DUMMY
c7hw-dun42rkq.pdf 1. 2. 3. 4. 5. 6. Student: Sam Henry Date: 06/14/21 Instructor: Course: Assignment: Evaluate the integral. ∫ − 11 − 10 dx x ∫ − 11 − 10 dx x = ln 10 11 Evaluate the integral. ∫ 4y3dy y4 − 3 ∫ 4y3dy y4 − 3 = ln + cy4 − 3 Evaluate .∫ dx 14 x + 14x ∫ dx 14 x + 14x = + c ln x + 1 7 Evaluate the integral .21 e dx∫ 7x + 3 21 e dx∫ 7x + 3 = 3 e + c7x + 3 Evaluate the integral .e dx∫ ln 8 ln 125 x 3 e dx∫ ln 8 ln 125 x 3 = 9 (Simplify your answer.) Evaluate the integral .15t e dt∫ 4 − t5 15t e dt∫ 4 − t5 = − 3 e + c− t5 7. 8. 9. 10. 11. Evaluate the integral .e ( − 6 sin (6x))dx∫ 6 cos (6x) e ( − 6 sin (6x))dx∫ 6 cos (6x) = + ce 6 cos (6x) 6 Evaluate the integral. dy∫ 7 e 7y 4 + e 7y dy∫ 7 e 7y 4 + e 7y = ln + Ce 7y + 4 Evaluate the integral. x2 dx∫ 1 3 x2 (Type an exact answer.)x2 dx∫ 1 3 x2 = 3 ln 2 Evaluate the following integral. (1 + ln x)x dx∫ 1 3 x (Type an exact answer.)(1 + ln x)x dx∫ 1 3 x = 26 Evaluate the integral .x dx∫ 0 9 2 + 1 2 x dx∫ 0 9 2 + 1 2 = 9 2 + 1 (Type an exact answer, using radicals as needed.) 12. 13. 14. 15. 16. 17. Evaluate the integral. dx∫ 0 8 log 64(x + 8) x + 8 dx∫ 0 8 log 64(x + 8) x + 8 = (Type an exact answer.) Evaluate the integral .∫ dx8x log 5x ∫ dx8x log 5x = Solve the initial value problem. , y( ) dy dt = 5 e sec 5π e − t 2 − t ln 4 = 6 / π y = − tan 5π e + 1 π − t 7 π (Type an exact answer, using as needed.) π Solve the initial value problem. , y(1) , (1) d2y dt2 = 1 − e 2t = − 1 y′ = 2 y = − e − 1 − t2 + e 2 + 2 t 2 1 4 2t e 2 + 6 4 Solve the initial value problem. ; ; = 5 sec x d2y dx2 2 y′ π 4 = 0 y(0) = 0 The solution is y .= − 5 ln − 5xcos (x) (Type an exact answer, using as needed.)π Find the length of the following curve. , y = − 4 ln x x2 32 5 ≤ x ≤ 30 The length of the curve is . (Type an exact answer.) 18. Instead of approximating near x 1, approximate near x 0 to obtain a simpler formula this way.ln x = ln ( )1 + x = (a) Find the linearization of at x 0.ln ( )1 + x = (b) Estimate the five decimal place error involved in replacing with x on the interval [0,0.1].ln ( )1 + x (c) Graph and x together for 0 x 0.5. Use different colors, if available. At what points does the approximation of seem the best? Least good? By reading coordinates from the graph, find as good an upper bound for the error as the graphing utility will allow. ln ( )1 + x ≤ ≤ ln ( )1 + x (a) The linearization of at x 0 is L(x) .ln ( )1 + x = ≈ (b) The approximation error is . (Round to five decimal places as needed.) (c) Choose the correct graph of y and y x shown both for 0 x 0.5.= ln ( )1 + x = ≤ ≤ A. -0.2 0.5 -0.5 0.5 x y B. -0.2 0.5 -0.5 0.5 x y C. -0.2 0.5 -0.5 0.7 x y D. -0.2 0.5 -0.5 0.5 x y Find the x-value in the interval 0 x 0.5 at which the approximation of seems the best. Choose the correct answer below. ≤ ≤ ln ( )1 + x A. 0.5 B. 0 C. 0.3 D. 0.2 Find the x-value in the interval 0 x 0.5 at which the approximation of seems the least good. Choose the correct answer below. ≤ ≤ ln ( )1 + x A. 0.2 B. 0.1 C. 0.5 D. 0.4 By reading the coordinates from the graph, the upper bound for the error is . (Round to four decimal places as needed.) y = x y = ln ( )1 + x y = ln ( )1 + x y = x y = x y = ln ( )1 + x y = ln ( )1 + x y = x 19. Which is larger, or ? Answer this question by completing parts (a) through (e) below.πe e π a. Find an equation for the line through the origin that is tangent to the graph of y .= ln x [ 3,6] by [ 3,3]− − The equation for the line through the origin tangent to the graph of y is y .= ln x = x e (Type an exact answer.) b. Give an argument based on the graphs of y and the tangent line to explain why for all positive x .= ln x ln x < x="" e="" ≠="" e="" to="" determine="" why="" for="" all="" positive="" x="" ,="" first,="" notice="" that="" when="" ,="" .ln="" x="">< x="" e="" ≠="" e="" x="e" ln="" x="x" e="Consider" how="" each="" function="" changes="" over="" time="" by="" finding="" the="" derivative="" of="" each.="" (simplify="" your="" answer.)ln="" x="" d="" dx="(Simplify" your="" answer.)="" d="" dx="" x="" e="For" ,="" the="" derivative="" of="" the="" function="" is="" (1)="" than="" the="" derivative="" of="" the="" function="" .="" therefore,="" as="" x="" decreases="" from="" ,="" decreases="" (2)="" than="" .="" 0="">< x="">< e="" x="" e="" ln="" x="" e="" ln="" x="" x="" e="" similarly,="" for="" ,="" the="" derivative="" of="" the="" function="" is="" (3)="" than="" the="" derivative="" of="" the="" function="" .="" therefore,="" as="" x="" increases="" from="" ,="" increases="" (4)="" than="" .="" e="">< x="" x="" e="" ln="" x="" e="" ln="" x="" x="" e="" based="" on="" the="" previous="" results,="" what="" can="" be="" concluded?="" a.="" the="" graph="" of="" y="" lies="" below="" the="" tangent="" line="" for="" all="" x="" .="ln" x=""> e B. The graph of y lies above the tangent line for all x .= ln x > e C. The graph of y lies above the tangent line for all positive x .= ln x ≠ e D. The graph of y lies below the tangent line for all positive x .= ln x ≠ e c. Show that for all positive x .ln < xxe="" ≠="" e="" ln="" (x)="">< x="" e="">< x="" multiply="" both="" sides="" by="" .e="" ln="">< x="" use="" the="" power="" rule="" (1)="" less="" greater="" (2)="" faster="" more="" slowly="" (3)="" greater="" less="" (4)="" faster="" more="" slowly="" d.="" conclude="" that="" for="" all="" positive="" x="" .="" using="" the="" result="" of="" the="" previous="" step="" how="" can="" this="" conclusion="" be="" drawn?="" select="" the="" correct="" choice="" below="" and,="" if="" necessary,="" fill="" in="" the="" answer="" box="" to="" complete="" your="" choice.="" x="">< ee="" x="" ≠="" e="" a.="" raise="" to="" the="" power="" of="" each="" side="" of="" the="" inequality="" ,="" and="" simplify.ln="">< xxe="" b.="" raise="" each="" side="" of="" the="" inequality="" to="" the="" power="" of="" ,="" and="" simplify.ln="">< xxe c. this cannot be concluded as the statement is false. e. so, which is larger, or ?πe e π e π π e you answered: 2x 20. check whether each of the following functions is a solution of the differential equation .3 + 7yy′ = 4 e − x (a) y = e − x (b) y = e − x + e − (7 / 3)x (c) y c= e − x + e − (7 / 3)x (a) find , y, and for y .3y′ 7 3 + 7yy′ = e − x 3y′ = − 3 e − x y7 = 7 e − x 3 + 7yy′ = 4 e − x is the function y a solution of ? choose the correct answer below.= e − x 3 + 7yy′ = 4 e − x no yes (b) find , y, and for y .3y′ 7 3 + 7yy′ = e + e− x − (7 / 3)x 3y′ = − 3 e − 7 e− x − (7 / 3)x y7 = 7 e + 7 e− x − (7 / 3)x 3 + 7yy′ = 4 e − x is the function y a solution of ? choose the correct answer below.= e + e− x − (7 / 3)x 3 + 7yy′ = 4 e − x no yes (c) find , y, and for y c .3y′ 7 3 + 7yy′ = e − x + e − (7 / 3)x 3y′ = − 3 e − 7c e− x − (7 / 3)x y7 = 7 e + 7c e− x − (7 / 3)x 3 + 7yy′ = 4 e − x is the function y c a solution of ? choose the correct answer below.= e − x + e − (7 / 3)x 3 + 7yy′ = 4 e − x no yes 21. 22. 23. check whether the function y is a solution of the differential equation y with the initial condition . = e tan 6 e− xxe="" c.="" this="" cannot="" be="" concluded="" as="" the="" statement="" is="" false.="" e.="" so,="" which="" is="" larger,="" or="" πe="" e="" π="" e="" π="" π="" e="" you="" answered:="" 2x="" 20.="" check="" whether="" each="" of="" the="" following="" functions="" is="" a="" solution="" of="" the="" differential="" equation="" .3="" +="" 7yy′="4" e="" −="" x="" (a)="" y="e" −="" x="" (b)="" y="e" −="" x="" +="" e="" −="" (7="" 3)x="" (c)="" y="" c="e" −="" x="" +="" e="" −="" (7="" 3)x="" (a)="" find="" ,="" y,="" and="" for="" y="" .3y′="" 7="" 3="" +="" 7yy′="e" −="" x="" 3y′="−" 3="" e="" −="" x="" y7="7" e="" −="" x="" 3="" +="" 7yy′="4" e="" −="" x="" is="" the="" function="" y="" a="" solution="" of="" choose="" the="" correct="" answer="" below.="e" −="" x="" 3="" +="" 7yy′="4" e="" −="" x="" no="" yes="" (b)="" find="" ,="" y,="" and="" for="" y="" .3y′="" 7="" 3="" +="" 7yy′="e" +="" e−="" x="" −="" (7="" 3)x="" 3y′="−" 3="" e="" −="" 7="" e−="" x="" −="" (7="" 3)x="" y7="7" e="" +="" 7="" e−="" x="" −="" (7="" 3)x="" 3="" +="" 7yy′="4" e="" −="" x="" is="" the="" function="" y="" a="" solution="" of="" choose="" the="" correct="" answer="" below.="e" +="" e−="" x="" −="" (7="" 3)x="" 3="" +="" 7yy′="4" e="" −="" x="" no="" yes="" (c)="" find="" ,="" y,="" and="" for="" y="" c="" .3y′="" 7="" 3="" +="" 7yy′="e" −="" x="" +="" e="" −="" (7="" 3)x="" 3y′="−" 3="" e="" −="" 7c="" e−="" x="" −="" (7="" 3)x="" y7="7" e="" +="" 7c="" e−="" x="" −="" (7="" 3)x="" 3="" +="" 7yy′="4" e="" −="" x="" is="" the="" function="" y="" c="" a="" solution="" of="" choose="" the="" correct="" answer="" below.="e" −="" x="" +="" e="" −="" (7="" 3)x="" 3="" +="" 7yy′="4" e="" −="" x="" no="" yes="" 21.="" 22.="" 23.="" check="" whether="" the="" function="" y="" is="" a="" solution="" of="" the="" differential="" equation="" y="" with="" the="" initial="" condition="" .="e" tan="" 6="">