En anglais

Area under the curve, panel by panel

Mathematics · Calculus & Analysis · ages 20-21

Name ______________________ Date ____________

Composite Simpson uses six equal subintervals with nodes x_0 through x_6. In the weighted sum multiplied by h/3, what coefficient multiplies f(x_3)?
Answer: ______________
On a trapezium panel of width W, use t=(x-a)/W. What coefficient of f(a) results from integrating its basis function 1-t on [0,1] and multiplying by W?
- W/2
- W
- W/3
- 2W/3
Use one trapezium panel to estimate the integral of f(x)=x² on [0,2]. What value do you get?
- 2
- 4
- 8
An adaptive quadrature rule must refine every subinterval by the same amount.
Circle one: True False
For f with a continuous fourth derivative on [0,2], suppose |f''''(x)|≤180. Composite Simpson uses four equal subintervals, so h=0.5. What absolute-error bound follows?
Answer: ______________
A Simpson panel has width 3. Let t=(x-a)/3 and p(t)=(2t²-3t+1)A+4t(1-t)B+(2t²-t)C, where A, B and C are arbitrary left, midpoint and right sample values. Integrate the basis functions. Which expression equals 3∫_0^1 p(t)dt?
- 3[A/3+B/3+C/3] = A+B+C
- 3[A/6+4B/3+C/6] = A/2+4B+C/2
- 3[A/6+2B/3+C/6] = A/2+2B+C/2
For f with a continuous second derivative on [0,3], suppose |f''(x)|≤6. With h=0.5, what absolute-error bound does the composite trapezium formula give?
- 0.75
- 0.375
- 0.1875
- 1.5
A recursive adaptive scheme gives each child half its parent's tolerance. If both nonnegative child error estimates meet their tolerances, their sum is at most the parent tolerance.
Circle one: True False
Apply the composite trapezium rule with two equal subintervals to f(x)=x² on [0,2]. What value do you get?
Answer: ______________
An adaptive calculation reaches its evaluation limit. Its error estimate is still above the requested tolerance, although recent floating-point results are identical. Which report is justified?
- Report tolerance met because repeated estimates now agree.
- Report an exact integral because the displayed value is stable.
- Report tolerance unmet; stop at the stated limit.
- Report tolerance met after rounding the error estimate down.

Answer key

For grown-ups. Fold this page away before handing over the rest.

Area under the curve, panel by panel W1-mt_fT5kRCUiD5-s1

4 · Composite Simpson gives coefficient 4 to odd-index samples. Since 3 is odd, f(x_3) has coefficient 4.
W/2 · The integral of 1-t on [0,1] is [t-t²/2]_0^1=1/2. Multiplying by W gives coefficient W/2.
4 · The width is 2 and the endpoint values are 0 and 4. The estimate is (2/2)(0+4)=4.
False · False. Adaptive refinement uses local error estimates to select where further work is needed; equal refinement is not required.
0.125 · The bound is (b-a)M_4 h⁴/180=2×180×0.5⁴/180=0.125.
3[A/6+2B/3+C/6] = A/2+2B+C/2 · The three basis integrals are 1/6, 2/3 and 1/6. Since dx=3 dt, the panel integral is 3[A/6+2B/3+C/6]=A/2+2B+C/2.
0.375 · Use (b-a)M_2 h²/12. Substitution gives 3×6×0.25/12=0.375.
True · True. If each child estimate is at most τ/2, their sum is at most τ/2+τ/2=τ. This concerns estimated error, not a proven true-error bound.
3 · Here h=1 and the values are 0, 1 and 4. The rule gives (1/2)[0+2×1+4]=3.
Report tolerance unmet; stop at the stated limit. · The error test still fails and the evaluation limit has been reached. Identical floating-point results do not establish accuracy, so the unmet tolerance must be reported.