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Application of Integrals - Applications in finding the area under simple curves (lines, circles, parabolas, ellipses)

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Area as a Definite Integral: The area of the region bounded by the curve y=f(x)y = f(x), the xx-axis, and the ordinates x=ax = a and x=bx = b is given by ∫aby dx\int_{a}^{b} y \, dx. Visually, this is equivalent to summing up the areas of infinitely thin vertical rectangular strips, each of height yy and width dxdx, placed side-by-side between the boundaries.

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Area with respect to the y-axis: For a curve defined as x=g(y)x = g(y), the area bounded by the curve, the yy-axis, and the horizontal lines y=cy = c and y=dy = d is given by ∫cdx dy\int_{c}^{d} x \, dy. Visually, this involves summing thin horizontal strips of length xx and thickness dydy moving vertically from cc to dd.

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Sign of the Area: If a portion of the curve lies below the xx-axis, the definite integral will result in a negative value. Since area is a physical quantity and must be positive, we take the absolute value ∣∫abf(x) dx∣|\int_{a}^{b} f(x) \, dx|. Visually, if the region is below the xx-axis, you reflect the calculation to ensure the magnitude is positive.

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Symmetry in Standard Curves: Standard figures like circles (x2+y2=a2x^2 + y^2 = a^2) and ellipses (x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1) are symmetric about both the xx and yy axes. To find the total area, it is often simpler to calculate the area of one quadrant (from x=0x=0 to x=ax=a) and multiply the result by 4.

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Points of Intersection: When finding the area bounded by two curves (like a line and a parabola), the first step is to solve the equations simultaneously to find the points of intersection. These intersection points provide the limits of integration for the bounded region.

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Visualizing the Region: A rough sketch of the curve is essential. For example, a parabola y2=4axy^2 = 4ax opens to the right, while x2=4ayx^2 = 4ay opens upwards. Identifying the shaded region helps determine whether to integrate with respect to xx or yy and which function is the 'upper' or 'lower' boundary.

📐Formulae

Area bounded by y=f(x)y=f(x) and x-axis: A=∫ab∣f(x)∣ dxA = \int_{a}^{b} |f(x)| \, dx

Area bounded by x=g(y)x=g(y) and y-axis: A=∫cd∣g(y)∣ dyA = \int_{c}^{d} |g(y)| \, dy

Area of a circle x2+y2=a2x^2 + y^2 = a^2: A=4∫0aa2−x2 dx=πa2A = 4 \int_{0}^{a} \sqrt{a^2 - x^2} \, dx = \pi a^2

Area of an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1: A=4∫0abaa2−x2 dx=πabA = 4 \int_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2} \, dx = \pi ab

Integration Formula: ∫a2−x2 dx=x2a2−x2+a22sin⁡−1(xa)+C\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C

💡Examples

Problem 1:

Find the area of the region bounded by the curve y2=xy^2 = x and the lines x=1x = 1, x=4x = 4 and the xx-axis in the first quadrant.

Solution:

  1. The given curve is a parabola y2=xy^2 = x. In the first quadrant, y=xy = \sqrt{x}.
  2. The limits of integration are x=1x = 1 to x=4x = 4.
  3. Area A=∫14y dx=∫14x dxA = \int_{1}^{4} y \, dx = \int_{1}^{4} \sqrt{x} \, dx.
  4. A=∫14x1/2 dx=[x3/23/2]14=23[x3/2]14A = \int_{1}^{4} x^{1/2} \, dx = [\frac{x^{3/2}}{3/2}]_{1}^{4} = \frac{2}{3}[x^{3/2}]_{1}^{4}.
  5. A=23(43/2−13/2)=23(8−1)=23(7)=143A = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{2}{3}(7) = \frac{14}{3} square units.

Explanation:

We use the standard formula for area under a curve with respect to the x-axis. Since only the first quadrant is mentioned, we only integrate the positive square root of xx.

Problem 2:

Find the area of the region bounded by the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1.

Solution:

  1. The equation is x242+y232=1\frac{x^2}{4^2} + \frac{y^2}{3^2} = 1. Here a=4a = 4 and b=3b = 3.
  2. Solve for yy: y=3416−x2y = \frac{3}{4}\sqrt{16 - x^2}.
  3. Total Area A=4×(Area in the first quadrant)A = 4 \times (\text{Area in the first quadrant}).
  4. A=4∫043416−x2 dx=3∫0416−x2 dxA = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx = 3 \int_{0}^{4} \sqrt{16 - x^2} \, dx.
  5. Using ∫a2−x2dx\int \sqrt{a^2 - x^2} dx: A=3[x216−x2+162sin⁡−1(x4)]04A = 3 [\frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}(\frac{x}{4})]_{0}^{4}.
  6. A=3[(0+8sin⁡−1(1))−(0+0)]=3[8×π2]=3×4π=12πA = 3 [ (0 + 8\sin^{-1}(1)) - (0 + 0) ] = 3 [ 8 \times \frac{\pi}{2} ] = 3 \times 4\pi = 12\pi square units.

Explanation:

The ellipse is symmetric about both axes, so we calculate the area in the first quadrant (from x=0x=0 to x=4x=4) and multiply by 4. This utilizes the standard integration formula for circular/elliptical arcs.

Applications in finding the area under simple curves (lines, circles, parabolas, ellipses)…