Section 13.2

Approximate Computation of Integrals

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Problem 5223

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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Problem 5224

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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Problem 5225

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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Problem 5226

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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Problem 5227

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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Problem 5228

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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Problem 5229

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

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Problem 5230

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

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Problem 5231

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

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Problem 5232

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

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Problem 5233

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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Problem 5234

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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Problem 5235

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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Problem 5236

Using Simpson's formula and the parametric equation of the ellipse, obtain the approximate formula for the length of the arc of a quarter of an ellipse with semi-axes aa and bb:

l=π12(a+b+8(a2+b2)). l = \frac{\pi }{12}\left(a+b+\sqrt{8(a^{2}+b^{2})}\right).
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Problem 5237

Investigate the degree of precision of the preceding formula (problem 5236) for b=0b=0, for b=ab=a, and for b=a2b = \dfrac {a}{2} (in the last case, compare with the result of Simpson's formula for n=4n=4).

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Problem 5238

Prove the following Gauss-type quadrature formula:

∫−11f(x) dx1−x2=πn∑ν=1nf(cos⁡(2ν−1)π2n). \int _{-1}^{1} \frac{f(x)\, dx}{\sqrt{1-x^{2}}} = \frac{\pi }{n} \sum _{\nu =1}^{n} f\left(\cos \frac{(2\nu -1)\pi }{2n}\right).
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Problem 5239

Prove the following Gauss-type quadrature formula:

∫−111−x2 f(x) dx=πn∑ν=1nsin⁡2νπn f(cos⁡νπn). \int _{-1}^{1} \sqrt{1-x^{2}}\, f(x)\, dx = \frac{\pi }{n} \sum _{\nu =1}^{n} \sin ^{2}\frac{\nu \pi }{n}\, f\left(\cos \frac{\nu \pi }{n}\right).
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Problem 5240

Prove the following Gauss-type quadrature formula (setting m=2n+1m=2n+1):

∫−111−x1+x f(x) dx=4πm∑ν=1nsin⁡2νπm f(cos⁡2πνm). \int _{-1}^{1} \sqrt{\frac{1-x}{1+x}}\, f(x)\, dx = \frac{4\pi }{m}\sum _{\nu =1}^{n}\sin ^{2}\frac{\nu \pi }{m}\, f\left(\cos \frac{2\pi \nu }{m}\right).
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Problem 5241

Prove the following Gauss-type quadrature formula:

∫01f(x) dxx(1−x)=πn∑ν=1nf(cos⁡2(2ν−1)π4n). \int _{0}^{1} \frac{f(x)\, dx}{\sqrt{x(1-x)}} = \frac{\pi }{n}\sum _{\nu =1}^{n} f\left(\cos ^{2}\frac{(2\nu -1)\pi }{4n}\right).
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Problem 5242

Prove the following Gauss-type quadrature formula:

∫01x(1−x) f(x) dx=π4n∑ν=1n−1sin⁡2νπn f(cos⁡2νπn). \int _{0}^{1} \sqrt{x(1-x)}\, f(x)\, dx = \frac{\pi }{4n}\sum _{\nu =1}^{n-1}\sin ^{2}\frac{\nu \pi }{n}\, f\left(\cos ^{2}\frac{\nu \pi }{n}\right).
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Problem 5243

Prove the following Gauss-type quadrature formula (setting m=2n+1m=2n+1):

∫011−xx f(x) dx=2πm∑ν=1nsin⁡2νπm f(cos⁡2νπm). \int _{0}^{1} \sqrt{\frac{1-x}{x}}\, f(x)\, dx = \frac{2\pi }{m}\sum _{\nu =1}^{n}\sin ^{2}\frac{\nu \pi }{m}\, f\left(\cos ^{2}\frac{\nu \pi }{m}\right).
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Remark.
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Formulas 5239 and 5241 are completely exact if f(x)f(x) is a polynomial of degree less than 2n−22n-2. The remaining formulas of this section (5238, 5240, 5242, 5243) are exact if f(x)f(x) is a polynomial of degree less than 2n2n.

Problem 5244

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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Problem 5245

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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Problem 5246

Prove the exactness of the formula

∫0∞e−xf(x) dx=2−24f(2+2)+2+24f(2−2), \int _{0}^{\infty } e^{-x} f(x)\, dx = \frac{2-\sqrt{2}}{4}f\left(2+\sqrt{2}\right) + \frac{2+\sqrt{2}}{4}f\left(2-\sqrt{2}\right),

where f(x)f(x) is a polynomial of degree less than four.

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Problem 5247

Prove the exactness of the formula

∫−∞∞e−x2f(x) dx=π6(f(−32)+4f(0)+f(32)), \int _{-\infty }^{\infty } e^{-x^{2}} f(x)\, dx = \frac{\sqrt{\pi }}{6}\left(f\left(-\sqrt{\frac{3}{2}}\right) + 4f(0) + f\left(\sqrt{\frac{3}{2}}\right)\right),

where f(x)f(x) is a polynomial of degree less than six.

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Problem 5248

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

Problem 5249

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}.

Problem 5250

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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