INTEGRATION BY PARTS
235
Find the volume of the solid obtained by revolving the region $1 of Problem 28.28 about the *-axis.
28.30
28.31
28.32
28.33
28.34
28.35
28.36
28.37
28.38
Change the variable to t = lnx and use Problem 28.1:
By the disk formula,
Use the solution to Problem 28.30 to establish the following bounds on e: 2.5 < e s 2.823.
dx = 2 — 5le. By Problem 24.59, lie is the maximum value
In Problem 28.30, it was shown that
Hence
{In general,
if M is an upper bound of
of In x/x.
f(x) on [a, b].} Thus, 0<2 -5/e<(e - l)/e
2 . The left-hand inequality gives e>2.5. The right-hand
inequality gives 2e
2 -5e
2 -6x + l=0 are (3±V7)/2,
e<(3 + V7)/2<2.823 (since V7< 2.646).
Let SI be the region under one arch of the curve y = sin x, above the x-axis, between x = 0 and x = -n.
Find the volume of the solid obtained by revolving 9? about the _y-axis.
By the cylindrical shell formula and Problem 28.5,
If n is a positive integer, find
(njt sin nx + cos nx).
Hence,
Let u = x, dv = cosnxdx, du = dx, v = (I In) sin nx. Then J" x cos nx
If n is a positive integer, find
Hence,
Find a reduction formula for J cos" x dx for n a 2.
Apply the reduction formula of Problem 28.35 to find J cos
6 x dx, using the result of Problem 28.12.
Find a reduction formula for / sin" x dx for n^2.
In the formula of Problem 28.35 replace x by tr/2 - x, to obtain:
Use the reduction formula of Problem 28.37 to find J sin
4 x dx, using the result of Problem 28.10.
x sin x dx = 2Tr{(ir + 0) - (0 + 0)] = 27r
: .
x cos nx dx.
sin nx dx =
(nx sin nx + cos nx)
x sin nx dx.
sin nx
x cos nx
(1-1) = 0.
Let M = x, dv = sin nx dx, du = dx, v = -(I In) cos nx. Then J A- sin «* cos nx +
Let M = cos" l x, dv = cosxdx, du = — (n — l)cos" 2xsinxdx, v = sin x. Then J cos" x dx =
sinxcos""
1 x + (n - 1) /cos""
2 xsin
2 x dx
=
sinx cos""
1 x + (n - 1) J cos"~
2 x(l - cos
2 x) dx
=
sin x cos""
1 x + (n — 1) f cos""
2 x dx — (n - 1) f cos" x d*. Solving for f cos" x dx,
235
Find the volume of the solid obtained by revolving the region $1 of Problem 28.28 about the *-axis.
28.30
28.31
28.32
28.33
28.34
28.35
28.36
28.37
28.38
Change the variable to t = lnx and use Problem 28.1:
By the disk formula,
Use the solution to Problem 28.30 to establish the following bounds on e: 2.5 < e s 2.823.
dx = 2 — 5le. By Problem 24.59, lie is the maximum value
In Problem 28.30, it was shown that
Hence
{In general,
if M is an upper bound of
of In x/x.
f(x) on [a, b].} Thus, 0<2 -5/e<(e - l)/e
2 . The left-hand inequality gives e>2.5. The right-hand
inequality gives 2e
2 -5e
2 -6x + l=0 are (3±V7)/2,
e<(3 + V7)/2<2.823 (since V7< 2.646).
Let SI be the region under one arch of the curve y = sin x, above the x-axis, between x = 0 and x = -n.
Find the volume of the solid obtained by revolving 9? about the _y-axis.
By the cylindrical shell formula and Problem 28.5,
If n is a positive integer, find
(njt sin nx + cos nx).
Hence,
Let u = x, dv = cosnxdx, du = dx, v = (I In) sin nx. Then J" x cos nx
If n is a positive integer, find
Hence,
Find a reduction formula for J cos" x dx for n a 2.
Apply the reduction formula of Problem 28.35 to find J cos
6 x dx, using the result of Problem 28.12.
Find a reduction formula for / sin" x dx for n^2.
In the formula of Problem 28.35 replace x by tr/2 - x, to obtain:
Use the reduction formula of Problem 28.37 to find J sin
4 x dx, using the result of Problem 28.10.
x sin x dx = 2Tr{(ir + 0) - (0 + 0)] = 27r
: .
x cos nx dx.
sin nx dx =
(nx sin nx + cos nx)
x sin nx dx.
sin nx
x cos nx
(1-1) = 0.
Let M = x, dv = sin nx dx, du = dx, v = -(I In) cos nx. Then J A- sin «* cos nx +
Let M = cos" l x, dv = cosxdx, du = — (n — l)cos" 2xsinxdx, v = sin x. Then J cos" x dx =
sinxcos""
1 x + (n - 1) /cos""
2 xsin
2 x dx
=
sinx cos""
1 x + (n - 1) J cos"~
2 x(l - cos
2 x) dx
=
sin x cos""
1 x + (n — 1) f cos""
2 x dx — (n - 1) f cos" x d*. Solving for f cos" x dx,