Unformatted text preview: MTH405 Wk 10 Lect 1 Integration by Parts
Our primary goal here is to develop a general method of solving
integrals of the form
f (x)g(x)dx. As a manifestation of the Product Rule in Dierentiation, we have
of the formula called Integration by Parts . This formula is given by
ˆ ˆ u dv = uv −
Example. v du. Use Integration by Parts to evaluate the function
ˆ x cos x dx. Solution. We make choices for u and dv for this integral. We
u=x du = dx dv = cos xdx
v = dv = cos xdx = sin x, then substituting, we have
ˆ ˆ u dv = uv − v du ˆ ˆ
x cos xdx = (x) (sin x) − sin xdx = x sin x − (− cos x)
= x sin x + cos x + c. 1 Exercise. ´
1. 2. 3. 4. ´
´ Evaluate the following integrals. xex dx.
x2 e−x dx.
ex cos xdx. Applications of Integration
Given below are some of the applications of Integration which we
will be covering in this course.
• Area between two curves. • Volumes by Disks and Washers 2 Area between Two Curves
Given two functions f and g continuous on the interval [a, b], and if
f (x) ≥ g(x) for all x ∈ [a, b], then the area of the region bounded is
b [f (x) − g(x)] dx. A=
a Note the condition that f (x) ≥ g(x) for all x ∈ [a, b]. If for
any part of [a, b] has g(x) ≥ f (x), then the area gets cancelled o
and we have the wrong answer.
Remark. Example. Find the area of the region bounded above by y = x + 6,
bounded below by y = x2 , and bounded on the sides by the lines
x = 0 and x = 2. Find the area of the region that is enclosed between the
curves y = x2 and y = x + 6.
Example. 3 Volumes by Disks and Washers
A solid of revolution is a solid that is generated by revolving a plane
region about a line that lies in the same plane as the region; the line
is called an axis of revolution . Many familiar solid are of this type. 4 Volume by Disks
Theorem. Let f be a continuous function and non-negative on [a, b], R be the region that is bounded above by y = f (x), below by
x − axis, and on the sides by the lines x = a and x = b. Then and let
the the volume of the solid of revolution that is generated by revolving
the region R about the x-axis is ˆ b [f (x)]2 dx. V =π
a 5 Volume by Washers f and g be continuous functions and non-negative over
the interval [a, b] with f (x) ≥ g(x) for all x ∈ [a, b]. Let R be the
region that is bounded above by f (x), below by g(x), and on the
sides by the lines x = a and x = b. Then the volume of the solid of
revolution that is generated by revolvng the region R about the x-axis
f (x)2 − g(x)2 dx.
Theorem. Let a 6 Volume of Disks and Washers about y -axis The method of disks and washers have analogs for regions revolved
about the y -axis. The formula of nding volumes using disks and
washers respectively, are given below.
V d [u(y)]2 dy = π (disks) c ˆ
u(y)2 − v(y)2 dy = π
c 7 (washers) Example.
1. 2. 3. Find the volume of the
√ solid that is obtained when the region
under the curve y = x over the interval [1, 4] is revolved about
[ Ans: V = 15π
Derive the formula for the volume of a sphere of radius r.
[Ans: V = 34 πr3 units3 ]
Find the√volume of the solid generated when the region enclosed
by y = x, y = 2 and x = 0 is revolved about the y -axis.
[Ans: V = 32π
5 8 ...
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- Spring '16
- Alveen Chand
- Region, dx, Two Curves, cos xdx