MTH405_Wk_10_Lect_1

# MTH405_Wk_10_Lect_1 - MTH405 Wk 10 Lect 1 Integration by...

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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 ˆ f (x)g(x)dx. As a manifestation of the Product Rule in Dierentiation, 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 have u=x du = dx dv = cos xdx ˆ ˆ 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. ln xdx 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 given by ˆ 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 is ˆ b   f (x)2 − g(x)2 dx. V =π 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. ˆ V d [u(y)]2 dy = π (disks) c ˆ V d   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 the x-axis. [ Ans: V = 15π units3 ] 2 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. units3 ] [Ans: V = 32π 5 8 ...
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