Integration And Area Calculation

To determine the value of , we start by evaluating the area of the region given by the set:

.

The area can be evaluated by integrating the top bounding curve within the specified limits of :

First, split the integral at because of the absolute value function:

Simplify the integrals:

1. For , , so:

2. For , , so:

Evaluating the above expression:

Area from to is .

Total Area =

We know that the total area is given as . Therefore,

Solving for , multiply both sides by :

Cancel terms and rearrange:

The value of is 5.

Step 1: Understanding the Question:
We need to find the area of a region in the xy-plane defined by a set of three inequalities:
1.
2.
3.
This involves sketching the region bounded by the curves , , and , and then setting up a definite integral to calculate its area.
Step 2: Key Formula or Approach:
1. First, we identify the boundary curves and find their points of intersection. The curves are , (a parabola), and (a hyperbola). The conditions imply the region is in the first quadrant, so we consider .
2. We can calculate the area by integrating either with respect to or with respect to . Let's analyze both approaches.
- Integrating with respect to : The upper boundary of the region changes at an intersection point, requiring the sum of two separate integrals.
- Integrating with respect to : The left and right boundaries are consistent throughout the region, requiring only a single integral. This approach is simpler. The formula is Area .
Step 3: Detailed Explanation:
Finding Intersection Points:
- Intersection of and : (since ). Point: (1, 1).
- Intersection of and : . Point: (27, 1).
- Intersection of and : . Then . Point: (3, 9).
Setting up the Integral with respect to y:
The region is defined by and . We can rewrite these in terms of :
From , we get (since ). This is the left boundary, .
From , we get . This is the right boundary, .
The condition must hold, so .
The condition gives the lower bound for integration.
Thus, the integration limits for are from 1 to 9.
The area is given by:
Now, we perform the integration:
Evaluate at the upper limit ( ):
Evaluate at the lower limit ( ):
The area is the difference:
Verification (Integrating w.r.t. x):
Area =



.
Both methods yield the same result.
Step 4: Final Answer:
The area of the region is .

Concept: A matrix is singular if its determinant is zero. Also, Thus maxima and minima in a closed interval occur at endpoints or critical points.

Step 1: {Find using determinant condition.} Expanding: Thus

Step 2: {Find .}

Step 3: {Find critical points.} Discriminant: Thus , so is increasing.

Step 4: {Find maximum and minimum.} Minimum at : Maximum at :

Step 5: {Compute required value.} Thus value is .

Step 1: Understanding the Concept:
First find using A.P. properties, find its minimum , then evaluate the definite integral.
: Key Formula or Approach:
Arithmetic Mean: .
AM-GM: for .
Step 2: Detailed Explanation:
.
Limits: and .
Integral .
Let .
When . When .
.
.
Step 3: Final Answer:
The integral value is .

Step 1: Understanding the Concept:
For an ellipse to have its major axis along the -axis, we must have . Since is a strictly decreasing function, implies .
Step 2: Key Formula or Approach:
1. Condition: .
2. Decreasing property: if .
Step 3: Detailed Explanation:
Given : Since is strictly decreasing: Rearranging: Factorizing the quadratic: The solution to this inequality is . This can be written as . Here, and . Calculate :
Step 4: Final Answer:
The value of is 40.