B U Departmet of Mathematics Math Calculus I Sprig 5 Fial Exam Calculus archive is a property of Boğaziçi Uiversity Mathematics Departmet. The purpose of this archive is to orgaise ad cetralise the distributio of the exam questios ad their solutios. This archive is a o-profit service ad it must remai so. Do ot let ayoe sell ad do ot buy this archive, or ay portio of it. Reproductio or distributio of this archive, or ay portio of it, without o-profit purpose may result i severe civil ad crimial pealties. Questio Fid the volume of the solid obtaied by revolvig the regio eclosed by the graphs of f(x) = ta x, g(x) = si x ad x = π about the x axis. 4 Solutio. ta x = si x holds for x =, kπ, k = ±, ±,... but oly for x = both graphs eclose a regio with x = π π/4 4. The V = π (ta x si x). Use ta x = sec x ad si cos x x =. V = π π/4 π/4 cos x ) cos x + ) π/4 si x x + (sec x = π (sec x 3 = π ta x 3 4 = π ta π 4 3 π si π/4 + 4 4 = π 3π 8 + 4 5 = π 4 3π 8 (ta + si 4 )
Questio Show that the fuctio f(x) = x 4 + x 3 has exactly oe zero i,. Solutio. First } ote that f(x) is everywhere cotiuous ad differetiable beig a polyomial. f() = f() = ad f() > hece f, beig cotiuous, has at least oe zero i (, ). f() = To have at least two zeros, f = must hold at some poit i (, ) as f is differetiable, ad further f must chage sig. Now check f = 4x 3 + 6x = x = or x = 3/. Hece f, which is also cotiuous, ever chages sig i (, ). Namely f is either always icreasig or decreasig. Now usig the fact that f() <, f() > we uderstad f is icreasig o,. Thus f has exactly oe zero i,. Questio 3 If x si πx = x x f(t)dt, where f is a cotiuous fuctio ad x >, fid f(). Solutio. We differetiate both sides by usig the Fudametal Theorem of Calculus: si πx + πx cos πx = f(x )x f( x) x. Substitutig x = : π cos π = f() f() π = 3 f() f() = π 3. Questio 4 Evaluate the followig itegrals: (a) x 3 + 4x Solutio. Let I = x 3 + 4x. x 3 + 4x = x(x + 4) = A x + Bx + C by partial fractios x + 4 A(x + 4) + (Bx + C)x = Ax + 4A + Bx + Cx = for each x A + B = C = 4A = 4 A = B = If u = x + 4, the x = du ad if x = ta θ, the = sec θ ad hece I = ( x x x + 4 + x + 4 ) = x du sec u + θdθ 4 sec θdθ = l x l u + θ + c = l x l() + arcta x + c = l x + arcta x + c.
x + x Solutio. Let I = x + x. Let x = ta θ. The = sec θ. sec θdθ I = ta θ sec θ = cos θ cos θ si θ dθ = csc θdθ = l csc θ + cot θ + c. x + csc θ = = si θ x cot θ = I = l x + + x x + c. x Questio 5 Evaluate the followig defiite itegrals: / arcsi x (a) + x Solutio. Call the itegral expressio I ad itegrate by parts. Let u = arcsi x ad dv = du = (valid sice x /; v = + x. So we get: x +x. The I = uv / / v du = + x arcsi x / 3 = π 6 / x = 6 π 6 + 4 x / = π + 4 6 4 = π + 4 4. 6 / + x x
l x x du =, v = x x = lim x l x A This is a improper itegral equal to = lim. So by the method of itegratio by parts: A + sice l A A ad A A x x = lim as A. A l A x A = lim l x. Let l x = u, dv =, the x x l A A A + = Questio 6 Determie whether the followig series are coverget or diverget: (a) = Take. Now lim = where ad. Hece both series coverge ad diverge = together. Sice is diverget, so is. = = = ( + )! 3 (!) This is typical ratio test: a = a + a = ( + )! 3 + ( + )!( + )! Sice < the series is coverget. ( + )! 3 (!) > 3!! ( + )! = + 3( + )( + ) as. (c) + ( ) =. If this series is coverget, fid its sum. This is sum of two geometric series oe with r =, the other with r = hece for both r: r < is satisfied; therefore the series is coverget. = = + ( ) = = = ( ) + = + + = = ( ) = ( ) 3 = 3 = 5 3. = + ( ) = = ( ) + + Questio 7 (a) Fid the Taylor series of f(x) = l(x + ) aroud the poit x =.
Taylor series of l( + x) = f() = f () = x + = x= f () = f () = = (x + ) = x= (x + ) 3 = x= f () () x! about x =. ( ) ( )! x! = = = x ( ) is the required Taylor series ear x =.. f () () = ( ) ( )! (x + ) = ( ) ( )! x= For which values of x is the Taylor series foud above coverget? First we apply absolute covergece test (ratio test): a + a = x + + x = x x as. + The series is coverget whe x < ad diverget whe x > by ratio test. For x =, i.e. x = ± this test is icoclusive. x = : ( ) ( ) = = = ( ) Hece at x = the series is diverget. = = }{{} diverget harmoic series x = +: () ( ) = ( ) = = Hece at x = the series is coverget. is a alteratig series with > ad as. Iterval of covergece is (,.