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### Understanding Area Under Curves Through Integration Techniques ###

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asher
45 diapositives · 589 Views · Sep 15, 2014

This document explores the concept of integration as a method for calculating the area under curves in distance-time and velocity-time graphs. It demonstrates the use of definite integrals to find areas beneath curves, using specific functions such as (y = x^2 + 4) and (y = 2x + 4), and discusses the importance of determining areas that may be negative due to the position of the curve relative to the x-axis. Various models, approximations (like the Trapezoidal Rule), and examples are provided, explaining integration's applications in real-world scenarios, such as calculating volume and areas of shapes. ###

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  1. Topic 17 Integration II

  2. Consider the following distance/time information and the corresponding velocity/time graph

  3. Area = 100 Area = the integral of y = 10 from x = 0 to x = 10

  4. Consider the following distance/time information and the corresponding velocity/time graph

  5. Area = 100 Area = the integral of y = 4 from x = 0 to x = 5 + the integral of y = 16 from x = 5 to x = 10

  6. Consider the following distance/time information and the corresponding velocity/time graph s = t2 v = ds/dt = 2t

  7. If we wanted to find the area under the curve y = x2 + 4 from x = 1 to x = 4

  8. If we wanted to find the area under the curve y = x2 + 4 from x = 1 to x = 4 1.5 1 2 2.5 1 2 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 The area is approximately 32.5 square units

  9. The exact value can be found by calculating the (definite) integral between the given values. The integral is actually the limit of the sum of the many rectangles as the width of each rectangle approaches 0. To find the area under y = x2 + 4 from x = 1 to x = 4

  10. The exact value can be found by calculating the (definite) integral between the given values. The integral is actually the limit of the sum of the many rectangles as the width of each rectangle approaches 0. To find the area under y = x2 + 4 from x = 1 to x = 4

  11. Model (3x-7)5 dx

  12. Integration Rules

  13. NewQ P127 Ex 4.2 1-12

  14. FM Page 608 Set 26.1 NEWQ P141 Ex 4.4 35 – 60

  15. To find the area under y = x3 from x = -2 to x = 2 ????? What the …..

  16. Clearly, there is some area under the curve from x = 0 to x = 2and there is an identical amountunder the curve from x = -2 to x = 0.

  17. Let’s calculate the area under y = x3 from x = 0 to x = 2 Hmmm

  18. N.B. Anytime that the area is below the x-axis, the calculated area will be negative. We then take the absolute value (the positive equivalent) of this calculated amount Let’s calculate the area under y = x3 from x = -2 to x = 0 The total area here is 4 + 4 =8

  19. FM Page 617 Set 26.3 NewQ P147 Ex 4.5 2a, 3, 6, 9, 11a, 13c, 20, 23, 24, 26

  20. Find the area enclosed by the curves y = x2+4 and y = 2x+4 First we need to sketch these curves …. Let’s go

  21. y = x2+4 ….(1)y = 2x+4 …(2)Equating (1) and (2) x2+4 = 2x+4x2 = 2xx2 – 2x = 0x(x-2) =0x = 0 or x = 2 We need to find where these curves intersect Or use your TI calculator

  22. Area = 2x+4 dx - x2+4 dx from x = 0 to x = 2 = 2x+4 – (x2+4) dx = 2x+4-x2-4 dx … 0n board

  23. FM Page 617 Set 26.3 NewQ P153 Ex 4.6 1

  24. Approximations for Area Trapezoidal Rule

  25. Area = ½ (8+7.3) x 1 + ½ (7.3+5.6) x 1 + ½ (5.6+3.5) x 1 + ½ (3.5+1.6) x 1 + ½ (1.6+0.5) x 1 + ½ (0.5+0.8 ) x 1 + ½ (0.8+3.1) x 1 = 24.85 Now the equation of this curve is y = .1x^3 - .8x^2 + 8 and the true area is 24.5583333…

  26. If we were to use more strips, we would achieve an even more accurate result This approximation can be written as A ≈ ½ w[E + 2M] Where: w = width of the strips E = sum of the length of 1st and last strips M = sum of the length of the middle strips

  27. Model 1: Find the approximate area of this shape which is divided into 5m wide strips. w = 5m E = 0.2 + 2.1 = 2.3m M = 10.1 + 12.2 + 8.6 + 5.0 + 7.9 + 12.4 = 56.2m A ≈½ w[E + 2M] = 0.5  5  (2.3 + 2  56.2) = 286.75m2

  28. NEWQ P 243 Ex 7.2 All

  29. Approximations for Volume Trapezoidal Rule

  30. Model 2: Find the amount of soil removed from the cutting shown on page 246 if the cross-sectional areas are 5m apart and have areas of 2.5m2, 8.4m2, 9.5m2 and 2.4m2 respectively. w = 5m E = 0 + 0 = 0m2 M = 2.5 + 8.4 + 9.5 + 2.4 m2 = 22.8m2 V ≈½ w[E + 2M] = 0.5  5  (0 + 2  22.8) = 114m3

  31. Model 3: Consider the mound below which is in the shape of a cone with the top cut off. The base of the mound has a diameter of 9m, the top 6m and the mound is 2.5m high. Use the trapezoidal rule with 5 sections to find an approximation for the volume of this mound. Step 2: Find how much the radius decreases by at each section. Step 1: The width of each section will be… w = 2.5  5 = 0.5m

  32. Step 2: Find how much the radius decreases by at each section. Step 1: The width of each section will be… w = 2.5  5 = 0.5m Step 3: Use the lists (stat→edit) on your TI as follows. W = 0.5m E = L2(1) + L2(6) = 91.8m → A M = sum(L2) – A = 178.1 V ≈½ w[E + 2M] = 0.5  0.5  (91.8 + 2  178.1) = 112m3

  33. Year 12Maths B End Semester IV Revision

  34. Find 9. 10. 11. A particle undergoing straight line motion has velocity v = m/s at time t sec, t0 If the position at time t=0 is s=0, find its position after 5 seconds.

  35. 12. A rectangular garden plot of 48m2 is to be fenced off in a backyard. Three sides will be fenced with strong wire mesh costing $18 per metre and the remaining side will be fenced with corrugated iron at $30 per metre. Find the dimensions of the garden plot which will minimise the cost of fencing. 13. If a TV rental company charges $25 per month it finds that it can hire out 1000 TV’s. It finds that for every $1/month decrease in price, an extra 80 TV’s are hired out. Costs associated with the hiring each month are $1 per TV and fixed costs of $1000. If the company charges a whole number of dollars for the hire, find the hiring fee which yields maximum income.