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Applications of Integration

7.1 Area of a Region Between Two Curves

Focus on upper minus lower, curves that exchange position, average vertical separation.

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Ideas in this lesson

01Upper minus lowerWorked steps · five questions02Curves that exchange positionWorked steps · five questions03Average vertical separationWorked steps · five questions
Idea 1 of 3

Upper minus lower

Find intersections, then integrate the upper curve minus the lower curve on each interval.

Worked example 1

Find the area between y=2xy = 2x and y=x2y = x^{2} on [0,2][0, 2].
Coordinate graph0.511.520.511.522.533.54xy
  1. 2x=x2  ⟹  x=0,22x = x^{2} \implies x = 0, 2; upper −- lower =2x−x2= 2x - x^{2}
  2. Area =∫022x−x2 dx=[2x22−x33]∣02=43≈1.33= \int_{0}^{2} 2x - x^{2}\,dx = [\dfrac{2x^{2}}{2} - \dfrac{x^{3}}{3}]\big|_{0}^{2} = \dfrac{4}{3} \approx 1.33
  3. Area =43≈1.33= \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Worked example 2

Find the area between y=3xy = 3x and y=x2y = x^{2} on [0,3][0, 3].
Coordinate graph0.511.522.532468xy
  1. 3x=x2  ⟹  x=0,33x = x^{2} \implies x = 0, 3; upper −- lower =3x−x2= 3x - x^{2}
  2. Area =∫033x−x2 dx=[3x22−x33]∣03=92≈4.5= \int_{0}^{3} 3x - x^{2}\,dx = [\dfrac{3x^{2}}{2} - \dfrac{x^{3}}{3}]\big|_{0}^{3} = \dfrac{9}{2} \approx 4.5
  3. Area =92≈4.5= \dfrac{9}{2} \approx 4.5

Answer: 92≈4.5\dfrac{9}{2} \approx 4.5

Idea 2 of 3

Curves that exchange position

Split at crossings so every area contribution is nonnegative. A signed integral alone may cancel areas.

Worked example 1

Find the area between y=xy = x and the xx-axis on [−2,2][-2, 2].
  1. x=0x = 0 is the crossing; ∣x∣=−x|x| = -x for x<0,xx < 0, x for x≥0x \ge 0
  2. A=∫−20−x dx+∫02x dx=42+42=4A = \int_{-2}^{0} -x\,dx + \int_{0}^{2} x\,dx = \dfrac{4}{2} + \dfrac{4}{2} = 4

Answer: 44

Worked example 2

Find the area between y=xy = x and the xx-axis on [−3,3][-3, 3].
  1. x=0x = 0 is the crossing; ∣x∣=−x|x| = -x for x<0,xx < 0, x for x≥0x \ge 0
  2. A=∫−30−x dx+∫03x dx=92+92=9A = \int_{-3}^{0} -x\,dx + \int_{0}^{3} x\,dx = \dfrac{9}{2} + \dfrac{9}{2} = 9

Answer: 99

Idea 3 of 3

Average vertical separation

Area divided by interval width gives the average distance between an upper and lower graph.

Worked example 1

Find the area between y=2xy = 2x and y=x2y = x^{2} on [0,2][0, 2].
Coordinate graph0.511.520.511.522.533.54xy
  1. 2x=x2  ⟹  x=0,22x = x^{2} \implies x = 0, 2; upper −- lower =2x−x2= 2x - x^{2}
  2. Area =∫022x−x2 dx=[2x22−x33]∣02=43≈1.33= \int_{0}^{2} 2x - x^{2}\,dx = [\dfrac{2x^{2}}{2} - \dfrac{x^{3}}{3}]\big|_{0}^{2} = \dfrac{4}{3} \approx 1.33
  3. Area =43≈1.33= \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Worked example 2

Find the area between y=3xy = 3x and y=x2y = x^{2} on [0,3][0, 3].
Coordinate graph0.511.522.532468xy
  1. 3x=x2  ⟹  x=0,33x = x^{2} \implies x = 0, 3; upper −- lower =3x−x2= 3x - x^{2}
  2. Area =∫033x−x2 dx=[3x22−x33]∣03=92≈4.5= \int_{0}^{3} 3x - x^{2}\,dx = [\dfrac{3x^{2}}{2} - \dfrac{x^{3}}{3}]\big|_{0}^{3} = \dfrac{9}{2} \approx 4.5
  3. Area =92≈4.5= \dfrac{9}{2} \approx 4.5

Answer: 92≈4.5\dfrac{9}{2} \approx 4.5

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.1,p7.1, p. 436436. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.