Question:

Area lying in the first quadrant and bounded by the circle x^2+y^2=4 and the lines x=0 and x=2 is A π B 2π​ C 3π​ D 4π

Answer:

Answer: B 2π

Question:

Find the area of the region bounded by the curves y=x^2+2,y=x,x=0 and x=3.

Answer:

Step 1: Graph the given curves.

Step 2: Identify the area bounded by the curves.

Step 3: Calculate the area of the region using the formula: Area = ∫ y₂ - y₁ dx

Step 4: Substituting the values, Area = ∫ (x2 + 2) - x dx

Step 5: Evaluate the integral from 0 to 3, Area = ∫ 3x² + 6x dx

Step 6: Solve the integral, Area = x³ + 3x² | from 0 to 3

Step 7: Substitute the limits, Area = 27 + 27 = 54

Question:

Find the area of the region bounded by the ellipse x^2​/4+y^2​/9=1.

Answer:

Step 1: Identify the equation of the ellipse given. The equation of the ellipse is x^2​/4+y^2​/9=1.

Step 2: Find the major and minor axes of the ellipse. The major axis is 4 and the minor axis is 9.

Step 3: Calculate the area of the ellipse using the formula A = πab, where a is the major axis and b is the minor axis. The area of the ellipse is A = π(4)(9) = 36π.

Question:

Find the area of the region bounded by the curve y^2=x and the lines x=1,x=4 and the x-axis in the first quadrant.

Answer:

Step 1: Draw the graph of the given equation to visualize the region.

Step 2: Calculate the area of the region by using the formula for area of a region bounded by two curves: Area = ∫y1−y2dx

Step 3: Substitute the given equations for y1 and y2 in the formula: Area = ∫√x−0dx

Step 4: Integrate the equation to get the area: Area = [x^(3/2)]/3

Step 5: Substitute the lower and upper limits of the integral: Area = [4^(3/2)−1^(3/2)]/3

Step 6: Evaluate the area: Area = (16−1)/3 = 5

Question:

Using integration find the area of the region bounded by the triangle whose vertices are (−1,0),(1,3) and (3,2).

Answer:

Step 1: Draw a diagram of the triangle.

Step 2: Identify the points of the triangle as A(-1, 0), B(1, 3) and C(3,2).

Step 3: Calculate the area of the triangle using Heron’s formula.

A = √s(s-a)(s-b)(s-c)

Where a, b, and c are the sides of the triangle and s = (a+b+c)/2

s = (1-3+2)/2 = 2

A = √2(2-1)(2-3)(2-2)

A = √2

Step 4: Calculate the area of the region bounded by the triangle using integration.

A = ∫[x1, x2] ∫[y1, y2] dxdy

A = ∫[-1, 3] ∫[0, 3x-2] dxdy

A = ∫[-1, 3] (3x-2)dx

A = [3x2/2 - 2x]|[-1, 3]

A = 9/2 -2 - (-3/2 -2)

A = 9/2 +3/2

A = 6

Therefore, the area of the region bounded by the triangle is 6.

Question:

Find the area of the region bounded by the ellipse x^2​/16+y^2​/9=1.

Answer:

Step 1: Identify the equation of the ellipse as x2​/16 + y2​/9 = 1.

Step 2: Find the area of the region bounded by the ellipse using the formula A = πab, where a and b are the semi-major and semi-minor axes of the ellipse respectively.

Step 3: Substitute the values of a and b in the equation. Here, a = 4 and b = 3.

Step 4: Calculate the area of the region. A = π43 = 36π.

Question:

Find the area of the region bounded by y^2=9x,x=2,x=4 and the x-axis in the first quadrant.

Answer:

Step 1: Find the points of intersection of the given curves.

The points of intersection are (2,0) and (4,0).

Step 2: Substitute the x-coordinates of the points of intersection into the equation of the curve to find the corresponding y-coordinates.

The corresponding y-coordinates are (2, 0) and (4, 0).

Step 3: Calculate the area of the region using the formula A = 1/2 (x2 - x1) (y2 - y1).

The area of the region is A = 1/2 (4 - 2) (0 - 0) = 0.

Question:

Find the area of the region bounded by the parabola y=x^2 and y=∣x∣.

Answer:

Step 1: Graph the two equations, y=x^2 and y=|x|, on the same coordinate plane.

Step 2: Identify the area of the region bounded by the two equations.

Step 3: Calculate the area of the region using integration.

Step 4: Set up the integral, ∫y=x^2 to y=|x|dx.

Step 5: Evaluate the integral using the limits of integration.

Step 6: Calculate the area of the region by solving the integral.

Answer: The area of the region bounded by the parabola y=x^2 and y=|x| is 4/3.

Question:

Find the area of the circle 4x^2+4y^2=9 which is interior to the parabola x^2=4y.

Answer:

1. First, we need to find the equation of the circle in the standard form. To do this, we need to complete the square for both the x and y terms.

2. The equation of the circle in standard form is (x-2)^2 + (y-1)^2 = 4.

3. Next, we need to find the points of intersection between the circle and the parabola. To do this, we can set the equations equal to each other and solve for x and y.

4. We get the points of intersection (2,1) and (-2,1).

5. Finally, we need to find the area of the circle interior to the parabola. To do this, we can use the formula for the area of a circle, A = pi*r^2, where r is the radius of the circle.

6. The radius of the circle is 2, so the area of the circle interior to the parabola is A = pi*2^2 = 4pi.

Question:

Area of the region bounded by the curve y^2=4x, y-axis and the line y=3 is A 2 B 9/4​ C 9/3​ D 9/2

Answer:

Answer: B

Step 1: Identify the equation of the given curve as y2 = 4x.

Step 2: Determine the area of the region bounded by the curve, y-axis and the line y = 3.

Step 3: Set up the integral to calculate the area by integrating y2 = 4x from y = 0 to y = 3.

Step 4: Solve the integral to get the area of the region as 9/4.

Hence, the answer is B.

Question:

​Find the area bounded by curves (x−1)^2+y^2=1 and x^2+y^2=1.

Answer:

Step 1: Find the points of intersection of the two curves.

The two curves intersect when (x−1)^2+y^2=1 and x^2+y^2=1.

Solving for x, we get x = 1 ± √2/2 and y = ± √2/2.

Therefore, the points of intersection are (1 + √2/2, √2/2) and (1 - √2/2, -√2/2).

Step 2: Find the area of the region bounded by the curves.

The area of the region can be found using the formula for the area of a circle.

A = πr^2

Therefore, the area of the region is equal to π(√2/2)^2 = π/2.

Question:

Find the area bounded by the curve x^2=4y and the line x=4y−2.

Answer:

Step 1: Draw a graph of the equation x2 = 4y and the line x = 4y – 2.

Step 2: Determine the points of intersection of the two curves.

Step 3: Calculate the area bounded by the two curves using the formula A = 1/2 ∫b a (x2 - (4y - 2)) dy.

Step 4: Substitute the points of intersection into the formula and calculate the area.

Question:

Using integration find the area of the triangular region whose sides have the equations y=2x+1,y=3x+1 and x=4

Answer:

Solution: Step 1: Find the vertices of the triangle The vertices of the triangle are (4,9), (0,1) and (2,5).

Step 2: Find the equation of the line joining the vertices The equation of the line joining (4,9) and (0,1) is y = -3x + 10

Step 3: Find the equation of the line joining the vertices The equation of the line joining (2,5) and (0,1) is y = 2x + 1

Step 4: Find the point of intersection of the two lines The point of intersection of the two lines is (2,5).

Step 5: Find the area of the triangle The area of the triangle is given by the formula A = 1/2 × base × height.

The base of the triangle is 4 units and the height is 4 units.

Therefore, the area of the triangle is A = 1/2 × 4 × 4 = 8 units2.

Question:

Smaller area enclosed by the circle x^2+y^2=4 and the line x+y=2 is A 2(π−2) B π−2 C 2π−1 D 2(π+2)

Answer:

Step 1: Identify the given equation.

The given equation is a circle with center at the origin and radius of 2, x2 + y2 = 4. The other equation is a line with equation x + y = 2.

Step 2: Find the intersection points of the two equations.

The intersection points of the two equations can be found by solving the system of equations.

x2 + y2 = 4 x + y = 2

Solving for x and y, we get (1,1) and (-1,3).

Step 3: Find the area enclosed by the circle and the line.

The area enclosed by the circle and the line can be found by subtracting the area of the triangle formed by the intersection points from the area of the circle.

The area of the triangle is ½ * base * height = ½ * 2 * 2 = 2.

The area of the circle is πr2 = π * 22 = 4π.

Therefore, the area enclosed by the circle and the line is 4π - 2 = 2π - 2.

Step 4: Choose the correct answer.

The correct answer is B, π - 2.

Question:

Area lying between the curve y2=4x and y=2x is A 2/3​ B 1/3​ C 1/4​ D 3/4

Answer:

Answer: B 1/3

Question:

The area between x=y^2 and x=4 is divided into two equal parts by the line x=a, find the value of a.

Answer:

1. Set the area of each part equal to each other: y^2 - a = 4 - a
2. Subtract a from both sides: y^2 - 4 = -2a
3. Divide both sides by -2: a = (y^2 - 4)/-2
4. Substitute the given values for x and y: a = (4 - 4)/-2
5. Simplify: a = 0

Question:

Find the area of the smaller part of the circle x^2+y^2=a^2 cut off by the line x=a​/√2.

Answer:

Step 1: Draw a diagram of the circle and the line.

Step 2: Determine the equation of the line. The equation of the line is x = a/√2

Step 3: Find the coordinates of the intersection points of the line and the circle. Substitute x = a/√2 into the equation of the circle to get: y^2 = a^2 - (a/√2)^2 y^2 = a^2 - a^2/2 y^2 = a^2/2 y = ±a/√2

Therefore, the coordinates of the intersection points are (a/√2, a/√2) and (a/√2, -a/√2).

Step 4: Calculate the area of the smaller part of the circle. The area of the smaller part of the circle is equal to the area of the circle minus the area of the larger part. The area of the circle is πa^2. The area of the larger part is the area of the triangle formed by the intersection points and the center of the circle. The height of the triangle is a and the base is a/√2, so the area of the triangle is (1/2)a(a/√2) = a^2/2√2. Therefore, the area of the smaller part of the circle is πa^2 - a^2/2√2.

Question:

Find the area of the region in the first quadrant enclosed by x-axis, line x=√3y and the circle x^2+y^2=4.

Answer:

Step 1: Draw the region in the first quadrant enclosed by x-axis, line x=√3y and the circle x^2+y^2=4.

Step 2: Use the formula for the area of a circle to calculate the area of the circle: A = πr^2, where r is the radius of the circle.

Step 3: Calculate the area of the triangle formed by the x-axis, line x=√3y and the circle x^2+y^2=4. Use the formula for the area of a triangle: A = 1/2bh, where b is the base and h is the height of the triangle.

Step 4: Add the area of the circle and the area of the triangle to get the total area of the region.

Question:

Find the area of the region bounded by x^2=4y,y=2,y=4 and the y-axis in the first quadrant.

Answer:

Step 1: Draw the graph of the given equation and the lines y = 2 and y = 4.

Step 2: Calculate the area of the region bounded by x^2=4y,y=2,y=4 and the y-axis in the first quadrant.

Step 3: The area of the region is given by the integral

A=∫2^4x^2dy

Step 4: Substitute y=4 in the equation x^2=4y to get x=2√y

Step 5: Substitute x=2√y in the integral and solve for A

A=∫2^4(2√y)^2dy

Step 6: Solve the integral to get

A=2[y^(3/2)]2^4−2[y^(3/2)]2^2

Step 7: Substitute y=2 and y=4 in the above equation to get

A=2[4^(3/2)]2^4−2[2^(3/2)]2^2

Step 8: Simplify the equation to get

A=256−32=224

Question:

Which rules did he break as a school boy?

Answer:

1. Identify the question: This is a closed-ended question that requires a specific answer.

2. Gather information: To answer this question, you will need to know what rules the school boy broke.

3. Analyze the information: Once you have the information, you will need to analyze it to determine which rules the school boy broke.

4. Draw a conclusion: Based on the analysis of the information, you can draw a conclusion about which rules the school boy broke.