Use the residue theorem to compute the next definite integral
please don't skip any steps and answer thoroughly
Use the residue theorem to compute the next definite integral please don't skip any steps and...
A. Express the limit as a definite integral on the given
interval.
B. Use the form of the definition of the integral to evaluate
the integral.
n Š lim n-> Xi Ax, [1, 3] (xi +13 * 2 i=1 3 6 (2x - x2) dx
step by step please, thank
you
(2) Use Stokes' Theorem to evaluate the integral F.dr, where F(x, y, z) =< -Y, I, z > and where S is the upper hemispherical surface defined by z = v1- 2 - y2. The boundary of S is the curve C defined by Cos (t) y= sin (t) 0t 27 Z=0
please prove part (b) use complex analysis and calculus of
residue
-dx neif a> 0 5. (a) x2+1 (b) For any real number a > 0, cos x dx ne"/a. a Hint: This is the real part of the integral obtained by replacing cos x by e
Problem 4: Use the surface integral in Stokes' theorem to evaluate F.dr for the hemisphere S : x2 + y2 + z2 = 9; z > 0, its bounding circle C: 2+9 and the field F-yi- xj. You only have to compute the surface integral, not the line integral. (20 points)
write legibly please!!
9. Evaluate the following definite integrals (1) >> 23 - 3x2 + 5da (2) Si cos xdx 4
1 9. Find the principal value of dc 25 +1 COS ac 10. Find the principal value of dr, a > 0, b>0. (.x2 +62)2
Show Sketch and all steps.
Problem 18 Use the Divergence Theorem to calculate the surface integral || FdS , F(x,y,z) =< x²yz,xy-z, xyz? > S is the surface of the box enclosed by the planes x = 0, x = a, y = 0, y = b, z = 0, and z = C, where a, b, c are positive numbers.
use residue theorem to evaluate the following
integrals
sin z 21) 20) Cosx dx (r? + 1) X 22) sin mx dx 2(x² + a²² (a > 0, b>0) 23) cos ex - cos bx -dx x?
1. Let F(x,y,z) =< 32, 5x, – 2y >. Use Stokes's Theorem to evaluate the integral Scurl F.ds, where S is the part of the paraboloid z = x² + y2 that lies below the plane z = 4 with upward- pointing normal vector.
Evaluate the following integral using residues: I = { cos(bx)-cos(ax) dx. x2 Let a and b: real constants such that a > b>0. Note: cos(bz)-cos(az) is well-behaved along the real axis (singularity at z = 0 is removable), ejbz-ejaz has a pole at the origin. Make sure to handle this point correctly 22