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Final Review Guide

Exercises pp789 790 1 36 integrate 1 r e 2 x cos x dx

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* Exercises: pp.789-790 1-36 · Integrate. 1. R e - 2 x cos x dx 2. R e x sin( πx ) dx 3. R ( θ + 1) sin( θ 2 + 2 θ ) 4. R tan 2 x sec 2 x dx 5. R sec 3 x tan x dx 6. R sin 1 / 3 x cos x dx 7. R cos x sin x dx 8. R cos x ln(sin x ) dx 9. R 5 0 θ ln( θ + 1) 10. R π \ 4 π \ 8 cot 2 θ dθ 11. R π \ 3 0 e θ sin e θ 12. R e π \ 2 1 cos(ln θ ) , dθ Chapter 6 Integration 6.5 Evaluating Definite Integrals * Properties of the Definite Integral: 1. Z a a f ( x ) dx = 0 2. Z b a f ( x ) dx = - Z a b f ( x ) dx 3. Z b a cf ( x ) dx = c Z b a f ( x ) dx ( c , a constant) 4. Z b a [ f ( x ) ± g ( x )] dx = Z b a f ( x ) dx ± Z b a g ( x ) dx 5. Z b a f ( x ) dx = Z c a f ( x ) dx + Z b c f ( x ) dx ( a < c < b ) * Be careful when implementing the method of substitution for definite integrals–your limits of integration must change consistently with your substitution. See examples 1,2,3 pp. 443-444 * The Average Value of a Function: Suppose f is integrable on [ a, b ]. Then the average value of f over [ a, b ] is 1 b - a Z b a f ( x ) dx * Exercises: p. 448 1-44 6.6 Area Between Two Curves * The Area Between Two Curves Formula: Let f and g be continuous functions such that f ( x ) g ( x ) on the interval [ a, b ]. Then, the area of the region bounded above by y = f ( x ) and below by y = g ( x ) on [ a, b ] is given by Z b a [ f ( x ) - g ( x )] dx * It is always a good idea to try to sketch the two curves. This will help you see which curve is above which. See examples 1,2,3 on pp. 454-455. * If one curve is not always above the other, break up the area between into the subregions where one function is always bigger on that subregion. Finding the point(s) of intersection is a helpful (sometimes necessary) step. See examples 4,5 on pp. 456-457. * Exercises: pp.459-461 1-42. 3
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Chapter 7 Additional Topics in Integration 7.1 Integration By Parts * Integration by Parts Formula: Z udv = uv - Z vdu * You need to decompose the integrand f ( x ) dx in question into two parts: u and dv –so f ( x ) dx = udv . Remember to always included dx in your dv term. * Once you choose u and dv , du and v are essentially already decided upon as well. * Try to choose u so that du is simpler than u , and dv such that dv is easy to integrate. The goal of integration by parts is to rewrite the integral in question as a simpler integral. * If the integrand involves e x dx or some variation, it is usually a good first try for dv . * Sometimes you can choose dx as your dv . For example, when computing R ln xdx . * There may be times when you need to implement integration by parts multiple times to solve an integral. * Exercises:p.488-489 1-32 1. Find the following antiderivatives using integration by parts. (a) Z x x + 2 dx (b) Z ( x + 1) 2 e x dx (c) Z ln(2 x ) dx (d) Z e x dx Hint: do a u-substitution first with u = x (e) Z ln(1 - x ) dx (f) Z cos(2 x ) sin( x - 1) dx 2. Calculate the following definite integrals using integration by parts. (a) Z e 1 x (ln x ) 2 (b) Z π 4 0 x sec 2 ( x ) dx 3. Why does Z π 0 x sec 2 ( x ) dx make no sense?
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