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Lesson: Estimating Integrals Using the Trapezoidal Rule

Worksheet • 5 Questions

Q1:

The following table shows how the midpoint and trapezoidal rule perform on estimating . The error is the difference from the actual value of the integral .

Intervals Mid Trap
2 0.790588 0.775
8 0.7857236 0.7847471
32 0.785418 0.785357
128 0.785399 0.785396 216.3397448
512 0.785398 16.33974482 0.785398 16.33974482

What appears to be true of the ratio of successive errors for the midpoint rule?

  • A The error is 16 times as much for intervals as for intervals.
  • B The error is 4 times as much for intervals as for intervals.
  • C The error is 16 times as much for intervals as for intervals.
  • D The error is 4 times as much for intervals as for intervals.
  • E The error is 16 times as much for intervals as for intervals.

What appears to be true of the ratio of successive errors for the trapezoidal rule?

  • AThe error is 16 times as much for intervals as for intervals.
  • BThe error is 4 times as much for intervals as for intervals.
  • CThe error is 16 times as much for intervals as for intervals.
  • DThe error is 4 times as much for intervals as for intervals.
  • EThe error is 16 times as much for intervals as for intervals.

The midpoint rule underestimates the integral and the trapezoidal rule overestimates it. What geometric property of the graph of explains this?

  • AThe graph is concave down.
  • B The graph is concave up.

For a fixed number of intervals , what appears to be the relation between the midpoint rule and the trapezoidal rule errors?

  • AThe trapezoidal error is times the midpoint error.
  • BThe trapezoidal error is times the midpoint error.
  • CThe trapezoidal error is times the midpoint error.
  • DThe trapezoidal error is times the midpoint error.
  • EThe trapezoidal error is 2 times the midpoint error.

Simpson’s rule can be expressed as the weighted average . Using the table above with , we get a Simpson’s error of . Using technology, find the actual error to 3 decimal places.

  • A
  • B0.59
  • C0.059
  • D
  • E

Q2:

For a fixed function on a given interval, let T r a p ( 𝑛 ) be the estimated integral using the trapezoidal rule with 𝑛 subintervals. Use the following diagram to relate the concavity of 𝑓 with how well T r a p ( 𝑛 ) estimates ο„Έ 𝑓 ( π‘₯ ) π‘₯ 𝑏 π‘Ž d .

  • A T r a p d ( 𝑛 ) ≀ ο„Έ 𝑓 ( π‘₯ ) π‘₯ 𝑏 π‘Ž if 𝑓 is concave down on [ π‘Ž , 𝑏 ] .
  • B T r a p d ( 𝑛 ) β‰₯ ο„Έ 𝑓 ( π‘₯ ) π‘₯ 𝑏 π‘Ž if 𝑓 is concave down on [ π‘Ž , 𝑏 ] .
  • CThere is no conclusion about the comparison between the estimate and actual values.
  • D T r a p d ( 𝑛 ) = ο„Έ 𝑓 ( π‘₯ ) π‘₯ 𝑏 π‘Ž if 𝑓 is concave down on [ π‘Ž , 𝑏 ] .

Q3:

Consider a function whose graph 𝑦 = 𝑓 ( π‘₯ ) is concave up on the interval 𝐹 𝐸 .

The line 𝐴 𝐡 that determines points 𝐴 and 𝐡 is the tangent line to 𝑦 = 𝑓 ( π‘₯ ) over the midpoint 𝑀 of segment 𝐹 𝐸 .

Which quadrilateral has the area given by T r a p ( 1 ) , the trapezoidal rule estimate of the integral ο„Έ 𝑓 ( π‘₯ ) π‘₯  οŒ₯ d ?

  • A 𝐢 𝐷 𝐸 𝐹
  • B 𝐢 𝐹 𝐸 𝐡 βˆ—
  • C 𝐴 𝐡 𝐸 𝐹 βˆ— βˆ—
  • D 𝐢 𝐹 𝐸 𝐡
  • E 𝐴 𝐡 𝐸 𝐹

Which quadrilateral has the area given by M i d ( 1 ) , the midpoint rule estimate of the integral ο„Έ 𝑓 ( π‘₯ ) π‘₯  οŒ₯ d ?

  • A 𝐴 𝐡 𝐸 𝐹 βˆ— βˆ—
  • B 𝐢 𝐹 𝐸 𝐡 βˆ—
  • C 𝐢 𝐷 𝐸 𝐹
  • D 𝐢 𝐹 𝐸 𝐡
  • E 𝐴 𝐡 𝐸 𝐹

Why is Area ( 𝐴 𝐡 𝐸 𝐹 ) = βˆ— βˆ— Area ( 𝐴 𝐡 𝐸 𝐹 ) ?

  • Abecause the triangles 𝑋 𝐴 𝐴 βˆ— and 𝑋 𝐡 𝐡 βˆ— are congruent
  • Bbecause 𝑋 𝐴 𝐴 βˆ— and 𝑋 𝐷 𝐡 βˆ— are equal
  • Cbecause 𝐴 𝐡 = 𝐴 𝐡 βˆ— βˆ—
  • Dbecause the quadrilaterals have the common base 𝐸 𝐹

What relationship can you deduce about the numbers M i d ( 1 ) , T r a p ( 1 ) , and ο„Έ 𝑓 ( π‘₯ ) π‘₯  οŒ₯ d in the case where the graph is concave up?

  • A M i d d T r a p ( 1 ) < ο„Έ 𝑓 ( π‘₯ ) π‘₯ < ( 1 )  οŒ₯
  • B M i d d T r a p ( 1 ) > ο„Έ 𝑓 ( π‘₯ ) π‘₯ > ( 1 )  οŒ₯
  • C M i d d T r a p ( 1 ) > ο„Έ 𝑓 ( π‘₯ ) π‘₯ < ( 1 )  οŒ₯
  • D M i d d T r a p ( 1 ) < ο„Έ 𝑓 ( π‘₯ ) π‘₯ > ( 1 )  οŒ₯
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