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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.

πŸ”‘Concepts

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The area AA of the region bounded by the curve y=f(x)y = f(x), the xx-axis, and the vertical lines x=ax = a and x=bx = b is given by A=∫ab∣f(x)βˆ£β€‰dxA = \int_{a}^{b} |f(x)| \, dx. If the curve lies entirely above the xx-axis in [a,b][a, b], the area is simply ∫abf(x) dx\int_{a}^{b} f(x) \, dx.

Diagram showing area under a curve f(x) between x=a and x=b.
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For curves bounded by the yy-axis and horizontal lines y=cy = c and y=dy = d, the area is calculated using A=∫cd∣g(y)βˆ£β€‰dyA = \int_{c}^{d} |g(y)| \, dy. This is particularly useful for horizontal parabolas of the form x=f(y)x = f(y).

Diagram showing area between a curve and the y-axis from y=c to y=d.
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Symmetry is a critical tool in integration. For symmetric 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), calculate the area of one quadrant and multiply by 4.

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When a region lies below the xx-axis, the integral result will be negative. Since area is always positive, we take the absolute value: A=∣∫abf(x) dx∣A = |\int_{a}^{b} f(x) \, dx|.

πŸ“Formulae

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

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

Area of a circle x2+y2=a2x^2 + y^2 = a^2: A=4∫0aa2βˆ’x2 dx=Ο€a2A = 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∫0abaa2βˆ’x2 dx=Ο€abA = 4 \int_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2} \, dx = \pi ab

Integration Formula: ∫a2βˆ’x2 dx=x2a2βˆ’x2+a22sinβ‘βˆ’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

πŸ’‘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(43/2βˆ’13/2)=23(8βˆ’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=3416βˆ’x2y = \frac{3}{4}\sqrt{16 - x^2}.
  3. Total Area A=4Γ—(AreaΒ inΒ theΒ firstΒ quadrant)A = 4 \times (\text{Area in the first quadrant}).
  4. A=4∫043416βˆ’x2 dx=3∫0416βˆ’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 ∫a2βˆ’x2dx\int \sqrt{a^2 - x^2} dx: A=3[x216βˆ’x2+162sinβ‘βˆ’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+8sinβ‘βˆ’1(1))βˆ’(0+0)]=3[8Γ—Ο€2]=3Γ—4Ο€=12Ο€A = 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.

Problem 3:

Find the area of the region bounded by the parabola y2=4xy^2 = 4x and the line x=3x = 3.

Graph of parabola y^2=4x and line x=3.

Solution:

  1. The parabola y2=4xy^2 = 4x is symmetric about the xx-axis.
  2. The required area is bounded by x=0x=0 to x=3x=3.
  3. Area =2Γ—βˆ«03y dx= 2 \times \int_{0}^{3} y \, dx
  4. Since y2=4xy^2 = 4x, y=2xy = 2\sqrt{x} in the first quadrant.
  5. Area =2∫032x dx=4∫03x1/2 dx= 2 \int_{0}^{3} 2\sqrt{x} \, dx = 4 \int_{0}^{3} x^{1/2} \, dx
  6. Area =4[x3/23/2]03=4Γ—23[33βˆ’0]= 4 [\frac{x^{3/2}}{3/2}]_{0}^{3} = 4 \times \frac{2}{3} [3\sqrt{3} - 0]
  7. Area =83Γ—33=83= \frac{8}{3} \times 3\sqrt{3} = 8\sqrt{3} square units.

Explanation:

We calculate the area in the first quadrant (from x=0x=0 to x=3x=3) and double it because the parabola is symmetric across the xx-axis.

Problem 4:

Find the area of the smaller region bounded by the circle x2+y2=4x^2 + y^2 = 4 and the line x=1x = 1.

Circle x^2+y^2=4 intersected by vertical line x=1.

Solution:

  1. The circle has radius r=2r=2 and center (0,0)(0,0).
  2. The line x=1x=1 intersects the circle at y=Β±3y = \pm\sqrt{3}.
  3. Required Area =2Γ—βˆ«124βˆ’x2 dx= 2 \times \int_{1}^{2} \sqrt{4 - x^2} \, dx
  4. Using the formula ∫a2βˆ’x2dx=x2a2βˆ’x2+a22sinβ‘βˆ’1(xa)\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}):
  5. Area =2[x24βˆ’x2+42sinβ‘βˆ’1(x2)]12= 2 [\frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2})]_{1}^{2}
  6. Area =2[(0+2sinβ‘βˆ’1(1))βˆ’(32+2sinβ‘βˆ’1(12))]= 2 [(0 + 2\sin^{-1}(1)) - (\frac{\sqrt{3}}{2} + 2\sin^{-1}(\frac{1}{2}))]
  7. Area =2[2(Ο€2)βˆ’32βˆ’2(Ο€6)]= 2 [2(\frac{\pi}{2}) - \frac{\sqrt{3}}{2} - 2(\frac{\pi}{6})]
  8. Area =2[Ο€βˆ’32βˆ’Ο€3]=2[2Ο€3βˆ’32]=4Ο€3βˆ’3= 2 [\pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3}] = 2 [\frac{2\pi}{3} - \frac{\sqrt{3}}{2}] = \frac{4\pi}{3} - \sqrt{3} square units.

Explanation:

The line x=1x=1 divides the circle into two segments. We integrate from x=1x=1 to the edge of the circle at x=2x=2 and double the result due to symmetry about the xx-axis.

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