Numerical quadrature by standard rules (trapezoidal, Simpson, Cotes, Chebyshev, Gauss)

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Compute by the trapezoidal method the integral ∫01dx1+x\displaystyle \int_{0}^{1} \dfrac {dx}{1+x}, taking the number of subintervals n=4n=4.

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Compute by the trapezoidal method the integral ∫13dxx+1\displaystyle \int_{1}^{3} \dfrac {dx}{x+1}, taking the number of subintervals n=4n=4.

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Compute by the trapezoidal method the integral ∫01dx1+x2\displaystyle \int_{0}^{1} \dfrac {dx}{1+x^{2}}, taking the number of subintervals n=4n=4.

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Compute by the trapezoidal method the integral ∫0π/2sin⁡xx dx\displaystyle \int_{0}^{\pi /2} \dfrac {\sin x}{x}\, dx, taking the number of subintervals n=4n=4.

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Compute by the trapezoidal method the integral ∫01ln⁡(1+x)1+x2 dx\displaystyle \int_{0}^{1} \dfrac {\ln (1+x)}{1+x^{2}}\, dx, taking the number of subintervals n=6n=6.

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Compute by the trapezoidal method the integral ∫01ln⁡(1+x)x dx\displaystyle \int_{0}^{1} \dfrac {\ln (1+x)}{x}\, dx, taking the number of subintervals n=6n=6.

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Compute the same integrals as in problems 5223--5228, for n=4n=4, by Simpson's formula.

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Compute the same integrals as in problems 5223--5228 by Cotes's formula.

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Compute the same integrals as in problems 5223--5228 by Chebyshev's method.

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Compute the same integrals as in problems 5223--5228 by Gauss's method.

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After the substitution x=sin⁡φx = \sin \varphi, compute by Simpson's method with n=4n=4 the integrals:

∫01x2 dx1−x2;∫01dx1−x4. \int _{0}^{1} \frac{x^{2}\, dx}{\sqrt{1-x^{2}}}; \qquad \int _{0}^{1} \frac{dx}{\sqrt{1-x^{4}}}.
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Compute by Simpson's method with n=4n=4 the integrals ∫01x2 dx1−x2\displaystyle \int_{0}^{1} \frac{x^{2}\, dx}{\sqrt{1-x^{2}}} and ∫01dx1−x4\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}} with the substitution 1−x=u21-x=u^{2}.

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After the substitution x=1tx = \dfrac {1}{t}, compute by Simpson's method with n=4n=4 the integral ∫1∞x2 dx1+x4\displaystyle \int_{1}^{\infty } \frac{x^{2}\, dx}{\sqrt{1+x^{4}}}.

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Using formula 5242 (see also the remark to problem 5243) with n=4n=4, compute the integral ∫01x(1−x2) dx\displaystyle \int_{0}^{1} \sqrt{x(1-x^{2})}\, dx.

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Using formula 5240 (see also the remark to problem 5243) with n=4n=4, compute the integral ∫−11dx1−x4 dx\displaystyle \int_{-1}^{1} \frac{dx}{\sqrt{1-x^{4}}}\, dx.

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Apply Gregory's formula to compute the integral ∫01dx1+x2\displaystyle \int_{0}^{1} \frac{dx}{1+x^{2}}, taking n=1n=1, h=0,1h=0{,}1, carrying out the computation with seven decimal digits.

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Remark.
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Gregory's formula (1670): 1h∫aa+nhy dx=12y0+y1+…+12yn+∑νAν(Δνyn−ν+(−1)νΔνy0)\dfrac {1}{h}\displaystyle \int_{a}^{a+nh} y\, dx = \dfrac {1}{2}y_{0} + y_{1} + \ldots + \dfrac {1}{2}y_{n} + \sum_{\nu } A_{\nu }\left(\Delta^{\nu }y_{n-\nu } + (-1)^{\nu }\Delta^{\nu }y_{0}\right), where A1=−112A_{1} = -\dfrac {1}{12}, A2=−124A_{2} = -\dfrac {1}{24}, A3=−19720A_{3} = -\dfrac {19}{720}, A4=−3160A_{4} = -\dfrac {3}{160}, A5=−86360 480,…A_{5} = -\dfrac {863}{60\, 480}, \ldots

The same for Gauss's formula.

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Remark.
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Gauss's formula: 1h∫aa+nhy dx=12y0+y1+…+12yn+∑νBν(Δ2ν+1yn−ν+Δ2ν+1yn−ν−12−Δ2ν+1y−ν+Δ2ν+1y−ν−12)\dfrac {1}{h}\displaystyle \int_{a}^{a+nh} y\, dx = \dfrac {1}{2}y_{0} + y_{1} + \ldots + \dfrac {1}{2}y_{n} + \sum_{\nu } B_{\nu }\left(\dfrac {\Delta^{2\nu +1}y_{n-\nu } + \Delta^{2\nu +1}y_{n-\nu -1}}{2} - \dfrac {\Delta^{2\nu +1}y_{-\nu } + \Delta^{2\nu +1}y_{-\nu -1}}{2}\right), where B0=−112B_{0} = -\dfrac {1}{12}, B1=11720B_{1} = \dfrac {11}{720}, B2=−19160 480B_{2} = -\dfrac {191}{60\, 480}.

Carry out an analogous computation (using the Gregory and Gauss formulas, problems 5248--5249) for the integral ∫100110dxx\displaystyle \int_{100}^{110} \frac{dx}{x}.

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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.