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Integration

4.2 Area

Focus on rectangle approximations, area as a limit, geometric versus signed area.

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Ideas in this lesson

01Rectangle approximationsWorked steps · five questions02Area as a limitWorked steps · five questions03Geometric versus signed areaWorked steps · five questions
Idea 1 of 3

Rectangle approximations

Partition the interval, choose sample points, and sum height times width. The choice of sample point changes a finite approximation.

Worked example 1

Approximate ∫02x2 dx\int_{0}^{2} x^{2}\,dx with 33 equal left rectangles.
xx000.66666666670.66666666671.3333333331.333333333
f(x)f(x)000.44444444440.44444444441.7777777781.777777778
  1. Δx=23≈0.67\Delta x = \dfrac{2}{3} \approx 0.67
  2. Sample points: xxᵢ =0,≈0.67,≈1.33= 0, \approx 0.67, \approx 1.33
  3. L3=23[02+≈0.672+≈1.332]=4027≈1.48L_3 = \dfrac{2}{3}[0^{2} + \approx 0.67^{2} + \approx 1.33^{2}] = \dfrac{40}{27} \approx 1.48

Answer: 4027≈1.48\dfrac{40}{27} \approx 1.48

Worked example 2

Approximate ∫03x2 dx\int_{0}^{3} x^{2}\,dx with 44 equal right rectangles.
xx0.750.751.51.52.252.2533
f(x)f(x)0.56250.56252.252.255.06255.062599
  1. Δx=34≈0.75\Delta x = \dfrac{3}{4} \approx 0.75
  2. Sample points: xxᵢ =0.75,1.5,2.25,3= 0.75, 1.5, 2.25, 3
  3. R4=34[0.752+1.52+2.252+32]=40532≈12.66R_4 = \dfrac{3}{4}[0.75^{2} + 1.5^{2} + 2.25^{2} + 3^{2}] = \dfrac{405}{32} \approx 12.66

Answer: 40532≈12.66\dfrac{405}{32} \approx 12.66

Idea 2 of 3

Area as a limit

Refining rectangles leads to an integral when the function is integrable. Simple geometry can verify some exact results.

Worked example 1

Estimate the area below f(x)=xf(x) = x on [0,2][0, 2] using one right-endpoint rectangle.
Coordinate graph0.511.520.511.52xy
  1. Interval width =2= 2
  2. Ar=2⋅2=4A_{r} = 2 \cdot 2 = 4

Answer: 44

Worked example 2

Estimate the area below f(x)=xf(x) = x on [0,3][0, 3] using two midpoint rectangles.
Coordinate graph0.511.522.530.511.522.53xy
  1. Interval width =3= 3
  2. AAₘ =32(0.75+2.25)=92≈4.5= \dfrac{3}{2}(0.75 + 2.25) = \dfrac{9}{2} \approx 4.5

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

Idea 3 of 3

Geometric versus signed area

An integral counts area below the axis negatively. Total geometric area requires absolute values or splitting at zeros.

Worked example 1

For f(x)=xf(x) = x on [−2,2][-2, 2], find the signed integral.
  1. ∫−20x dx=−42\int_{-2}^{0} x\,dx = -\dfrac{4}{2}; ∫02x dx=42\int_{0}^{2} x\,dx = \dfrac{4}{2}
  2. Signed area =−42+42=0= -\dfrac{4}{2} + \dfrac{4}{2} = 0

Answer: 00

Worked example 2

For f(x)=xf(x) = x on [−3,3][-3, 3], find the total geometric area.
  1. ∫−30x dx=−92\int_{-3}^{0} x\,dx = -\dfrac{9}{2}; ∫03x dx=92\int_{0}^{3} x\,dx = \dfrac{9}{2}
  2. Geometric area =92+92=9= \dfrac{9}{2} + \dfrac{9}{2} = 9

Answer: 99

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §4.2,p4.2, p. 254254. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.