Area Under Simple Curves

The given problem requires finding the area of a region defined by the set of inequalities:

Step 1: Understand the inequalities

1. The inequality implies two conditions:
   • represents the upper half of a circle centered at the origin with a radius of .

Step 2: Intersecting the inequalities

1. The region of integration is the intersection of these two conditions, hence lies within the quarter circle and between the lines and .

Step 3: Calculate the area

The area bounded by these inequalities is calculated by integrating over the quarter-circle. This involves converting the Cartesian coordinates to polar coordinates for simplification.

Step 4: Use symmetry and geometry

1. The area enclosed by the arc of the circle from to is a sector of or radians.
2. Calculate the area of the sector:

1. The area below the lines is given by the triangular area under the line over the range :

Step 5: Combine areas

The area of the region is:

Conclusion

Thus the area of the given region is .

Let's solve the differential equation given by:

We are provided that the solution curve passes through the points and , where .

Step 1: Solve the Differential Equation

We can use the variable separable method to solve this differential equation.

Rearrange the given equation:

Separating variables, we have:

Integrating both sides:

Step 2: Put the initial point

The curve passes through the point . Substitute and into the equation to find the constant :

Solving gives:

Hence, .

Step 3: Check the condition at

Now, the solution curve passes through , so substitute and :

After simplifying both sides, you get the condition:

This matches the correct option given in the question.

Conclusion: The correct answer is:

The correct answer is 1 , not there in the options

∴ Coordinates of centre of circle is

r = 1

BC = 2
Apply Pythagoras theorem in ΔABC, we get
AC² + AB² = BC²
⇒ AC² = 4-2 = 2

Area of ΔABC = × AB × AC
sq. unit

Computing the area:

The correct answer is : 42

A = required area

Step 1: Analyzing the regions.
We are given two areas and . The region is bounded by the curve , which represents a semicircle with center at and radius 2. The region is bounded by the line and the same semicircle.
The area corresponds to the area of the semicircle, and the area consists of the area of a sector of the circle minus the area of the triangle formed by the points , , and the origin .
Step 2: Area of A.
The area of the semicircle is: Thus, the area of is , as the area of the triangular portion is subtracted, where represents the area of the right-angled triangle. Step 3: Area of B. The area of consists of the area of the sector of the circle minus the area of the triangle: The area of the triangle is: The area of the sector is: Thus, the area of is: Step 4: Ratio of the areas. The ratio of the area of to the area of is: Thus, the correct answer is option (3).

Step 1: The intersection point of cos x - sin x = sin x is given by:

tan x = 1/2

So, let ψ = tan^-1(1/2)

Hence, sin ψ = 1/√5 and cos ψ = 2/√5

Step 2: The graph of the equation is shown below:

Step 3: Now, the area between the curves can be calculated using the integral:

Area = ∫_ψ^π/2 (sin x - |cos x - sin x|) dx

Step 4: Break the integral into two parts:

= ∫_ψ^π/4 (sin x - (cos x - sin x)) dx + ∫_π/4^π/2 (sin x - (sin x - cos x)) dx

Step 5: Simplify the integrals:

= ∫_ψ^π/4 (2 sin x - cos x) dx + ∫_π/4^π/2 cos x dx

Step 6: Now, evaluate the integrals:

= [-2 cos x - sin x]_ψ^π/4 + [sin x]_π/4^π/2

Step 7: Substituting the limits:

= -√2 ¹/_√2 + 2 cos ψ + sin ψ + (1 - 1/√2)

Step 8: Final evaluation:

= -√2 ¹/_√2 + 2 (2/√5) + (1/√5) + (1 - 1/√2)

Conclusion: Therefore, the area is:

√5 - 2√2 + 1

We are asked to find the area of the region .
The given region involves two curves: and .
1. Step 1: Understanding the curves:
The curve is a standard parabola opening upwards.
The curve represents two different cases based on the value of :
When , .
When , .
2. Step 2: Setting up the area integral:
The total area can be calculated by integrating the difference between the two curves.
The required area is: 3. Step 3: Calculating the integrals:
First integral: Second integral:

To find the area , we need to evaluate the integral over the region bounded by and from to .

Step 1. Set up the area as the sum of two integrals based on the intersection point of and , which is :**

Step 2. Evaluate each integral:
- For :

- For :


Step 3. Combine the results:


Step 4. Calculate :
12A = = 164

The Correct Answer is:

Simplify step-by-step:

Combine and simplify further:

Final answer:

The correct answer is : 3.

Solution for the Area Between the Circle and Parabola

We are given the equation of a circle and a parabola. The task is to find the area of the region that lies inside the circle and outside the parabola. The equations are:

Circle:

Parabola:

Step 1: Understanding the Geometry

The circle is centered at with radius . The parabola opens to the right with the equation , which is in the standard form where .

Step 2: Finding the Points of Intersection

To find the points where the circle and the parabola intersect, we substitute into the equation of the circle.

Substitute from the parabola equation into the circle equation:

Expanding the terms: Simplify: Factoring out : So, the solutions for are and . These correspond to the points where the circle and parabola intersect.

Step 3: Setting up the Integral for the Area

The area of the region inside the circle and outside the parabola can be computed by finding the area of the circle segment and subtracting the area under the parabola from it.

The area under the circle, from to , can be obtained by integrating the upper half of the circle's equation: The area under the parabola can be computed by integrating: Therefore, the total area of the region inside the circle and outside the parabola is given by: This integral evaluates to .

Step 4: Final Answer

Thus, the area of the region inside the circle and outside the parabola is:

We are asked to find the value of given that the area of the region defined by is equal to , where .

Concept Used:

The area of a region bounded by curves can be found using definite integration. The area between two curves (upper curve) and (lower curve) from to is given by:

We must first analyze the inequalities, especially the one involving the absolute value, to determine the exact boundaries and the intervals of integration.

Step-by-Step Solution:

Step 1: Analyze the inequalities to define the region.

The region is defined by the following conditions:

From the first inequality, we must have . This can be written as:

This gives two separate inequalities:

• , which is always true.
• .

Since we are given , the condition implies . So, the region only exists for .

Step 2: Split the region of integration based on the absolute value function.

The expression inside the absolute value, , changes sign at . We will split the integration at this point.

Case 1:

In this interval, , so . The inequality becomes .

The bounds on are . The condition is automatically satisfied since .

Case 2:

In this interval, , so . The inequality becomes , or .

The bounds on are . We also have the condition . This means the upper bound is . Since , , so the upper bound is .

The bounds become . For a region to exist, we must have , which means , or . So this case applies for .

Step 3: Set up and calculate the integral for the first part of the area ( ).

For the interval , the area is:

Step 4: Set up and calculate the integral for the second part of the area ( ).

For the interval , the area is:

Step 5: Calculate the total area.

Total Area :

Final Computation & Result:

We are given that the area is . We equate this with our calculated area.

To match the form, we can write our result as:

By comparing the two expressions, we can identify and :

Both and are natural numbers, as required. The question asks for the value of .

The value of is 22.

To determine the area of the region defined by the inequalities , , , we solve step-by-step:

1. Inequality Analysis:

• simplifies to .
• can be split into two cases:
  • Case 1: .
  • Case 2: .
• can be split into two linear equations:
  • Region above line for .
  • Region above line for .

2. Intersection of curves:

• Curve and lines need to be checked for intersections within bounds.
• Set equations equal for intersections:
  • .
  • , no intersection.
  • Check intersection with .
  • .

3. Bounded region:

• Analyze valid region for integration satisfying all inequalities.
• Bound by lines , , applicable constraints within bounded values.
• Using symmetry, find valid intersections for bounding y between curves.

4. Area Calculation:

• Split region into sectors for integration: clockwise from to .
• Calculate separated by above-below line segments, sum bound integrals.
• .
• Simplified yields direct calculation of area .

5. Completion:

By further finding two entire areas surrounded and symmetry, total area . Therefore, the answer is 12