Lesson Worksheet: Numerical Integration: The Trapezoidal Rule Mathematics • Higher Education

In this worksheet, we will practice approximating definite integrals using the trapezoidal rule and estimate the error when using it.

Q1:

The following table shows how the midpoint and trapezoidal rule perform on estimating ๏„ธ11+๐‘ฅ๐‘ฅ๏Šง๏Šฆ๏Šจd. The error is the difference from the actual value of the integral ๐œ‹4.

IntervalsMidErrร—10๏ŠฎTrapErrร—10๏Šฎ
20.790588โˆ’518,983.66030.7751,039,816.34
80.7857236โˆ’32,543.660260.784747165,106.33974
320.785418โˆ’1,983.6602550.7853574,116.339745
1280.785399โˆ’83.660255170.785396216.3397448
5120.78539816.339744820.78539816.33974482

What appears to be true of the ratio of successive errors ErrErr(๐‘›)(4๐‘›) for the midpoint rule?

  • AThe error is 4 times as much for 4๐‘› intervals as for ๐‘› intervals.
  • BThe error is 16 times as much for ๐‘› intervals as for 4๐‘› intervals.
  • CThe error is 4 times as much for ๐‘› intervals as for 4๐‘› intervals.
  • DThe error is 16 times as much for 4๐‘› intervals as for ๐‘› intervals.
  • EThe error is 16 times as much for 4๐‘› intervals as for 4๐‘› intervals.

What appears to be true of the ratio of successive errors ErrErr(๐‘›)(4๐‘›) for the trapezoidal rule?

  • AThe error is 16 times as much for 4๐‘› intervals as for ๐‘› intervals.
  • BThe error is 4 times as much for 4๐‘› intervals as for ๐‘› intervals.
  • CThe error is 16 times as much for ๐‘› intervals as for 4๐‘› intervals.
  • DThe error is 16 times as much for 4๐‘› intervals as for 4๐‘› intervals.
  • EThe error is 4 times as much for ๐‘› intervals as for 4๐‘› intervals.

The midpoint rule underestimates the integral and the trapezoidal rule overestimates it. What geometric property of the graph of ๐‘“(๐‘ฅ)=11+๐‘ฅ๏Šช explains this?

  • AThe graph is concave up.
  • BThe graph is concave down.

For a fixed number of intervals ๐‘›, what appears to be the relation between the midpoint rule and the trapezoidal rule errors?

  • AThe trapezoidal error is 12times the midpoint error.
  • BThe trapezoidal error is โˆ’2 times the midpoint error.
  • CThe trapezoidal error is โˆ’4 times the midpoint error.
  • DThe trapezoidal error is โˆ’12 times the midpoint error.
  • EThe trapezoidal error is 2 times the midpoint error.

Simpsonโ€™s rule can be expressed as the weighted average 2+3MidTrap. Using the table above with ๐‘›=8, we get a Simpsonโ€™s error of 0.03ร—10๏Šฑ๏Šฎ. Using technology, find the actual error to 3 decimal places.

  • Aร—10๏Šฑ๏Šฎ
  • B0.059
  • C0.59ร—10๏Šฑ๏Šฎ
  • D0.059ร—10๏Šฑ๏Šฎ
  • E0.59

Q2:

For a fixed function on a given interval, let Trap(๐‘›) be the estimated integral using the trapezoidal rule with ๐‘› subintervals. Use the following diagram to relate the concavity of ๐‘“ with how well Trap(๐‘›) estimates ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบd.

  • ATrapd(๐‘›)โ‰ค๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบ if ๐‘“ is concave down on [๐‘Ž,๐‘].
  • BThere is no conclusion about the comparison between the estimate and actual values.
  • CTrapd(๐‘›)โ‰ฅ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบ if ๐‘“ is concave down on [๐‘Ž,๐‘].
  • DTrapd(๐‘›)=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบ if ๐‘“ is concave down on [๐‘Ž,๐‘].

Q3:

Consider a function whose graph ๐‘ฆ=๐‘“(๐‘ฅ) is concave up on the interval ๐น๐ธ.

The line ๐ด๐ต that determines points ๐ด and ๐ต is the tangent line to ๐‘ฆ=๐‘“(๐‘ฅ) over the midpoint ๐‘€ of segment ๐น๐ธ.

Which quadrilateral has the area given by Trap(1), the trapezoidal rule estimate of the integral ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œค๏Œฅd?

  • A๐ถ๐น๐ธ๐ต
  • B๐ด๐ต๐ธ๐นโˆ—โˆ—
  • C๐ถ๐น๐ธ๐ตโˆ—
  • D๐ถ๐ท๐ธ๐น
  • E๐ด๐ต๐ธ๐น

Which quadrilateral has the area given by Mid(1), the midpoint rule estimate of the integral ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œค๏Œฅd?

  • A๐ถ๐ท๐ธ๐น
  • B๐ถ๐น๐ธ๐ต
  • C๐ถ๐น๐ธ๐ตโˆ—
  • D๐ด๐ต๐ธ๐น
  • E๐ด๐ต๐ธ๐นโˆ—โˆ—

Why is Area(๐ด๐ต๐ธ๐น)=โˆ—โˆ— Area(๐ด๐ต๐ธ๐น)?

  • Abecause the quadrilaterals have the common base ๐ธ๐น
  • Bbecause ๐‘‹๐ด๐ดโˆ— and ๐‘‹๐ท๐ตโˆ— are equal
  • Cbecause ๐ด๐ต=๐ด๐ตโˆ—โˆ—
  • Dbecause the triangles ๐‘‹๐ด๐ดโˆ— and ๐‘‹๐ต๐ตโˆ— are congruent

What relationship can you deduce about the numbers Mid(1), Trap(1), and ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œค๏Œฅd in the case where the graph is concave up?

  • AMiddTrap(1)>๏„ธ๐‘“(๐‘ฅ)๐‘ฅ<(1)๏Œค๏Œฅ
  • BMiddTrap(1)<๏„ธ๐‘“(๐‘ฅ)๐‘ฅ>(1)๏Œค๏Œฅ
  • CMiddTrap(1)<๏„ธ๐‘“(๐‘ฅ)๐‘ฅ<(1)๏Œค๏Œฅ
  • DMiddTrap(1)>๏„ธ๐‘“(๐‘ฅ)๐‘ฅ>(1)๏Œค๏Œฅ

Q4:

Calculate the trapezoidal rule estimate of ๏„ธ๐‘ฅ+2๐‘ฅ๏Šช๏Šฆ๏Šจd with ๐‘›=2 subintervals. Is the result an overestimate or underestimate of the actual value?

  • A28, an overestimate
  • B48, an overestimate
  • C16, an underestimate
  • D28, an underestimate
  • E32, an overestimate

Q5:

For a fixed function on a given interval, let Trap(๐‘›) be the estimated integral using the trapezoidal rule with ๐‘› subintervals. Use the following diagram to relate the concavity of ๐‘“ with how well Trap(๐‘›) estimates ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบd.

  • ATrapd(๐‘›)=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบ if ๐‘“ is concave up on [๐‘Ž,๐‘].
  • BThere is no conclusion about the comparison between the estimate and actual values.
  • CTrapd(๐‘›)โ‰ค๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบ if ๐‘“ is concave up on [๐‘Ž,๐‘].
  • DTrapd(๐‘›)โ‰ฅ๏„ธ๐‘“(๐‘ฅ)๐‘ฅ๏Œป๏Œบ if ๐‘“ is concave up on [๐‘Ž,๐‘].

Q6:

Use the trapezoidal rule to estimate ๏„ธโˆš๐‘ฅ+1๐‘ฅ๏Šจ๏Šฆ๏Žขd using five subintervals. Round your answer to three decimal places.

Q7:

Use the trapezoidal rule to estimate ๏„ธ๐‘ฅ๐‘ฅ๏Šจ๏Šฆ๏Šฉd using four subintervals.

  • A258
  • B174
  • C98
  • D172
  • E94

Q8:

Estimate ๏„ธโˆš๐‘ฅ+2๐‘ฅโˆ’4๐‘ฅ๏Šฉ๏Šจ๏Šฉd using the trapezoidal rule with four subintervals. Approximate your answer to two decimal places.

Q9:

Estimate ๏„ธ๐‘’๐‘ฅ๐‘ฅ๏Šจ๏Šง๏—d using the trapezoidal rule with four subintervals. Approximate your answer to two decimal places.

Q10:

Use the trapezoidal rule to estimate ๏„ธโˆš๐‘ฅ+1๐‘ฅ๏Šง๏Šฆd using four subintervals. Round your answer to three decimal places.

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