Approximate integration formulas (rectangle, trapezoid, Simpson)

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Evaluate the integral

I=∫01/2sin⁡xx dx I = \int _{0}^{1/2} \frac{\sin x}{x}\, dx

to an accuracy of two decimal places; that is, find a number I∗I^{*} such that ∣I−I∗∣<0.005\left|I-I^{*}\right| < 0.005.

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Suppose the area S=S(x)S=S(x) of the cross-section of a solid perpendicular to the OxOx axis varies according to the following quadratic law:

S(x)=Ax2+Bx+C(a⩽x⩽b), S(x)=Ax^{2}+Bx+C\quad (a\leqslant x\leqslant b) ,

where AA, BB, and CC are constants.

Prove that the volume of this solid equals

V=H6[S(a)+4S(a+b2)+S(b)], V=\dfrac {H}{6}\left[S(a)+4S\left(\dfrac {a+b}{2}\right)+S(b)\right] ,

where H=b−aH=b-a (Simpson's formula).

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Using the rectangle formula (n=12n=12), approximately compute

∫02πxsin⁡xdx \int _{0}^{2\pi }x\sin x dx

and compare the result with the exact value.

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Use the trapezoid formula to compute the following integral and estimate its error:

∫01dx1+x(n=8). \int _{0}^{1}\dfrac {dx}{1+x}\quad (n=8) .
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Use the trapezoid formula to compute the following integral and estimate its error:

∫01dx1+x3(n=12). \int _{0}^{1}\dfrac {dx}{1+x^{3}}\quad (n=12) .
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Use the trapezoid formula to compute the following integral and estimate its error:

∫0π21−14sin⁡2xdx(n=6). \int _{0}^{\frac{\pi }{2}}\sqrt{1-\dfrac {1}{4}\sin ^{2}x} dx\quad (n=6) .
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Use Simpson's formula to compute the following integral:

∫19xdx(n=4). \int _{1}^{9}\sqrt{x} dx\quad (n=4) .
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Use Simpson's formula to compute the following integral:

∫0π3+cos⁡xdx(n=6). \int _{0}^{\pi }\sqrt{3+\cos x} dx\quad (n=6) .
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Use Simpson's rule to compute the following integral:

∫0π2sin⁡xxdx(n=10). \int _{0}^{\frac{\pi }{2}}\dfrac {\sin x}{x} dx\quad (n=10) .
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Use Simpson's rule to compute the following integral:

∫01xdxln⁡(1+x)(n=6). \int _{0}^{1}\dfrac {x dx}{\ln (1+x)}\quad (n=6) .
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Taking n=10n=10, compute the Gauss constant (Catalan constant)

C=∫01arctan⁡xxdx. C=\int _{0}^{1}\dfrac {\arctan x}{x} dx .
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Compute approximately the integral JJ: a) by the rectangle rule (n=1)(n = 1); b) by the trapezoidal rule (n=1)(n = 1); c) by Simpson's rule (n=2)(n = 2), and find the difference between the exact and approximate values, if:

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(1)

J=∫12dxx2J = \int_{1}^{2} \frac{d x}{x^{2}};

(2)

J=∫0π/2sin⁡xdxJ = \int_{0}^{\pi / 2} \sin x d x.

Find δ=∣J−J∗∣\delta = \left|J-J^{*}\right|, where JJ is the exact value of the integral and J∗J * is its approximate value, computed with step h=2h = 2 by the formula: a) rectangle rule; b) trapezoidal rule; c) Simpson's rule. In each case find the difference between the right-hand side Δ0\Delta_{0} in the estimates (5), (9), (12) (k=4)(k = 4) respectively, and δ(2),(3)\delta (2),(3): J=∫15x4dxJ = \int_{1}^{5} x^{4} d x.

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Find δ=∣J−J∗∣\delta = \left|J-J^{*}\right|, where JJ is the exact value of the integral and J∗J * is its approximate value, computed with step h=2h = 2 by the formula: a) rectangle rule; b) trapezoidal rule; c) Simpson's rule. In each case find the difference between the right-hand side Δ0\Delta_{0} in the estimates (5), (9), (12) (k=4)(k = 4) respectively, and δ(2),(3)\delta (2),(3): J=∫−22exdxJ = \int_{-2}^{2} e^{x} d x.

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Compute approximately, with n=12n = 12 (computing to four decimal places), the integral ∫02πxsin⁡xdx\int_{0}^{2 \pi } x \sin x d x:

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(1)

using the rectangle rule;

(2)

using the trapezoidal rule;

(3)

using Simpson's rule.

Find approximately, with n=6n = 6 (computing to four decimal places), the integral ∫01sin⁡xxdx\int_{0}^{1} \frac{\sin x}{x} d x:

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(1)

using the rectangle rule;

(2)

using the trapezoidal rule;

(3)

using Simpson's rule.

Compute approximately, with n=10n = 10, the integral ∫011+xdx\int_{0}^{1} \sqrt{1+x} d x:

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(1)

using the rectangle rule (to three decimal places);

(2)

using the trapezoidal rule (to three decimal places);

(3)

using Simpson's rule (to six decimal places).

Estimate the error of the result.

Compute approximately, using the rectangle rule with step hh, the integrals:

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(1)

∫02x4dx,h=0,2\int_{0}^{2} x^{4} d x, \quad h = 0,2;

(2)

∫0πsin⁡xdx,h=π6\int_{0}^{\pi } \sin x d x, \quad h = \frac{\pi }{6};

(3)

∫15dxx,h=0,4;\int_{1}^{5} \frac{d x}{x}, \quad h = 0,4 ;

(4)

∫15dx1+x4,h=0,5\int_{1}^{5} \frac{d x}{1+x^{4}}, \quad h = 0,5

(5)

∫02ex2dx,h=0,2\int_{0}^{2} e^{x^{2}} d x, \quad h = 0,2

(6)

∫01ln⁡(1+x)xdx,h=0,2\int_{0}^{1} \frac{\ln (1+x)}{x} d x, h = 0,2.

Compute approximately, using the trapezoidal formula with step hh, the integrals:

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(1)

∫29dxx−1,h=1\int_{2}^{9} \frac{d x}{x-1}, \quad h = 1

(2)

∫27dxx+2,h=1\int_{2}^{7} \frac{d x}{\sqrt{x+2}}, \quad h = 1

(3)

∫01x1−x2dx,h=0,1\int_{0}^{1} x \sqrt{1-x^{2}} d x, \quad h = 0,1;

(4)

∫021+x4dx,h=18\int_{0}^{2} \sqrt{1+x^{4}} d x, \quad h = \frac{1}{8};

(5)

∫01dx1+x3,h=0,1\int_{0}^{1} \frac{d x}{1+x^{3}}, \quad h = 0,1;

(6)

∫25ln⁡2xdx,h=0,5\int_{2}^{5} \ln^{2} x d x, \quad h = 0,5;

(7)

∫0π/21−0,25sin⁡2xdx,h=π12\int_{0}^{\pi / 2} \sqrt{1-0,25 \sin^{2} x} d x, \quad h = \frac{\pi }{12}

Compute approximately, using Simpson's formula with step hh, the integrals:

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(1)

∫13x+1xdx,h=0,5\int_{1}^{3} \sqrt{\frac{x+1}{x}} d x, \quad h = 0,5;

(2)

∫011−x3dx,h=0,1\int_{0}^{1} \sqrt{1-x^{3}} d x, \quad h = 0,1;

(3)

∫02e−x2dx,h=0,5\int_{0}^{2} e^{-x^{2}} d x, \quad h = 0,5

(4)

∫02ex2dx,h=0,2\int_{0}^{2} e^{x^{2}} d x, \quad h = 0,2

(5)

∫12ln⁡xx−1dx,h=0,25\int_{1}^{2} \frac{\ln x}{x-1} d x, \quad h = 0,25;

(6)

∫0π3+cos⁡xdx,h=π6\int_{0}^{\pi } \sqrt{3+\cos x} d x, \quad h = \frac{\pi }{6}

(7)

∫01xdxln⁡(1+x),h=16\int_{0}^{1} \frac{x d x}{\ln (1+x)}, \quad h = \frac{1}{6}.

Compute approximately, using Simpson's formula with n=10n = 10:

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(1)

the integral ∫0π/2sin⁡xxdx\int_{0}^{\pi / 2} \frac{\sin x}{x} d x

(2)

Catalan's constant G=∫01arctg⁡xxdxG = \int_{0}^{1} \frac{\operatorname {arctg} x}{x} d x.

Compute approximately, using Simpson's formula with n=4n = 4 and n=8n = 8, the integral:

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(1)

∫01e−x2dx;\int_{0}^{1} e^{-x^{2}} d x ;

(2)

∫03e−x2dx\int_{0}^{3} e^{-x^{2}} d x.

Compute approximately, using Simpson's formula with n=4n = 4 and n=8n = 8, the integral:

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(1)

∫0π/21−k2sin⁡2xdx\int_{0}^{\pi / 2} \sqrt{1-k^{2} \sin^{2} x} d x, if: а) k2=0,5k^{2} = 0,5; б) k2=0,25k^{2} = 0,25

(2)

∫0π/2dx1−k2sin⁡2x\int_{0}^{\pi / 2} \frac{d x}{\sqrt{1-k^{2} \sin^{2} x}}, if: а) k=0,5k = 0,5; б) k=0,8k = 0,8; в) k=0,9k = 0,9.

For the given integral, estimate the error of the rectangle rule with step hh:

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(1)

∫02x4dx,h=0,2\int_{0}^{2} x^{4} d x, \quad h = 0,2

(2)

∫15dxx,h=0,4\int_{1}^{5} \frac{d x}{x}, \quad h = 0,4

(3)

∫0πsin⁡xdx,h=π6\int_{0}^{\pi } \sin x d x, \quad h = \frac{\pi }{6}

(4)

∫15dx1+x4,h=0,5\int_{1}^{5} \frac{d x}{1+x^{4}}, \quad h = 0,5

(5)

∫02ex2dx,h=0,1\int_{0}^{2} e^{x^{2}} d x, \quad h = 0,1

(6)

∫05x+1xdx,h=0,25\int_{0}^{5} \sqrt{\frac{x+1}{x}} d x, \quad h = 0,25.

For the given integral, estimate the error of the trapezoidal rule with step hh:

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(1)

∫29dxx−1,h=1\int_{2}^{9} \frac{d x}{x-1}, \quad h = 1

(2)

∫27dxx+2,h=1\int_{2}^{7} \frac{d x}{\sqrt{x+2}}, \quad h = 1;

(3)

∫01dx1+x3,h=13\int_{0}^{1} \frac{d x}{1+x^{3}}, \quad h = \frac{1}{3};

(4)

∫25ln⁡2xdx,h=0,5;\int_{2}^{5} \ln^{2} x d x, \quad h = 0,5 ;

(5)

∫021+x4,h=18\int_{0}^{2} \sqrt{1+x^{4}}, \quad h = \frac{1}{8}.

For the given integral, estimate the error of Simpson's rule with step hh:

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(1)

∫0π/2cos⁡x2dx,h=π8;\int_{0}^{\pi / 2} \cos \frac{x}{2} d x, \quad h = \frac{\pi }{8} ;

(2)

∫13ln⁡2xdx,h=13\int_{1}^{3} \ln 2 x d x, \quad h = \frac{1}{3};

(3)

∫06xln⁡(1+x)dx,h=14\int_{0}^{6} x \ln (1+x) d x, \quad h = \frac{1}{4};

(4)

∫25dxln⁡x,h=0,5\int_{2}^{5} \frac{d x}{\ln x}, \quad h = 0,5

(5)

∫011+x4dx,h=0,1\int_{0}^{1} \sqrt{1+x^{4}} d x, \quad h = 0,1;

(6)

∫0π/3cos⁡xdx,h=0,1\int_{0}^{\pi / 3} \sqrt{\cos x} d x, \quad h = 0,1.

Find the step hh for which the error of the approximate value of the given integral by the rectangle rule is not more than ε\varepsilon:

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(1)

∫02x4dx,ε=10−2\int_{0}^{2} x^{4} d x, \quad \varepsilon = 10^{-2};

(2)

∫0π/3cos⁡2xdx,ε=10−3\int_{0}^{\pi / 3} \cos 2 x d x, \quad \varepsilon = 10^{-3};

(3)

∫03ln⁡(1+x2)dx,ε=10−4\int_{0}^{3} \ln \left(1+x^{2}\right) d x, \quad \varepsilon = 10^{-4};

(4)

∫011+x3dx,ε=10−3\int_{0}^{1} \sqrt{1+x^{3}} d x, \quad \varepsilon = 10^{-3}.

Find the step hh for which the error of the approximate value of the integral by the trapezoidal rule is not more than ε\varepsilon:

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(1)

∫03dxx+2,ε=10−1\int_{0}^{3} \frac{d x}{x+2}, \quad \varepsilon = 10^{-1}

(2)

∫01,2exdx,ε=10−3\int_{0}^{1,2} e^{x} d x, \quad \varepsilon = 10^{-3};

(3)

∫13sin⁡xdx,ε=10−3\int_{1}^{3} \sin x d x, \quad \varepsilon = 10^{-3};

(4)

∫00,5arcsin⁡xdx,ε=10−4\int_{0}^{0,5} \arcsin x d x, \quad \varepsilon = 10^{-4}.

Find the step hh for which the error of the approximate value of the integral by Simpson's rule is not more than ε\varepsilon:

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(1)

∫0π/2cos⁡x2dx,ε=10−3\int_{0}^{\pi / 2} \cos \frac{x}{2} d x, \quad \varepsilon = 10^{-3};

(2)

∫13ln⁡2xdx,ε=10−4\int_{1}^{3} \ln 2 x d x, \quad \varepsilon = 10^{-4};

(3)

∫12dxx2,ε=10−4\int_{1}^{2} \frac{d x}{x^{2}}, \quad \varepsilon = 10^{-4};

(4)

∫02arctg⁡xdx,ε=10−4\int_{0}^{2} \operatorname {arctg} x d x, \quad \varepsilon = 10^{-4}.

Compute, using the rectangle rule, the integral

∫0π/21−0,64sin⁡2xdx \int _{0}^{\pi / 2} \sqrt{1-0,64 \sin ^{2} x} d x

for n=4n = 4 and find the error of the result.

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Compute approximately, using the rectangle rule with error not exceeding 10−210^{-2}:

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(1)

∫12x3dx\int_{1}^{2} x^{3} d x

(2)

∫12dxx2\int_{1}^{2} \frac{d x}{x^{2}}

(3)

∫15dx1+ln⁡x\int_{1}^{5} \frac{d x}{1+\ln x}

(4)

∫0π/2sin⁡x1+xdx\int_{0}^{\pi / 2} \frac{\sin x}{1+x} d x

(5)

∫01e−x1+xdx\int_{0}^{1} \frac{e^{-x}}{1+x} d x

(6)

∫011+x2/3dx\int_{0}^{1} \sqrt{1+x^{2 / 3}} d x.

Compute approximately, using the trapezoidal rule with error not exceeding 10−210^{-2}:

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(1)

∫25dxx\int_{2}^{5} \frac{d x}{x}

(2)

∫01,2exdx\int_{0}^{1,2} e^{x} d x

(3)

∫01,2sin⁡xdx\int_{0}^{1,2} \sin x d x

(4)

∫13ln⁡2xdx\int_{1}^{3} \ln 2 x d x

(5)

∫02e−x2dx\int_{0}^{2} e^{-x^{2}} d x

(6)

∫14dx1+4x3\int_{1}^{4} \frac{d x}{1+4 x^{3}}.

Compute approximately, using Simpson's rule with error not exceeding ε\varepsilon:

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(1)

∫15dxx,ε=10−2\int_{1}^{5} \frac{d x}{x}, \quad \varepsilon = 10^{-2};

(2)

∫0πsin⁡xdx,ε=10−3\int_{0}^{\pi } \sin x d x, \quad \varepsilon = 10^{-3};

(3)

∫13xdx,ε=10−3\int_{1}^{3} \sqrt{x} d x, \quad \varepsilon = 10^{-3};

(4)

∫011+x2dx,ε=10−4\int_{0}^{1} \sqrt{1+x^{2}} d x, \quad \varepsilon = 10^{-4};

(5)

∫01dx1+x2,ε=10−4\int_{0}^{1} \frac{d x}{\sqrt{1+x^{2}}}, \quad \varepsilon = 10^{-4};

(6)

∫25ln⁡xdx,ε=10−4\int_{2}^{5} \ln x d x, \quad \varepsilon = 10^{-4};

(7)

∫416dxx+1,ε=10−4\int_{4}^{16} \frac{d x}{\sqrt{x}+1}, \quad \varepsilon = 10^{-4}.

Compute approximately, with error not exceeding ε\varepsilon, the integral:

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(1)

∫12xdx1+x2,ε=10−4\int_{1}^{2} \frac{x d x}{1+x^{2}}, \quad \varepsilon = 10^{-4};

(2)

∫021+4x2dx,ε=10−3\int_{0}^{2} \sqrt{1+4 x^{2}} d x, \quad \varepsilon = 10^{-3};

(3)

∫02dx(1+x2)2,ε=10−3\int_{0}^{2} \frac{d x}{\left(1+x^{2}\right)^{2}}, \quad \varepsilon = 10^{-3};

(4)

∫19dx1+x3,ε=10−3\int_{1}^{9} \frac{d x}{\sqrt{1+x^{3}}}, \quad \varepsilon = 10^{-3};

(5)

∫13x2+1xdx,ε=10−3\int_{1}^{3} \frac{\sqrt{x^{2}+1}}{x} d x, \quad \varepsilon = 10^{-3};

(6)

∫12ln⁡xxdx,ε=5⋅10−3\int_{1}^{2} \frac{\ln x}{x} d x, \quad \varepsilon = 5 \cdot 10^{-3};

(7)

∫12e−xxdx,ε=10−4\int_{1}^{2} \frac{e^{-x}}{x} d x, \quad \varepsilon = 10^{-4};

(8)

∫0π/2cos⁡x1+xdx,ε=10−3\int_{0}^{\pi / 2} \frac{\cos x}{1+x} d x, \quad \varepsilon = 10^{-3}

(9)

∫0π/2cos⁡x2dx,ε=10−3\int_{0}^{\sqrt{\pi / 2}} \cos x^{2} d x, \quad \varepsilon = 10^{-3};

(10)

∫0πsin⁡(sin⁡x)dx,ε=10−3\int_{0}^{\pi } \sin (\sin x) d x, \quad \varepsilon = 10^{-3}.

Find the surface area of a hemisphere of radius RR, using Simpson's formula with n=2n = 2.

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Prove that Simpson's formula with n=2n = 2 gives an exact result when computing the volume of:

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(1)

a sphere;

(2)

a cone;

(3)

a cylinder;

(4)

a spherical segment.

Using Simpson's formula, compute, with error not exceeding 5⋅10−25 \cdot 10^{-2}, the volume of the solid formed by rotating the curve y=sin⁡xy = \sin x, 0⩽x⩽π0 \leqslant x \leqslant \pi, around the axis OxO x.

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Find approximately, using Simpson's formula, the length of the arc of the given curve with error not exceeding 10−810^{-8}:

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(1)

y=x2,−2⩽x⩽2y = x^{2}, \quad -2 \leqslant x \leqslant 2;

(2)

y2=4x,1⩽x⩽5y^{2} = 4 x, \quad 1 \leqslant x \leqslant 5

(3)

y=ln⁡x,1⩽x⩽5y = \ln x, \quad 1 \leqslant x \leqslant 5.

Compute the length of an ellipse with eccentricity ε=0,5\varepsilon = 0,5 and semi-axis a=1a = 1, using Simpson's formula with n=6n = 6. Estimate the error of the result.

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Find approximately, with error not exceeding 5⋅10−25 \cdot 10^{-2}, the length of an ellipse with semi-axes a=10a = 10 and b=6b = 6.

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Find approximately, with error not exceeding 0.03, the surface area of the ellipsoid formed by rotating the ellipse x2+4y2=4x^{2} + 4 y^{2} = 4 around the axis OxO x (use Simpson's formula).

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Find approximately, with error not exceeding 0.01, the volume of the solid formed by rotating around the axis OxO x the figure 0⩽y⩽1/(1+x2),−2⩽x⩽20 \leqslant y \leqslant 1 /(1+x^{2}), \quad -2 \leqslant x \leqslant 2.

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Prove inequality (5) for the rectangle rule.

Here [a;b][a ; b] is divided into nn equal parts, h=(b−a)/nh = (b-a) / n, xi=a+(i+1/2)hx_{i} = a+(i+1 / 2) h, 0⩽i⩽n−10 \leqslant i \leqslant n-1, the rectangle rule is ∫abf(x)dx≈h∑i=0n−1f(xi)\int_{a}^{b} f(x) d x \approx h \sum_{i = 0}^{n-1} f\left(x_{i}\right) and Δ=∫abf(x)dx−h∑i=0n−1f(xi)\Delta = \int_{a}^{b} f(x) d x-h \sum_{i = 0}^{n-1} f\left(x_{i}\right) is its error. Inequality (5): if ff has a piecewise continuous second derivative on [a;b][a ; b], then ∣Δ∣⩽b−a24h2sup⁡[a;b]∣f′′(x)∣\left|\Delta \right| \leqslant \frac{b-a}{24} h^{2} \sup_{[a ; b]}\left|f^{\prime \prime }(x)\right|.

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Prove that inequality (5) is sharp, i.e. there exists a function having a piecewise continuous second derivative such that

∣∫abf(x)dx−h∑i=0n−1f(xi)∣=b−a24h2sup⁡[a;b]∣f′′(x)∣. \left|\int _{a}^{b} f(x) d x-h \sum _{i = 0}^{n-1} f\left(x_{i}\right)\right| = \frac{b-a}{24} h^{2} \sup _{[a ; b]}\left|f^{\prime \prime }(x)\right|.

Here [a;b][a ; b] is divided into nn equal parts, h=(b−a)/nh = (b-a) / n, xi=a+(i+1/2)hx_{i} = a+(i+1 / 2) h, 0⩽i⩽n−10 \leqslant i \leqslant n-1, and Δ=∫abf(x)dx−h∑i=0n−1f(xi)\Delta = \int_{a}^{b} f(x) d x-h \sum_{i = 0}^{n-1} f\left(x_{i}\right) is the error of the rectangle rule ∫abf(x)dx≈h∑i=0n−1f(xi)\int_{a}^{b} f(x) d x \approx h \sum_{i = 0}^{n-1} f\left(x_{i}\right). Inequality (5): if ff has a piecewise continuous second derivative on [a;b][a ; b], then ∣Δ∣⩽b−a24h2sup⁡[a;b]∣f′′(x)∣\left|\Delta \right| \leqslant \frac{b-a}{24} h^{2} \sup_{[a ; b]}\left|f^{\prime \prime }(x)\right|.

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Prove inequality (9) for the error of the trapezoidal rule.

Here [a;b][a ; b] is divided into nn equal parts, h=(b−a)/nh = (b-a) / n, xi=a+ihx_{i} = a+i h, 0⩽i⩽n0 \leqslant i \leqslant n, the trapezoidal rule is ∫abf(x)dx≈h(12(f(x0)+f(xn))+∑i=1n−1f(xi))\int_{a}^{b} f(x) d x \approx h\left(\frac{1}{2}\left(f\left(x_{0}\right)+f\left(x_{n}\right)\right)+\sum_{i = 1}^{n-1} f\left(x_{i}\right)\right), and Δ\Delta is the difference between the integral and the right-hand side. Inequality (9): if ff has a piecewise continuous second derivative on [a;b][a ; b], then ∣Δ∣⩽b−a12h2sup⁡[a;b]∣f′′(x)∣\left|\Delta \right| \leqslant \frac{b-a}{12} h^{2} \sup_{[a ; b]}\left|f^{\prime \prime }(x)\right|.

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Prove that Simpson's formula is an exact equality (exact) for any polynomial up to the third degree inclusive.

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Prove that if an approximate integration formula

∫x0x2f(x)dx≈p0f(x0)+p1f(x1)+p2f(x2) \int _{x_{0}}^{x_{2}} f(x) d x \approx p_{0} f\left(x_{0}\right)+p_{1} f\left(x_{1}\right)+p_{2} f\left(x_{2}\right)

where xi=x0+jh,h=1,2x_{i} = x_{0}+j h, h = 1,2, is an exact equality for any polynomial up to degree two inclusive, then this is Simpson's formula, i.e. p0=p2=h/3,p1=4h/3p_{0} = p_{2} = h / 3, p_{1} = 4 h / 3.

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For a function ff having four continuous derivatives on [a;b][a ; b], estimate the error of the formula obtained in problem 41, taking as known that for the error of Chebyshev's formula (22) the estimate

∣Δ∣⩽1135sup⁡[−1;1]∣f(4)(x)∣ \left|\Delta \right| \leqslant \frac{1}{135} \sup _{[-1 ; 1]}\left|f^{(4)}(x)\right|
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Using the Maclaurin series for f(x)f(x), compute the integral ∫01f(x)dx\int_{0}^{1} f(x) d x to within 10−310^{-3}, if:

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(1)

f(x)=sin⁡xxf(x) = \frac{\sin x}{x}

(2)

f(x)=x3cos⁡xf(x) = \sqrt[3]{x} \cos x

(3)

f(x)=cos⁡x2f(x) = \cos x^{2}

(4)

f(x)=sh⁡xxf(x) = \frac{\operatorname {sh} x}{x}

(5)

f(x)=sin⁡x2f(x) = \sin x^{2}

(6)

f(x)=11+x4f(x) = \frac{1}{\sqrt{1+x^{4}}}.

Compute to within 10−310^{-3}:

?
(1)

∫12exxdx\int_{1}^{2} \frac{e^{x}}{x} d x

(2)

∫24e1xdx\int_{2}^{4} e^{\frac{1}{x}} d x

(3)

∫01/2arctg⁡xxdx\int_{0}^{1 / 2} \frac{\operatorname {arctg} x}{x} d x;

(4)

∫01/3dx1−x23\int_{0}^{1 / 3} \frac{d x}{\sqrt[3]{1-x^{2}}};

(5)

∫510ln⁡(1+x2)xdx\int_{5}^{10} \frac{\ln \left(1+x^{2}\right)}{x} d x;

(6)

∫01/2arcsin⁡xxdx\int_{0}^{1 / 2} \frac{\arcsin x}{x} d x;

(7)

∫2+∞dx1+x3dx\int_{2}^{+\infty } \frac{d x}{1+x^{3}} d x;

(8)

∫01xxdx\int_{0}^{1} x^{x} d x.

References

[deSi04] de Souza P.N., Silva J. Berkeley Problems in Mathematics. 3rd ed., Springer, 2004.
[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.