Question
Question: Find the area of the region\(\left\\{ \left( x,y \right):{{x}^{2}}\le y\le \left| x \right| \right\\...
Find the area of the region\left\\{ \left( x,y \right):{{x}^{2}}\le y\le \left| x \right| \right\\}.
Solution
Hint: First draw the graph of y=x2. Now, y≥x2 is the region inside the parabola. Then draw the graph of y=∣x∣. Now, y≤∣x∣ is the region outside the graph y=∣x∣. To find the limits of integration, solve the equation: x=∣x∣ and determine the values of ‘x’. The smaller value of x will be the lower limit and greater value of x will be the upper limit. When we will draw the graph of these functions, we will see that it is symmetric about the y-axis. Therefore, there will be two similar regions of area, one in the 1st quadrant and the other in the 2nd quadrant. We have to determine the area of the region in the 1st quadrant and multiply it by 2 to get the total bounded area. To find the area, use the formula: A=a∫b[f(x)−g(x)]dx, where A is the area bounded between the curves, ‘a’ is the lower limit and ‘b’ is the upper limit. In the formula of area, f(x) is the graph of a function having higher value than the function g(x) in the range [a,b].
Complete step-by-step answer:
We have been provided with two inequalities:
y≥x2......................(i)
y≤∣x∣.....................(ii)
Let us solve these two equations by removing the inequality sign to determine the point of intersection.
Removing the inequality sign, we get,
y=x2......................(iii)
y=∣x∣.....................(iv)
Equating the value of ‘y’ from equation (iii) in equation (iv), we get,
∣x∣=x2
On squaring the two sides to remove modulus, we get,
x2=x4⇒x4−x2=0⇒x2(x2−1)=0
Substituting each term equal to 0, we get,
x2=0 and x2−1=0⇒x=0 and x2=1⇒x=0 and x=±1
Hence, the lower limit is (x = -1) and the upper limit is (x = 1).
Let us draw the graph of the two functions.
To draw the graph of y=∣x∣, we have to consider two different situations: