Calculus (Math Problem Sample)

CACULUS 1 Name: Institution: QUESTION 1: find the value of the integrals * 0ᴨ2cosxsin3xdx Solution Let u=cosx, du=-sinx, dv= sin3x dx then V= sin3x dx = -13sin2xcos x+23sin2xdx = 0ᴨ2cosxsin3xdx=uv-vdu =-cosx3{sin2xcosx+2sin2xdx)-vdu vdu= {-sinx (-13sin2xcosx+23sin2xdx}dx =13sin3xcos x dx+ 23sin2xsin2x But sin2vdy=12 sinxcosx+12sinx dx Thus 0ᴨ2cosx sin3dx= -cosx2(sin2xcosx+sinxcosx- sinx dx-130ᴨ2sin3xcos x dx + 0ᴨ2{23sin2xsin2x dx} dx This simplifies to 230ᴨ2-cos2 x sin2x3+ sinx cos2x3+ cos2x 3 130ᴨ2sin3xcosx dx=16-0ᴨ2cosxcos(2x) dx 0ᴨ2sin3xcosx dx=12-30ᴨ2cosxcos2xdx * (2x-3)√(x2 -3x+5) dx Solution Let u=x2 -3x+5, du=2x-3dx Substituting udu= u12du =23u32 Let u=x2+3x+5 Implying (2x-3)√(x2-3x+5) dx =23(x2+3x=5)32 QUESTION 2: Evaluate the integrals * (x4+e2x-1x+1+cos3xdx Solution Taking integrals separately =x4dx+e2xdx+1x+1+cos3x dx =15x5+12e2x-lnx+1+cos3x dx =15x5+12e2x-lnx+1+13sin3x * 12xdx+21√(1-x2dx =12lnx+ … Let x=sin u dx=cosudu =1(1-sin2 u) cosu du But, sin2x+cos2x=1 Therefore The expression can be rewritten as =1√cos2xcosu du =cosucosudu =du=u Since x=sinu, then sin-1x=u =12lnx+2sin-1x =12lnx+2arcsinx+c * 1 x2+1+4.(3)xdx d) (3x+1x4-4)dx Solution =1x2+1dx+43xdx =arctanx+43xdx =tan-1x+43xdx Let y=3x elny= eyln3 y=1ln3exln3 =tan-1x+4ln3exln3 Solution =x13dx+x-4dx-4dx =32x32-x-33-4x QUESTION 3: find the following limits, if exists. Otherwise, write DNE, ∞, −∞ limx-0(1-3x)4x Solution Let y=limx-0(1-3x)4x Then lny= limx-04xln1-3x =limx-04x(ln1lnx) But ln1=0, thus lny=0 Limit = 0 QUESTION 4: show that limx-∞ln(1+1x)x=1 Solution let y=limx-∞(1+1x)x ey=limx-∞(1+1x)ex but x approaches infitity, it tends to zero while as1x approaches infinity, it tends to zero Therefore limx-∞(1+1x)x= 1 QUESTION 5: Find the derivative of the following functions * fx=ex. sinxcosx11+t2dt Solution Integrating the function =extan-1t between sinx andcosx =extan-1(cosx)-tan-1sinx fx= ex{cosec2(cosx)sin-cosec2(sinx)cos(x) * fx=x+1x2+3xtant dt Solution fx=lnsecx from x+1 to x2+3x=lnsecx2+3x-sec⁡(x+1)lnsec(x2+3x)sec(x+1)Using quotient rule =cosec x2+3xtan(x2+3x)secx+1-secx2+3xsecx+1tan(x+1){sec(x+1)}2 QUESTION 6: Find the derivative of the following functions * gx=tan-1x+ tan-1(1x) b) gx=sin-1(1-x6)+3cos(x3-5x) Solution g'x=-cosec2x+2x2cosec2(1x) =-cosec2x+2x2cosec2(1x) Note sin-1x=11-x2 g'x= 11-x6 ×121-x6-12×6×5 +cos(x3-5x)ln3 =-3x51-x6 (1-x6)-32+-sinx3-5xln3 =-3x5(1-x6)-321-x6 -3x2-5sinx3-5xln3 QUESTION 7: by using the Right endpoint rule, Rn with n=5 rectangles, approximate the integral 01ex2dx Round your answer to 4 decimal places. Solution x 0 0.2 0.4 0.6 0.8 1.0 F(x) 1 1.04 1.17 1.43 2.27 2.73 01ex2 =12(1.04+1.17+1.43+2.27+2.71) =4.315 QUESTION 8: Use newton’s method to estimate the zero of the function f(x)=x4+x-3.Use x0=1 to find x4. Round your answer to 4 decimal places Solution xn=1= xn-fxnf'x0 , for n=0,1, 2… x1= x0-fx0f'(x0) f'x=4x3+x f'1=4+1=5 f1=1+1-3=-1 x1=1- -15 Answer = 65 or 1.2 QUESTION 9: Consider the following implicit equationꞥ x2cos2y-siny-x=0. Find dydx at the point (1,ᴨ) Solution dydx=2xcos2y-siny-1 At (1, ᴨ) dydx=2cos2ᴨ-sinᴨ-1 sinᴨ=0 dydx=-2-1=-3 QUESTION 10: Find dydx by using implicit differentiation: 2y4-xy+cos⁡(yx)=7 Solution dydx=d{2y4-xy+cosyx-7}dx =2y4-y+sinyx1x2-0 =2y4-y+1x2sin(yx) =y2y3-1+1x2sin(yx) QUESTION 11: Find the total area of the region between the function (...