Class 12 — Mathematics (NCERT)

Application of Integrals: Area Under Curves

Class 12

✓By the end of this lesson students will be able to understand how definite integrals are used to find the area under simple curves.
✓By the end of this lesson students will be able to calculate the area of the region bounded by a curve, the x-axis, and two ordinates.
✓By the end of this lesson students will be able to calculate the area of the region bounded by a curve, the y-axis, and two abscissae.
✓By the end of this lesson students will be able to find the area of the region bounded by two curves.
✓By the end of this lesson students will be able to apply the concept of area under curves to solve problems involving lines, circles, parabolas, and ellipses.

Key concepts

Area under a Curve (with respect to x-axis)

If y = f(x) is a continuous function and f(x) ≥ 0 for all x in [a, b], then the area A of the region bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by the definite integral. If the curve lies below the x-axis (i.e., f(x) ≤ 0), the area is the absolute value of the integral. If the curve crosses the x-axis, the area must be calculated by summing the absolute values of the integrals over sub-intervals where the function does not change sign.

A = \int_a^b y \, dx = \int_a^b f(x) \, dx. If f(x) < 0, A = |\int_a^b f(x) \, dx|.

Area under a Curve (with respect to y-axis)

If x = g(y) is a continuous function and g(y) ≥ 0 for all y in [c, d], then the area A of the region bounded by the curve x = g(y), the y-axis, and the abscissae y = c and y = d is given by the definite integral. If the curve lies to the left of the y-axis (i.e., g(y) ≤ 0), the area is the absolute value of the integral. Similar to the x-axis case, if the curve crosses the y-axis, the area must be calculated by summing the absolute values of the integrals over sub-intervals.

A = \int_c^d x \, dy = \int_c^d g(y) \, dy. If g(y) < 0, A = |\int_c^d g(y) \, dy|.

Area between Two Curves

If y = f(x) and y = g(x) are two continuous functions such that f(x) ≥ g(x) for all x in [a, b], then the area A of the region bounded by these curves and the lines x = a and x = b is given by the integral of the difference between the 'upper' curve and the 'lower' curve. Similarly, if integrating with respect to y, and x = f(y) and x = g(y) are such that f(y) ≥ g(y) (right curve is greater than left curve) for all y in [c, d], the area is the integral of their difference.

A = \int_a^b [f(x) - g(x)] \, dx \quad \text{or} \quad A = \int_c^d [f(y) - g(y)] \, dy

Key facts to remember

1The area is always a non-negative quantity. If the integral yields a negative value, take its absolute value.
2The area bounded by y = f(x), the x-axis, x = a, and x = b is given by \int_a^b y \, dx.
3The area bounded by x = g(y), the y-axis, y = c, and y = d is given by \int_c^d x \, dy.
4The area between two curves y = f(x) and y = g(x) from x = a to x = b, where f(x) \ge g(x), is given by \int_a^b [f(x) - g(x)] \, dx.
5Always draw a neat sketch of the region to be able to correctly set up the integral and identify the limits.
6If a curve crosses the axis (x-axis or y-axis) within the given interval, the area must be calculated by splitting the integral at the points of intersection and summing the absolute values of the integrals over each sub-interval.
7Symmetry of the region can be used to simplify calculations by finding the area of a part of the region and multiplying it by an appropriate factor.

Worked examples

Example 1

Find the area of the region bounded by the curve y = x^2, the x-axis, and the lines x = 1 and x = 3.

IStep 1: Sketch the region. The curve y = x^2 is a parabola opening upwards. The region is in the first quadrant, bounded by x=1 and x=3.

IIStep 2: Identify the formula for area. Since the curve is above the x-axis in the given interval [1, 3], the area A is given by A = \int_a^b y \, dx.

IIIStep 3: Set up the integral. Here, y = x^2, a = 1, and b = 3. So, A = \int_1^3 x^2 \, dx.

IVStep 4: Evaluate the integral. A = [\frac{x^3}{3}]_1^3.

VStep 5: Apply the limits. A = (\frac{3^3}{3}) - (\frac{1^3}{3}) = (\frac{27}{3}) - (\frac{1}{3}) = 9 - \frac{1}{3} = \frac{27-1}{3} = \frac{26}{3}.

Answer

The required area is \frac{26}{3} square units.

Always ensure the curve is above the x-axis (or to the right of the y-axis) for a direct positive integral result. Otherwise, take the absolute value.

Example 2

Determine the area of the region bounded by the curve y = x^2 - 4, the x-axis, and the lines x = 0 and x = 2.

IStep 1: Sketch the region. The curve y = x^2 - 4 is a parabola opening upwards with vertex at (0, -4). For x in [0, 2], y = x^2 - 4 is negative (e.g., at x=0, y=-4; at x=2, y=0). Thus, the region lies below the x-axis.

IIStep 2: Identify the formula for area. Since the curve is below the x-axis, the area A is given by A = |\int_a^b y \, dx|.

IIIStep 3: Set up the integral. Here, y = x^2 - 4, a = 0, and b = 2. So, A = |\int_0^2 (x^2 - 4) \, dx|.

IVStep 4: Evaluate the integral. \int_0^2 (x^2 - 4) \, dx = [\frac{x^3}{3} - 4x]_0^2.

VStep 5: Apply the limits. [\frac{x^3}{3} - 4x]_0^2 = (\frac{2^3}{3} - 4(2)) - (\frac{0^3}{3} - 4(0)) = (\frac{8}{3} - 8) - 0 = \frac{8 - 24}{3} = -\frac{16}{3}.

VIStep 6: Take the absolute value. A = |-\frac{16}{3}| = \frac{16}{3}.

Answer

The required area is \frac{16}{3} square units.

When the curve is below the x-axis, the definite integral yields a negative value. The area, being a physical quantity, must always be positive, hence we take the absolute value.

Example 3

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

IStep 1: Sketch the region. Draw the parabola y = x^2 and the line y = x.

IIStep 2: Find the points of intersection. Set the equations equal to each other: x^2 = x. This gives x^2 - x = 0, so x(x - 1) = 0. The intersection points are x = 0 and x = 1. Corresponding y-values are y = 0^2 = 0 and y = 1^2 = 1. So, the points are (0, 0) and (1, 1).

IIIStep 3: Identify the 'upper' and 'lower' curves. In the interval (0, 1), pick a test point, say x = 0.5. For y = x, y = 0.5. For y = x^2, y = (0.5)^2 = 0.25. Since 0.5 > 0.25, the line y = x is the upper curve and the parabola y = x^2 is the lower curve.

IVStep 4: Set up the integral. The area A is given by A = \int_a^b (y_{upper} - y_{lower}) \, dx. Here, a = 0, b = 1, y_{upper} = x, and y_{lower} = x^2. So, A = \int_0^1 (x - x^2) \, dx.

VStep 5: Evaluate the integral. A = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1.

VIStep 6: Apply the limits. A = (\frac{1^2}{2} - \frac{1^3}{3}) - (\frac{0^2}{2} - \frac{0^3}{3}) = (\frac{1}{2} - \frac{1}{3}) - 0 = \frac{3 - 2}{6} = \frac{1}{6}.

Answer

The required area is \frac{1}{6} square units.

Carefully determine which function is 'upper' and which is 'lower' in the interval of integration. If the curves cross multiple times, the integral needs to be split into multiple parts.

Example 4

Find the area of the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

IStep 1: Sketch the ellipse. The ellipse is symmetric about both the x-axis and the y-axis, centered at the origin. Its x-intercepts are (±a, 0) and y-intercepts are (0, ±b).

IIStep 2: Use symmetry. Due to symmetry, the total area of the ellipse is four times the area of the region in the first quadrant.

IIIStep 3: Express y in terms of x for the upper half of the ellipse in the first quadrant. From \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, we get \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2}. So, y^2 = \frac{b^2}{a^2}(a^2 - x^2). Taking the positive square root for the first quadrant, y = \frac{b}{a}\sqrt{a^2 - x^2}.

IVStep 4: Set up the integral for the area in the first quadrant. The limits of integration for x are from 0 to a. Area in first quadrant = \int_0^a y \, dx = \int_0^a \frac{b}{a}\sqrt{a^2 - x^2} \, dx.

VStep 5: Evaluate the integral. Area in first quadrant = \frac{b}{a} \int_0^a \sqrt{a^2 - x^2} \, dx. We use the standard integral formula: \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}).

VIStep 6: Apply the limits. Area in first quadrant = \frac{b}{a} [\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})]_0^a.

VIIStep 7: Substitute the limits. = \frac{b}{a} [ (\frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}(\frac{a}{a})) - (\frac{0}{2}\sqrt{a^2 - 0^2} + \frac{a^2}{2}\sin^{-1}(\frac{0}{a})) ].

VIIIStep 8: Simplify. = \frac{b}{a} [ (0 + \frac{a^2}{2}\sin^{-1}(1)) - (0 + 0) ] = \frac{b}{a} [\frac{a^2}{2} \cdot \frac{\pi}{2}] = \frac{b}{a} \cdot \frac{\pi a^2}{4} = \frac{\pi ab}{4}.

9Step 9: Calculate the total area. Total Area = 4 \times (Area in first quadrant) = 4 \times \frac{\pi ab}{4} = \pi ab.

Answer

The area of the ellipse is \pi ab square units.

Recognising symmetry can significantly simplify the calculation by reducing the integration range and multiplying the result by an appropriate factor.

Common mistakes

✗Forgetting to take the absolute value of the integral when the curve lies below the x-axis or to the left of the y-axis, leading to a negative area.
✗Incorrectly identifying the 'upper' and 'lower' curves (or 'right' and 'left' curves) when finding the area between two curves.
✗Errors in finding the points of intersection of curves, which are crucial for setting the limits of integration.
✗Not sketching the region, which often leads to misinterpretation of the problem and incorrect setup of the integral.
✗Algebraic or calculus errors during the integration process or while applying the limits of integration.
✗Using the wrong variable of integration (e.g., integrating with respect to x when integration with respect to y would be simpler or necessary).

Exam tips

★Always begin by drawing a clear and accurate sketch of the region whose area is to be found. This helps in visualising the problem and correctly setting up the integral.
★Accurately determine the points of intersection of the curves involved. These points usually define the limits of integration.
★Choose the appropriate variable of integration (dx or dy) based on the shape of the region and the ease of expressing one variable in terms of the other.
★Look for symmetry in the region. If the region is symmetric, calculate the area of one symmetric part and multiply it by the number of such parts to simplify calculations.
★Double-check all calculations, especially the evaluation of definite integrals and algebraic simplifications, to avoid silly errors. State the final answer with 'square units'.

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