Integrate 1 sqrt x 2

• [PDF File]1 Evaluating an integral with a branch cut

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  1/x−1 when we approach from the upper half plane and is −i p 1/x−1 when we approach from the lower half plane. Thus the boundary values of f(z) are f(x+i0) = 1 ±ix q 1− 1 x = 1 ±i p x(1−x). Take a curve C going around the interval 0 ≤ x ≤ 1 counterclockwise. We can replace C by such a curve that goes around the interval and ...

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• [PDF File]GAUSSIAN INTEGRALS

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  x = rcosθ, y = rsinθ (3) so that r2 = x2 +y2 (4) The element of area in polar coordinates is given by rdrdθ, so that the double integral becomes I2 = Z ∞ 0 Z 2π 0 e−r2 rdrdθ (5) Integration over θ gives a factor 2π. The integral over r can be done after the substitution u = r2, du = 2rdr: Z ∞ 0 e−r2 rdr = 1 2 Z ∞ 0 e−u du = 1 ...

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• [PDF File]Calculus II, Section6.2, #34 Volumes Set up an integral ...

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  ∆x Using the input Integrate[Pi*(Sqrt[1-x^2])^2 - Pi*(x^2)^2, {x, -Sqrt[(-1+Sqrt[5])/2], Sqrt[(-1+Sqrt[5])/2]}] WolframAlpha gives us Thus, the volume of the solid obtained by revolving the region bounded by y = x2 and x2+y2 = 1 for y ≥ 0 about the x-axis is ≈ 3.54459. Calculus II

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• [PDF File]RESIDUE CALCULUS, PART II

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  ln(x2 +1) x2 +1 Take principal branch of log. Branch cut ΓR dz ln ( z + i ) ( z + 1 ) 2 Im z Re z i −i R Γ R Consider •By residue theorem I ΓR ln(z +i) z2 +1 dz = 2πi Resz=if = 2πi ln2i 2i = π ln2 + iπ 2 . •By Jordan lemma Z SR →0 for R → ∞ . • Z R −R ln(x +i) x2 + 1 dx = Z R 0 ln(−x+ i) x2 +1 dx+ Z R 0 ln(x+ i) x2 +1 ...

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• [PDF File]MA 104 Graded Homework 2 Solutions

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  Plot[2 + Sqrt[9 - x^2], {x, -3, 0}] Integrate[2 + Sqrt[9 - x^2], {x, -3, 0}] (2) (5 Points) Evaluate the following definite integral Z 4 0 e √ x √ x dx. This problem is done with Mathematica using the following command. Integrate[Exp[Sqrt[x]]/Sqrt[x], {x, 0, 4}] Manually, we set u = √ x and get du = dx 2 √ x or 2du = dx √ x. We have ...

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• [PDF File]Table of Integrals

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  1 2 secxtanx+ 1 2 ln|secxtanx| (76)!secxtanxdx=secx (77)!sec2xtanxdx= 1 2 sec2x (78)!secnxtanxdx= 1 n secnx, n!0 2 (79)!cscxdx=ln|cscx"cotx| (80)!csc2xdx="cotx (81)!csc3xdx=" 1 2 cotxcscx+ 1 2 ln|cscx"cotx| (82)!cscnxcotxdx=" 1 n cscnx, n!0 (83)!secxcscxdx=lntanx TRIGONOMETRIC FUNCTIONS WITH xn (84)!xcosxdx=cosx+xsinx (85)!xcos(ax)dx= 1 a2 ...

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• [PDF File]Techniques of Integration - Whitman College

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  (1−x 2)) p 1− x dx. This looks messy, but we do now have something that looks like the result of the chain rule: the function 1 − x2 has been substituted into −(1/2)(1 − x) √ x, and the derivative 8.1 Substitution 165 of 1−x2, −2x, multiplied on the outside. If we can find a function F(x) whose derivative is −(1/2)(1− x) √

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• [PDF File]THE GAUSSIAN INTEGRAL

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  2 KEITH CONRAD Instead of using polar coordinates, set x= ytin the inner integral (yis xed). Then dx= ydtand (2.1) J2 = Z 1 0 Z 1 0 e 2y2(t2+1)ydt dy= Z 1 0 Z 1 0 ye y2(t +1) dy dt; where the interchange of integrals is justi ed by Fubini’s theorem for improper Riemann integrals.

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• [PDF File]Maxima by Example: Ch.7: Symbolic Integration

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  x(b2 ¡x2)¡1=2 dx: (%i3) integrate (x/ sqrt (bˆ2 - xˆ2), x); 2 2 (%o3) - sqrt(b - x ) (%i4) diff(%,x); x (%o4) -----2 2 sqrt(b - x ) Example 3 The definite integral can be related to the ”area under a curve” and is the more accessible concept, while the integral is simply a function whose first derivative is the original integrand.

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• [PDF File]Table of Integrals

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  Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=

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• [PDF File]Table of Useful Integrals

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  2 0 ∞ ∫ x2n+1e−ax2dx= n! 2an+1 0 ∞ ∫ xne−axdx= n! an+1 0 ∞ ∫ Integration by Parts: UdV a b ∫="#UV$% a b −VdU a b ∫ U and V are functions of x. Integrate from x = a to x = b sin ...

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• [PDF File]Math 104: Improper Integrals (With Solutions)

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  1 (x−1)2/3 dx, if it converges. Solution: We might think just to do Z 3 0 1 (x−1)2/3 dx= h 3(x− 1)1/3 i 3 0, but this is not okay: The function f(x) = 1 (x−1)2/3 is undefined when x= 1, so we need to split the problem into two integrals. Z 3 0 1 (x− 1)2/ 3 dx= Z 1 0 1 (x− 1)2/ dx+ Z 3 1 1 (x− 1)2/3 dx. The two integrals on the ...

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• [PDF File]Monte Carlo Integration with R - UMD

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  Consider: f(x)=x^2*exp(-x), k(x)=x^2, g(x)=exp(-x) I(f)=Int_{0}^{infinity}(f(x))=Gamma(3)=2 k

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• [PDF File]Integration by substitution

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  1+x2 dx = 2 3 (1+x2)3/2 +c We have completed the integration by substitution. Let us analyse this example a little further by comparing the integrand with the general case f(g(x))g′(x). Suppose we write g(x) = 1+x2 and f(u) = √ u Then we note that the composition1of the functions f and g is f(g(x)) = √ 1+x2. 1when finding the composition ...

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• [PDF File]Techniques of Integration

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  204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Evaluate Z sin6 xdx. Use sin2 x = (1 − cos(2x))/2 to rewrite the function: Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x+3cos2 2x− cos3 2xdx. Now we have four integrals to evaluate: Z 1dx = x and Z

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• [PDF File]Table of Basic Integrals Basic Forms

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  a 2+ x dx= 1 2 x2 1 2 a2 lnja2 + x2j (13) Z 1 ax2 + bx+ c dx= 2 p 4ac b2 tan 1 2ax+ b p 4ac b2 (14) Z 1 (x+ a)(x+ b) dx= 1 b a ln a+ x b+ x; a6=b (15) Z x (x+ a)2 dx= a a+ x + lnja+ xj (16) Z x ax2 + bx+ c dx= 1 2a lnjax2+bx+cj b a p 4ac 2b2 tan 1 2ax+ b p 4ac b Integrals with Roots (17) Z p x adx= 2 3 (x a)3=2 (18) Z 1 p x a dx= 2 p x a (19) Z ...

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• [PDF File]Integral Calculus Formula Sheet

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  Fundamental Theorem of Calculus: x a d F xftdtfx dx where f t is a continuous function on [a, x]. b a f xdx Fb Fa, where F(x) is any antiderivative of f(x). Riemann Sums: 11 nn ii ii ca c a 111 nnn ii i i iii ab a b 1

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• [PDF File]NUMERICAL INTEGRATION: ANOTHER APPROACH

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  ,x2 = 1 sqrt(3) This yields the formula Z 1 ... −1 x4 dx= 2 5 f µ −1 sqrt(3) ¶ + f µ 1 sqrt(3) ¶ = 2 9 Thus (1) has degree of precision exactly 3. EXAMPLE Integrate Z 1 −1 dx 3+x =log2. =0.69314718 The formula (1) yields 1 3+x1 + 1 3+x2 =0.69230769 Error= .000839. THE GENERAL CASE We want to find the weights {wi} and nodes {xi} so as ...

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• [PDF File]9 De nite integrals using the residue theorem

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  1 + x 2 Z 1+ 0 dt Me a(x 1+x 2) Again, clearly this last expression goes to 0 as x 1 and x 2 go to 1. The argument for C 3 is essentially the same as for C 1, so we leave it to the reader. The proof for part (b) is the same. You need to keep track of the sign in the exponentials

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