2: Finding the Bounds of Integration:
First, we find the intersection points of the two curves by setting their y-values equal:
\[ 1+x^2 = 3-x \]
\[ x^2 + x - 2 = 0 \]
Factoring the quadratic equation:
\[ (x+2)(x-1) = 0 \]
The points of intersection are at \(x=-2\) and \(x=1\). In this interval, the line \(3-x\) is above the parabola \(1+x^2\).
Step 3: Calculating the Total Area:
The area A is given by the integral of the upper curve minus the lower curve, from \(x=-2\) to \(x=1\).
\[ A = \int_{-2}^{1} ((3-x) - (1+x^2)) dx = \int_{-2}^{1} (2-x-x^2) dx \]
\[ A = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{1} \]
\[ A = \left(2(1) - \frac{1^2}{2} - \frac{1^3}{3}\right) - \left(2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3}\right) \]
\[ A = \left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right) = \left(\frac{7}{6}\right) - \left(-6 + \frac{8}{3}\right) \]
\[ A = \frac{7}{6} - \left(\frac{-10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2} \]
Step 4: Calculating the Sub-Areas:
The line \(x=-1\) divides the area into two parts, \(A_1\) and \(A_2\).
\(A_1\) is the area from \(x=-2\) to \(x=-1\).
\[ A_1 = \int_{-2}^{-1} (2-x-x^2) dx = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{-1} \]
\[ A_1 = \left(2(-1) - \frac{(-1)^2}{2} - \frac{(-1)^3}{3}\right) - \left(2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3}\right) \]
\[ A_1 = \left(-2 - \frac{1}{2} + \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right) = \left(\frac{-13}{6}\right) - \left(\frac{-10}{3}\right) = \frac{-13+20}{6} = \frac{7}{6} \]
\(A_2\) is the area from \(x=-1\) to \(x=1\). We can find it by subtracting \(A_1\) from the total area.
\[ A_2 = A - A_1 = \frac{9}{2} - \frac{7}{6} = \frac{27}{6} - \frac{7}{6} = \frac{20}{6} = \frac{10}{3} \]
Step 5: Finding m + n:
The ratio of the areas is \(\frac{m}{n} = \frac{A_1}{A_2} = \frac{7/6}{10/3} = \frac{7}{6} \times \frac{3}{10} = \frac{7}{20}\).
So, we have \(m=7\) and \(n=20\). The HCF of (7, 20) is 1.
We need to find the value of \(m+n\).
\[ m+n = 7 + 20 = 27 \]
