Prove the following equations by induction. In each case, n is a positive integer. a. 1 C 4 C 7 C C .3n 2/ D n.3n1/ 2 . b. 1 3 C 2 3 C C n 3 D n 2.nC1/2 4 . c. 9 C 9 10 C 9 100 C C 9 10n1 D 10n 1. d. 1 12 C 1 23 C C 1 n.nC1/ D 1 1 nC1 . e. 1 C x C x 2 C x 3 C C x n D .1 x nC1 /=.1 x/. You should assume x 6D 1. What is the correct right hand side when x D 1? The next parts are for those who have studied calculus. f. lim x!1 x n e x D 0: g. n D Z 1 0 x n e x dx: h. The nth derivative of x n is n; that is d n dxn x n D n: 2
Reece Witcher 2/27/18 Improper Integrals ● Improper integrals are integrals who have a bound that goes to infinity or when an integral has an asymptote at a bound ○ If there is an asymptote, you should separate the integral into two other integral and substitute the asymptote value with a limit on the appropriate side ○ ● The infinite bound can be pull out by substituting it for a variable and taking the limit of the variable as it goes to infinity 2 ○ lim a →infinitydx a ● You should then complete the integration ○ NOTE: make sure to use t with the solution ● Plug in the bounds, then plug in the limit ● If the bound value is infinite or undefined, the integral is divergent ● If the value is finite (a normal number like 1 or 800), the number is convergent ● Tips: ○ This strategy is useful for solving problems with asymptotes ■ Such as 1/x1, which has an asymptotes at x=1 ○ Don’t forget to solve for the bounds when you solve an integral using usub Arc Length b ● The formula for arc length is ∫a√❑ ● If you are given the f(x) formula, simply take the derivative and plug it into the equation ● Tips: ○ This formula is can be used to solve for the length along any curve ○ If you are calculating a shape that is symmetrical, simply set the bounds at
