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Areas Between Two Curves: Integral of Difference Equals Difference of Integrals

Areas Between Two Curves: Integral of Difference Equals Difference of Integrals
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2020-10-19 18:17
FunctionCalculusIntegralSlider AnimationGraphing
①This example dynamically demonstrates a fundamental calculus result: the area between two curves equals the difference of their definite integrals, which is also equal to the definite integral of the difference function. ②Key mathematical concepts include the linearity of definite integrals, function graphs, interval partitioning, Riemann sum approximation, and area computation. ③Construction details: GeoGebra commands such as Intersect find curve intersections, Sequence generates partition points, Textfield and InputBox allow user-defined limits, IntegralBetween computes the area directly, while sliders and buttons animate the rectangular approximation process. ④Suitable for senior high school or early university calculus students to intuitively grasp the equivalence, and useful for classroom demonstrations.

GeoGebra Commands & Parameters

  1. Intersect(f, g) → A, B
  2. Sequence(x(A) + (ii / numRec * (x(B) - x(A))), ii, 0, numRec, 1) → xVals
  3. Sequence(Min[abs(g(xVals(ii)) - f(xVals(ii))), abs(g(xVals(ii + 1)) - f(xVals(ii + 1)))], ii, 1, Length[xVals] - 1) → heights
  4. Sequence(If[abs(g(xVals(ii)) - f(xVals(ii))) < abs(g(xVals(ii + 1)) - f(xVals(ii + 1))), Min[f(xVals(ii)), g(xVals(ii))], Min[f(xVals(ii + 1)), g(xVals(ii + 1))]], ii, 1, numRec) → lower
  5. Sequence(Polygon[(xVals(ii), (lower(ii) * (100 - base) / 100)), (xVals(ii + 1), (lower(ii) * (100 - base) / 100)), (xVals(ii + 1), (lower(ii) * (100 - base) / 100) + heights(ii)), (xVals(ii), (lower(ii) * (100 - base) / 100) + heights(ii))], ii, 1, numRec) → list1
  6. Textfield(f) → InputBox1
  7. Textfield(g) → InputBox2
  8. IntegralBetween(If[f(x) < g(x), (f(x) * (100 - base) / 100), (g(x) * (100 - base) / 100)], If[f(x) < g(x), (f(x) * (100 - base) / 100), (g(x) * (100 - base) / 100)] + abs(h(x)), x(A), x(B)) → d

AI Drawing Prompt Reference

Plot two function curves in the same coordinate system and mark their intersection points. Divide the x-axis interval between the intersection points into equal segments, then draw vertical line segments from each division point to create a series of rectangles approximating the area between the two curves. The height of each rectangle equals the absolute difference of the function values, and the lower boundary takes the smaller of the two function values. Include sliders to adjust the number of divisions and rectangle width parameters, dynamically showing how the area approximation improves as the number of rectangles increases. Also display the definite integral calculation result in the diagram, demonstrating the fundamental calculus theorem that the difference of two integrals equals the integral of their difference.
The prompt above was compiled from the original drawing commands and is for reference only.