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

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 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 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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern