Dabour Hub EGEgyptian Curriculum 2026/2027
3-4 (a) Area of a plane region

3-4 (a) Area of a plane region

Point 10 of 12 — Lesson 1: Integration Techniques (Substitution and By Parts) and Definite IntegralsQuestion bank: 2 questions (2 with pictures)

10 3-4 (a) Area of a plane region

Rule

  • If f is continuous on [a,b] and f(x)≥0, the area bounded by the curve y=f(x), the x-axis and the lines x=a, x=b is A=∫abf(x) dx
  • In general the area is A=∫ab∣f(x)∣ dx.
  • The area is measured in square units, and it is never negative.

Method when the curve crosses the x-axis

  • Find the zeros of f by putting f(x)=0.
  • Study the sign of f on each sub-interval between the zeros and the limits.
  • Integrate on each sub-interval separately, taking the absolute value of each integral where f<0.
  • Add the areas.
✍️ Worked examples

Example 19: Find the area bounded by f(x)=1+4x−x2, the x-axis and the lines x=1 and x=4.

Solution:

  • f(x)>0 on [1,4] (its zeros are 2±5, i.e. about −0.24 and 4.24, both outside [1,4]).
  • A=∫14(1+4x−x2) dx=[x+2x2−x33]14.
  • =(4+32−643)−(1+2−13).
  • =443−83=12 square units.

Application: cost of covering an arch

✍️ Worked examples

Example 23: A hotel entrance arch has equation y=−12(x−1)(x−7) with x in metres. Find the cost of covering it with glass at 1500 LE per square metre.

Solution:

  • Cost = area × 1500.
  • The zeros are x=1 and x=7, and y>0 between them.
  • A=∫17−12(x2−8x+7) dx=−12[x33−4x2+7x]17.
  • […]17=(3433−196+49)−(13−4+7)=−983−103=−36.
  • A=−12(−36)=18 square metres.
  • Cost =18×1500=27 000 LE.

📝 Question bank for this point

2 questions linked to this point only, 2 of them with pictures.