The area of the region bounded by the curve x = 2y + 3 and the lines y = 1, y = -1 and y-axis is
The y-axis is the vertical axis in a Cartesian coordinate system. It represents the dependent variable and is used to plot values of \( y \). Points on the y-axis have an x-coordinate of zero. It helps in graphing equations and analyzing relationships between variables in mathematics physics and engineering.
Class 12 Maths Chapter 8 Applications of Integrals is an important topic for the CBSE Exam 2024-25. It focuses on finding areas enclosed by curves and calculating volumes of solids using integration. These concepts have real-life applications in physics and engineering. Understanding them is essential for solving practical problems and higher studies.
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In order to discover the area of the region bounded by the curve x = 2y + 3, the lines y = 1, y = -1, and the y-axis, we should set up the definite integral.
Step 1: Determine the x-intercepts of the curve
The curve is defined by the equation x = 2y + 3. For the y-axis, we let x = 0:
0 = 2y + 3
Solve for y:
So the curve crosses the y-axis at y = -3/2.
Step 2: Put the integral
The area is bounded by y = 1, y = -1, and the curve. We’ll have to integrate the function w.r.t to y in the range from y = -1 to y = 1 for finding the area,
A = ∫₋₁¹ (2y + 3) dy
Step 3: Evaluate the integral
Integrate the expression 2y + 3 first.
∫ (2y + 3) dy = y² + 3y
Now, substitute these values into this from y = -1 to y = 1:
A = [y² + 3y]₋₁¹
When y = 1:
1² + 3(1) = 1 + 3 = 4
When y = -1:
(-1)² + 3(-1) = 1 – 3 = -2
Thus the area is,
A = 4 – (-2) = 4 + 2 = 6
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