Use (11.6) and (11.14) to (11.16) to evaluate the following integrals. Warning hint: See comments just after (11.6) and (11.16) about the range of integration. 15. (c) / e*s'(z) dz
Use (11.6) and (11.14) to (11.16) to evaluate the following integrals. Warning hint: See comments just after (11.6) and (11.16) about the range of integration. 15. (c) / e*s'(z) dz
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please solve question 15
![[ «95(2 – to) dt = { dlto), a<to <b,
$(t)8(t – to) dt = {
(11.6)
0,
otherwise.
Derivatives of the o Function To see that we can attach a meaning to the
derivative of 8(x – a), we write $(x)8'(x – a) dx and integrate by parts to get
(11.14) (x)8'(x-a) dr = 0(x)d(x-a) -| (x)(x- a) dr = -o'(a).
The integrated term is zero at too, and we evaluated the integral using equation
(11.6). Thus, just as 8(x – a) “picks out" the value of o(x) at x = a [see equation
(11.6)], so d'(x - a) picks out the negative of o (x) at r = a. Integrating by parts
twice (Problem 14), we find
(11.15)
I «(x)8"(x – a) dr = ¢" (a).
Repeated integrations by parts gives the formula for the derivative of any order of
the d function (Problem 14):
(11.16)
/ $(x)6{») (x – a) dz = (-1)"ø(") (a).
Use (11.6) and (11.14) to (11.16) to evaluate the following integrals. Warning hint:
See comments just after (11.6) and (11.16) about the range of integration.
15.
(e) L *
xp (x),8 =g](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfe83958-fab2-41e0-b954-baff63ecab2e%2Fc3350605-92ed-4028-9531-824cc0f4497d%2F1yegwd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[ «95(2 – to) dt = { dlto), a<to <b,
$(t)8(t – to) dt = {
(11.6)
0,
otherwise.
Derivatives of the o Function To see that we can attach a meaning to the
derivative of 8(x – a), we write $(x)8'(x – a) dx and integrate by parts to get
(11.14) (x)8'(x-a) dr = 0(x)d(x-a) -| (x)(x- a) dr = -o'(a).
The integrated term is zero at too, and we evaluated the integral using equation
(11.6). Thus, just as 8(x – a) “picks out" the value of o(x) at x = a [see equation
(11.6)], so d'(x - a) picks out the negative of o (x) at r = a. Integrating by parts
twice (Problem 14), we find
(11.15)
I «(x)8"(x – a) dr = ¢" (a).
Repeated integrations by parts gives the formula for the derivative of any order of
the d function (Problem 14):
(11.16)
/ $(x)6{») (x – a) dz = (-1)"ø(") (a).
Use (11.6) and (11.14) to (11.16) to evaluate the following integrals. Warning hint:
See comments just after (11.6) and (11.16) about the range of integration.
15.
(e) L *
xp (x),8 =g
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