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CIE A-Level Maths Study Notes

1.8.4 Areas under Curves

Integration is more than just the reverse of differentiation; it represents the accumulation of quantities. When integrating a function, it is essentially adding up an infinite number of infinitesimally small areas under the curve to find a total area. This concept is fundamental in understanding how to calculate areas under curves. In this section, the focus is on the application of integration to find areas bounded by curves, lines, and between multiple curves.

Area Under a Single Curve

area under the curve

Image courtesy of Cuemath

Area Bounded by the Curve and the x-axis

  • Concept: Finding the area between a curve and the x-axis.
  • Method: Integrate with respect to dx.
  • Procedure: Express y as a function of x, integrate from a to b.

abydx\int_a^b y \, \mathrm{dx}

  • Example: For the curve y=x2y = x^2 between x=1x = 1 and ( x = 3 ), the area is 13x2dx\int_1^3 x^2 \, \mathrm{dx}.
area bounded by the curve and the x-axis

Area Bounded by the Curve and the y-axis

  • Concept: Finding the area between a curve and the y-axis.
  • Method: Integrate with respect to dy.
  • Procedure: Make x the subject, integrate from a to b.

abxdy \int_a^b x \, \mathrm{dy}

  • Example: For the curve x=y2x = y^2 from y=2y = 2 to y=4y = 4, the area is 24y2dy.\int_2^4 y^2 \, \mathrm{dy}.
areas bounded by the curve and the x-axis

Example: Specific Area Calculation

Calculate the area under the curve (24(3x34x2+2x+5)dx.\int_2^4 (3x^3 - 4x^2 + 2x + 5) \, \mathrm{dx}.

Solution:

1. Integrate each term, including the constant as 5x05x^0.

24(3x34x2+2x+5)dx=[3x444x33+x2+5x]24\int_2^4(3x^3 - 4x^2 + 2x + 5) \, \mathrm{dx} = \left[\frac{3x^4}{4} - \frac{4x^3}{3} + x^2 + 5x \right]_2^4

2. Substitute the limits to get 3823\frac{382}{3}.

area under the curve

Area Between Two Curves

area between two curves

Image courtesy of Cuemath

With Respect to x

  • Concept: Calculating the area between two curves along the x-axis.
  • Method: Subtract the integral of the lower curve from the upper curve.

ab(y1y2)dx \int_a^b (y_1 - y_2) \, \mathrm{dx}

  • Example: Between curves y1=x2y_1 = x^2 and y2=xy_2 = x from x=0x = 0 to x=1x = 1, calculate 01(x2x)dx\int_0^1 (x^2 - x) \, \mathrm{dx}.
area between two curves

With Respect to y

  • Method: Similar approach, but with respect to dy.

ab(x1x2)dy \int_a^b (x_1 - x_2) \, \mathrm{dy}

  • Example: Between curves x1=yx_1 = y and x2=y2x_2 = y^2 from y=0y = 0 to y=1y = 1, calculate 01(yy2)dy\int_0^1 (y - y^2) \, \mathrm{dy}.
area between two curves

Application Example:

Problem Statement

Given the curve y=4x+1+94x+1y = \sqrt{4x + 1} + \frac{9}{\sqrt{4x + 1}} and its minimum point M:

  1. Find derivatives and integrals.
  2. Determine coordinates of M.
  3. Calculate the area of the region bounded by the curve, the y-axis, and the line through M parallel to the x-axis.

Solution:

  • Differentiation and Integration: Apply the Chain Rule and reverse chain rule.
  • Finding M's Coordinates:

Set dydx=0\frac{\mathrm{dy}}{\mathrm{dx}} = 0, solve for x, then find y.

  • Area Calculation: Integrate between the curve and the line y=6y = 6 from x=0x = 0 to x=2x = 2.

Practice Questions

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Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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