Mastering Area Between Curves: Application of Integrals Ex 8.2 (Class 12 Maths NCERT)

Understanding Area Between Two Curves

Exercise 8.2 of NCERT Class 12 Maths focuses exclusively on the application of definite integrals to calculate the area of a region bounded by two (or more) curves. Unlike finding the area under a single curve from the x-axis, here we are interested in the space that exists between two different functions, y = f(x) and y = g(x), over a specified interval [a, b]. The core idea is to subtract the area under the 'lower' curve from the area under the 'upper' curve. Mathematically, if f(x) ≥ g(x) for all x in [a, b], the area A is given by A = ∫[a, b] (f(x) - g(x)) dx.

Similarly, if the curves are defined as x = f(y) and x = g(y) and we need to find the area bounded by y = c and y = d, we use A = ∫[c, d] (f(y) - g(y)) dy, where f(y) is the 'rightmost' curve and g(y) is the 'leftmost' curve. A crucial first step is always to sketch the given curves to visualize the region and correctly identify the upper/lower or right/left functions, as well as the limits of integration. This visualization helps prevent common errors in setting up the integral.

Key Definitions for Area Between Curves

Area Between Two Curves

The region enclosed by two continuous functions, say y = f(x) and y = g(x), over an interval [a, b], calculated by A = ∫[a, b] |f(x) - g(x)| dx. If one function is consistently above the other, the absolute value sign can be removed.

Points of Intersection

The coordinates where two or more curves meet. These points are found by setting the equations of the curves equal to each other (f(x) = g(x)) and solving for x (or y). These often determine the limits of integration.

Vertical Strip (dx)

An infinitesimally thin rectangle of width dx and height (y_upper - y_lower). Used when integrating with respect to x, especially when y can be easily expressed as a function of x and the 'upper' and 'lower' curves remain consistent throughout the region.

Horizontal Strip (dy)

An infinitesimally thin rectangle of width (x_right - x_left) and height dy. Used when integrating with respect to y, typically when x can be easily expressed as a function of y and the 'right' and 'left' curves remain consistent.

Step-by-Step Method to Find Area Between Curves

1. Step 1: Sketch the Curves — Graph both given functions on the same coordinate plane. This visual representation is crucial to understand the region whose area you need to find. Identify the points where the curves intersect.
2. Step 2: Find Points of Intersection — Set the equations of the curves equal to each other (e.g., f(x) = g(x) or f(y) = g(y)) and solve for the variable. These intersection points will often serve as your limits of integration. If the problem specifies limits, verify if these points lie within those limits.
3. Step 3: Choose Integration Method (dx or dy) — Decide whether to integrate with respect to x (vertical strips) or y (horizontal strips). Choose the method that simplifies the functions and avoids splitting the region into multiple integrals. If integrating with respect to x, express y in terms of x for both curves. If with respect to y, express x in terms of y.
4. Step 4: Set Up the Integral(s) — Based on your sketch, identify which curve is 'upper' and which is 'lower' (for dx) or which is 'right' and 'left' (for dy). Formulate the definite integral: ∫[a, b] (y_upper - y_lower) dx or ∫[c, d] (x_right - x_left) dy. If the 'upper/lower' or 'right/left' function changes over the region, you'll need multiple integrals.
5. Step 5: Evaluate the Integral(s) — Calculate the definite integral(s) using the Fundamental Theorem of Calculus. Remember that area must always be a non-negative value. If your calculation yields a negative result, it often indicates you've incorrectly identified the 'upper'/'lower' or 'right'/'left' function, or need to take the absolute value of the integral over that sub-region.

Worked Examples: Finding Area Between Curves

• Example 1: Area bounded by a parabola and a line Find the area of the region bounded by y = x² and y = x. Step 1: Sketch the curves. y = x² is a parabola opening upwards with its vertex at the origin. y = x is a straight line passing through the origin with a slope of 1. Step 2: Find points of intersection. Set y = x² and y = x equal: x² = x. x² - x = 0 x(x - 1) = 0 So, x = 0 or x = 1. When x = 0, y = 0 (Point: (0,0)) When x = 1, y = 1 (Point: (1,1)) Step 3: Choose integration method. From the sketch, it's clear that for x between 0 and 1, the line y = x is above the parabola y = x². Integrating with respect to x is straightforward. Step 4: Set up the integral. The upper curve is y_upper = x and the lower curve is y_lower = x². The limits of integration are x = 0 to x = 1. Area A = ∫[0, 1] (x - x²) dx Step 5: Evaluate the integral. A = [x²/2 - x³/3] from 0 to 1 A = (1²/2 - 1³/3) - (0²/2 - 0³/3) A = (1/2 - 1/3) - 0 A = (3 - 2)/6 = 1/6 Final Answer: The area of the region bounded by y = x² and y = x is 1/6 square units.
• Example 2: Area bounded by two parabolas Find the area of the region bounded by y² = 4x and x² = 4y. Step 1: Sketch the curves. y² = 4x is a parabola opening to the right. x² = 4y is a parabola opening upwards. Step 2: Find points of intersection. From y² = 4x, we have x = y²/4. Substitute this into x² = 4y: (y²/4)² = 4y y⁴/16 = 4y y⁴ - 64y = 0 y(y³ - 64) = 0 So, y = 0 or y³ = 64 which gives y = 4. When y = 0, x = 0 (Point: (0,0)) When y = 4, x = 4²/4 = 4 (Point: (4,4)) Step 3: Choose integration method. From the sketch, between x=0 and x=4, y = 2√x (from y² = 4x) is the upper curve, and y = x²/4 (from x² = 4y) is the lower curve. Integrating with respect to x is suitable. Step 4: Set up the integral. The upper curve y_upper = 2√x = 2x^(1/2) The lower curve y_lower = x²/4 The limits of integration are x = 0 to x = 4. Area A = ∫[0, 4] (2x^(1/2) - x²/4) dx Step 5: Evaluate the integral. A = [2 (x^(3/2) / (3/2)) - (1/4) (x³/3)] from 0 to 4 A = [(4/3)x^(3/2) - x³/12] from 0 to 4 A = [(4/3)(4)^(3/2) - (4)³/12] - [(4/3)(0)^(3/2) - (0)³/12] A = [(4/3) * 8 - 64/12] A = [32/3 - 16/3] A = 16/3 Final Answer: The area of the region bounded by y² = 4x and x² = 4y is 16/3 square units.

YoLearn.ai Exam Tips for Application of Integrals (Ex 8.2)

To ace questions from Exercise 8.2, keep these critical points in mind:

1. Always Sketch the Region: This is the most crucial step. A clear, labelled sketch helps you correctly identify the points of intersection, determine which function is 'upper'/'lower' or 'right'/'left', and establish the correct limits of integration. Misidentifying these can lead to incorrect setup and a wrong answer.
2. Find ALL Points of Intersection: Don't stop at the first intersection point you find. Ensure you've found all points where the curves intersect within the region of interest. These points define the boundaries of your integration intervals.
3. Correctly Identify y_upper - y_lower or x_right - x_left: If f(x) is above g(x), use f(x) - g(x). If g(x) is above f(x), use g(x) - f(x). If the relative positions change, you might need to split the integral into multiple parts. The same logic applies to dx vs dy integration. A common mistake is simply subtracting f(x) - g(x) without considering which is larger, which can lead to a negative area.
4. Choose the Appropriate Variable of Integration (dx or dy): Sometimes integrating with respect to x is easier, other times y. If a curve's equation is simpler in terms of x (e.g., x = f(y)), consider integrating with respect to y. Sometimes, switching variables can avoid splitting the integral.
5. Be Careful with Modulus and Symmetry: If the area needs to be calculated in multiple quadrants or if the function changes sign, remember that area is always positive. You might need to use |f(x) - g(x)| or utilize symmetry (e.g., 2 * ∫[0, a] ... dx) if the region is symmetric about an axis.
6. Accuracy in Integration: Even with the correct setup, errors in basic integration or arithmetic can cost marks. Double-check your calculations, especially with fractional powers or trigonometric functions.

Practice Questions with Solutions

• Q: Find the area of the region bounded by y = x² and y = 2x. A: Step 1: Sketch the curves y = x² (parabola) and y = 2x (line). Step 2: Find points of intersection: Set x² = 2x => x² - 2x = 0 => x(x - 2) = 0. This gives x = 0 and x = 2. The points of intersection are (0,0) and (2,4). Step 3: Observe the graph. Between x = 0 and x = 2, the line y = 2x is above the parabola y = x². Step 4: Set up the integral: Area A = ∫[0, 2] (2x - x²) dx. Step 5: Evaluate the integral: A = [2(x²/2) - x³/3] from 0 to 2 = [x² - x³/3] from 0 to 2 = (2² - 2³/3) - (0² - 0³/3) = (4 - 8/3) - 0 = (12 - 8)/3 = 4/3. Final answer: The area is 4/3 square units.
• Q: Determine the area of the region bounded by y = x and y = x³. A: Step 1: Sketch y = x (line) and y = x³ (cubic curve). Step 2: Find points of intersection: Set x³ = x => x³ - x = 0 => x(x² - 1) = 0 => x(x - 1)(x + 1) = 0. This gives x = -1, 0, 1. Step 3: Observe the graph. Between x = -1 and x = 0, y = x³ is above y = x. Between x = 0 and x = 1, y = x is above y = x³. Due to symmetry, we can calculate the area from 0 to 1 and multiply by 2. Step 4: Set up the integral: Area A = ∫[-1, 0] (x³ - x) dx + ∫[0, 1] (x - x³) dx. Using symmetry, A = 2 ∫[0, 1] (x - x³) dx. Step 5: Evaluate the integral: A = 2 [x²/2 - x⁴/4] from 0 to 1 = 2 [(1²/2 - 1⁴/4) - (0²/2 - 0⁴/4)] = 2 [(1/2 - 1/4) - 0] = 2 * (1/4) = 1/2. Final answer: The area is 1/2 square units.
• Q: Find the area bounded by x = y² and x = y. A: Step 1: Sketch the curves x = y² (parabola opening right) and x = y (line passing through origin with slope 1). Step 2: Find points of intersection: Set y² = y => y² - y = 0 => y(y - 1) = 0. This gives y = 0 and y = 1. The points of intersection are (0,0) and (1,1). Step 3: Observe the graph. Between y = 0 and y = 1, the line x = y is to the right of the parabola x = y². Integration with respect to y is suitable. Step 4: Set up the integral: Area A = ∫[0, 1] (y - y²) dy. Step 5: Evaluate the integral: A = [y²/2 - y³/3] from 0 to 1 = (1²/2 - 1³/3) - (0²/2 - 0³/3) = (1/2 - 1/3) - 0 = (3 - 2)/6 = 1/6. Final answer: The area is 1/6 square units.
• Q: Calculate the area bounded by the curves y = sin(x) and y = cos(x) between x = 0 and x = π/2. A: Step 1: Sketch y = sin(x) and y = cos(x) from x = 0 to x = π/2. You'll notice they intersect once in this interval. Step 2: Find point of intersection: Set sin(x) = cos(x). Dividing by cos(x) (assuming cos(x) ≠ 0) gives tan(x) = 1. In the interval [0, π/2], this occurs at x = π/4. Step 3: From x = 0 to x = π/4, cos(x) ≥ sin(x). From x = π/4 to x = π/2, sin(x) ≥ cos(x). Step 4: Set up integrals: Area A = ∫[0, π/4] (cos(x) - sin(x)) dx + ∫[π/4, π/2] (sin(x) - cos(x)) dx. Step 5: Evaluate the integrals: For the first integral: [sin(x) + cos(x)] from 0 to π/4 = (sin(π/4) + cos(π/4)) - (sin(0) + cos(0)) = (1/√2 + 1/√2) - (0 + 1) = 2/√2 - 1 = √2 - 1. For the second integral: [-cos(x) - sin(x)] from π/4 to π/2 = (-cos(π/2) - sin(π/2)) - (-cos(π/4) - sin(π/4)) = (0 - 1) - (-1/√2 - 1/√2) = -1 - (-2/√2) = -1 + √2. Total Area A = (√2 - 1) + (√2 - 1) = 2√2 - 2. Final answer: The area is (2√2 - 2) square units.

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