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Application of Integrals - Area between any of the two above said curves

Grade 12CBSE

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

πŸ”‘Concepts

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Identifying the Bounded Region: The area between two curves, y=f(x)y = f(x) and y=g(x)y = g(x), is the physical space enclosed between their graphs from a starting point x=ax = a to an ending point x=bx = b. Visually, this involves identifying the 'upper' curve and the 'lower' curve within the given interval to ensure a positive area calculation.

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Finding Points of Intersection: To determine the limits of integration (aa and bb), set the two functions equal to each other (f(x)=g(x)f(x) = g(x)) and solve for xx. Visually, these points are the coordinates where the two graphs cross or touch, marking the boundaries of the shaded region.

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Vertical Elementary Strips: When curves are expressed as functions of xx, we imagine thin vertical rectangles of width dxdx and height (yupperβˆ’ylower)(y_{upper} - y_{lower}). The total area is the sum of these strips. Visually, this corresponds to integrating along the xx-axis from the leftmost intersection to the rightmost intersection.

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Horizontal Elementary Strips: If curves are more easily expressed as functions of yy (e.g., x=f(y)x = f(y)), we use horizontal strips of thickness dydy and width (xrightβˆ’xleft)(x_{right} - x_{left}). Visually, this involves integrating along the yy-axis from the lowest point of the region to the highest point.

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Area of Crossing Curves: If the curves intersect at an intermediate point cc within the interval [a,b][a, b], the 'upper' and 'lower' curves swap positions. In such cases, the area must be split into two separate integrals: ∫ac∣f(x)βˆ’g(x)∣dx+∫cb∣g(x)βˆ’f(x)∣dx\int_{a}^{c} |f(x) - g(x)| dx + \int_{c}^{b} |g(x) - f(x)| dx. Visually, this accounts for the change in which graph is on top.

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Utilizing Symmetry: Many standard curves like circles (x2+y2=r2x^2 + y^2 = r^2) or parabolas (y2=4axy^2 = 4ax) are symmetric about the axes. If the region is symmetric, you can calculate the area of one-half or one-quarter of the region and multiply by the appropriate factor to find the total area, simplifying the integration process.

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Geometric Visualizations: When sketching the region, always plot the intersection points and test a value between the limits to determine which function is greater. For example, if comparing y=xy = x and y=x2y = x^2 between 00 and 11, testing x=0.5x = 0.5 shows 0.5>0.250.5 > 0.25, meaning the line is the upper boundary.

πŸ“Formulae

A=∫ab[f(x)βˆ’g(x)]dxA = \int_{a}^{b} [f(x) - g(x)] dx, where f(x)β‰₯g(x)f(x) \geq g(x) for all x∈[a,b]x \in [a, b]

A=∫cd[xrightβˆ’xleft]dyA = \int_{c}^{d} [x_{right} - x_{left}] dy, where x=f(y)x = f(y) and x=g(y)x = g(y) are functions of yy

A=∫ab∣f(x)βˆ’g(x)∣dxA = \int_{a}^{b} |f(x) - g(x)| dx, used when the relative positions of curves change over the interval

Area of circle x2+y2=a2x^2 + y^2 = a^2 is Ο€a2\pi a^2 (derived via 4∫0aa2βˆ’x2dx4 \int_{0}^{a} \sqrt{a^2 - x^2} dx)

Area of ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 is Ο€ab\pi ab

πŸ’‘Examples

Problem 1:

Find the area of the region bounded by the curves y=x2y = x^2 and y=xy = x.

Solution:

  1. Find points of intersection: Set x2=xβ‡’x2βˆ’x=0β‡’x(xβˆ’1)=0x^2 = x \Rightarrow x^2 - x = 0 \Rightarrow x(x - 1) = 0. So, x=0x = 0 and x=1x = 1.
  2. Identify the upper curve: For x∈(0,1)x \in (0, 1), x>x2x > x^2 (e.g., at x=0.5x = 0.5, 0.5>0.250.5 > 0.25). Thus, y=xy = x is the upper curve and y=x2y = x^2 is the lower curve.
  3. Set up the integral: A=∫01(xβˆ’x2)dxA = \int_{0}^{1} (x - x^2) dx
  4. Integrate: A=[x22βˆ’x33]01=(12βˆ’13)βˆ’(0βˆ’0)=3βˆ’26=16A = [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^{1} = (\frac{1}{2} - \frac{1}{3}) - (0 - 0) = \frac{3-2}{6} = \frac{1}{6}
  5. The area is 16\frac{1}{6} square units.

Explanation:

We first solve for the limits of integration by finding where the parabola and the line meet. By observing the interval [0,1][0, 1], we determine the line is above the parabola, so we subtract the parabola's function from the line's function.

Problem 2:

Find the area of the region bounded by the parabolas y2=4xy^2 = 4x and x2=4yx^2 = 4y.

Solution:

  1. Find intersection points: From x2=4yx^2 = 4y, we get y=x24y = \frac{x^2}{4}. Substitute into y2=4xy^2 = 4x: (x24)2=4xβ‡’x416=4xβ‡’x4=64xβ‡’x(x3βˆ’64)=0(\frac{x^2}{4})^2 = 4x \Rightarrow \frac{x^4}{16} = 4x \Rightarrow x^4 = 64x \Rightarrow x(x^3 - 64) = 0. Intersection at x=0x = 0 and x=4x = 4.
  2. Upper curve identification: For x∈(0,4)x \in (0, 4), y=4xy = \sqrt{4x} is the upper boundary and y=x24y = \frac{x^2}{4} is the lower boundary.
  3. Set up integral: A=∫04(2xβˆ’x24)dxA = \int_{0}^{4} (2\sqrt{x} - \frac{x^2}{4}) dx
  4. Integrate: A=[2β‹…23x3/2βˆ’14β‹…x33]04=[43x3/2βˆ’x312]04A = [2 \cdot \frac{2}{3}x^{3/2} - \frac{1}{4} \cdot \frac{x^3}{3}]_{0}^{4} = [\frac{4}{3}x^{3/2} - \frac{x^3}{12}]_{0}^{4}
  5. Evaluate: A=(43(4)3/2βˆ’4312)βˆ’0=(43β‹…8βˆ’6412)=323βˆ’163=163A = (\frac{4}{3}(4)^{3/2} - \frac{4^3}{12}) - 0 = (\frac{4}{3} \cdot 8 - \frac{64}{12}) = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}
  6. The area is 163\frac{16}{3} square units.

Explanation:

This problem involves two parabolas. We find the intersection points (0,0)(0,0) and (4,4)(4,4). Since the first parabola y2=4xy^2=4x is 'wider' towards the x-axis in this quadrant, it acts as the upper boundary (y=2xy = 2\sqrt{x}) while x2=4yx^2=4y acts as the lower boundary.

Area between any of the two above said curves Class 12 Notes & Examples