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Area under Curves

Area under Curves. 5.2. velocity. time. Consider an object moving at a constant rate of 3 ft/sec. Since rate . time = distance:. If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line. After 4 seconds, the object has gone 12 feet.

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Area under Curves

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  1. Area under Curves 5.2

  2. velocity time Consider an object moving at a constant rate of 3 ft/sec. Since rate . time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line. After 4 seconds, the object has gone 12 feet.

  3. Approximate area: If the velocity is not constant, we might guess that the distance traveled is still equal to the area under the curve. (The units work out.) Example: We could estimate the area under the curve by drawing rectangles touching at their left corners. This is called the Left-hand Rectangular Approximation Method (LRAM).

  4. Approximate area: We could also use a Right-hand Rectangular Approximation Method (RRAM).

  5. Approximate area: Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM). In this example there are four subintervals. As the number of subintervals increases, so does the accuracy.

  6. Approximate area: The exact answer for this problem is . With 8 subintervals: width of subinterval

  7. Circumscribed rectangles are all above the curve: Inscribed rectangles are all below the curve:

  8. We will be learning how to find the exact area under a curve if we have the equation for the curve. Rectangular approximation methods are still useful for finding the area under a curve if we do not have the equation. Area = ((b-a)/2)[f(x0)+f(x1)+…+f(xn)]

  9. Trapezoidal Rule

  10. What if: We could split the area under the curve into a lot of thin trapezoids. It seems reasonable that the distance will equal the area under the curve. Area = .5(b1+b2)h

  11. Example • Find the area under x3 using 4 subintervals using: left, right, midpoint and trapezoidal methods from [2, 3] 2 3

  12. Example • Find the area under x3 using 4 subintervals using: left, right, midpoint and trapezoidal methods from [2, 3] 2 3

  13. Example • Find the area under x3 using 4 subintervals using: left, right, midpoint and trapezoidal methods from [2, 3] 2 3

  14. Example • Find the area under x3 using 4 subintervals using: left, right, midpoint and trapezoidal methods from [2, 3] 2 3 Actual Area = 16.25

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