![]() |
VOOZH | about |
Areas of Bounded Regions" typically refers to the mathematical calculation of the area enclosed by curves or lines on a coordinate plane. This concept is fundamental in calculus, particularly in the context of definite integrals.
Solution:
From the question it is given that,
Lines, x = 0, y = 1, y = 4
Parabola y = 4x2 … [equation (i)]
So, equation (i) represents a parabola with vertex (0, 0) and axis as y – axis. x = 0 is y – axis and y = 1, y = 4 are line parallel to x – axis passing through (0, 1) and (0, 4) respectively, as shown in the rough sketch below,
👁 ImageNow, we have to find the area of ABCDA,
Then, the area can be found by taking a small slice in each region of width Δy,
And length = x
The area of sliced part will be as it is a rectangle = x Δy
So, this rectangle can move horizontal from y = 1 to x = 4
The required area of the region bounded between the lines = Region ABCDA
Given, y = 4x2
x =
On integration, we get,
Now, applying limits we get,
Therefore, the required area is square units.
Solution:
From the question it is given that,
Region in first quadrant bounded by y = 1, y = 4
Parabola x2 = 16y … [equation (i)]
So, equation (i) represents a parabola with vertex (0, 0) and axis as y-axis, as shown in the rough sketch below,
👁 ImageNow, we have to find the area of ABCDA,
Then, the area can be found by taking a small slice in each region of width Δy,
And length = x
The area of sliced part will be as it is a rectangle = x Δy
So, this rectangle can move horizontal from y = 1 to x = 4
The required area of the region bounded between the lines = Region ABCDA
Given, x2 = 16y
On integrating we get,
Given, x2 = 16y
On integrating we get,
Now, applying limits we get,
Therefore, the required area is square units.
Solution:
We have to find the area of the region bounded by x2 = 4ay
👁 ImageThen,
Area of the region =
On integrating we get,
=
Now applying limits,
Therefore, the area of the region is square units.
Solution:
We have to find the area of the region bounded by x2 + 16y = 0
👁 ImageThen,
Area of the region =
On integrating we get,
=
Now applying limits,
Therefore, the area of the region is square units.
Solution:
We have to find the area of the region bounded by curve ay2 = x3, and lines y = a and y = 2a.
👁 ImageThen,
Area of the region =
On integrating we get, =
Now applying limits we get, =
1. Find the area of the region bounded by the parabola y = x^2 - 4x + 3 and the x-axis.
2. Calculate the area enclosed by the curve y = sin x and the x-axis from x = 0 to x = π.
3. Determine the area of the region bounded by the curve y = |x - 1| and the x-axis from x = -1 to x = 3.
4. Find the area of the region bounded by the curve y = x^3 - 3x + 2 and the x-axis.
5. Calculate the area enclosed by the curve y = e^x and the x-axis from x = 0 to x = ln 2.
6. Determine the area of the region bounded by the curve y = √(4 - x^2) and the x-axis.
7. Find the area of the region bounded by the curve y = cos x and the x-axis from x = -π/2 to x = π/2.
8. Calculate the area enclosed by the curve y = x^2 - 5x + 6 and the x-axis.
9. Determine the area of the region bounded by the curve y = ln x and the x-axis from x = 1 to x = e.
10. Find the area of the region bounded by the curve y = x^3 - x and the x-axis.
Chapter 21 of RD Sharma's Class 12 mathematics textbook, specifically Exercise 21.2, focuses on calculating the areas of regions bounded by curves and the x-axis.
Key points covered in this exercise include:
1. Using definite integrals to calculate areas
2. Identifying the points where a curve intersects the x-axis
3. Handling curves that cross the x-axis multiple times
4. Dealing with absolute value functions in area calculations
5. Applying the concept of even and odd functions to simplify calculations
6. Understanding the significance of positive and negative areas
7. Utilizing symmetry to reduce computational complexity
8. Interpreting the geometrical meaning of the calculated areas
This exercise builds on the fundamental concepts of integration and applies them to practical problems involving area calculations. It's crucial for understanding many applications in physics, engineering, and advanced mathematics.