Exercises
1.Evaluate :
Int dx/[(x+2)(3-x) from 1 to -1
2.Find int 3^[(2x+1)^1/2]dx
Show that the surface x^2-2yz+y^3= 4 is perpendicular to any member of the family of surfaces x^2+1=(2-4a)y^2+az^2 at the point of intersection (1,-1,2)!
Evaluate
Int dx/ (5+3cos x) by using the substitution tan x/2= u
Prove that 1/1.3 +1/3.5 + 1/5.7+ … =sigma nequals 1 to infinite 1/(2n-1)(2n+1) converge and find its sum.
Answered By Heni Ayu Pertiwi:
1.Method 1
Int dx/[(x+2)(3-x)]^1/2 = int dx/ (6+x-x^2)^1/2 = int dx /[ 6- (x^2-x)]^1/2 = int dx / [ 25/4 –(x-1/2)^2]^1/2
Letting x-1/2= u, this becomes
Int du/ (25/4 – u^2)^1/2 = sin ^-1 u/ (5/2) + c = sin^-1 (2x-1/5) + c
Then
Int dx/ [(x+2)(3-x)]^1/2 from -1 to 1 = sin^-1 (2x-1/5) from -1 to 1 = sin^-1(1/5) – sin^-1(3/5) = sin^-1 . 2 + sin^-1. 6
Method 2
Let x-1/2=u, as in method 1, now whwn x= -1, u=-3/2, and whwn x=1, u=1/2. Thus,
Int dx/[(x+2)(3-x)]^1/2 from -1 to 1 = int dx / [ 25/4 –(x-1/2)^2]^1/2 from -1 to 1= int du/[25/4 – u^2] from -3/2 to ½ = sin^-1 u/(5/2) from -3/2 to 1/2 = sin^-1 . 2 + sin^-1. 6
2.Let (2x+1)^1/2 = y , 2x+1= y^2 then dx = y. dy and the integral becomes
Int 3^y. ydy integrate by parts, letting u= y, dv= 3^y
Then du=dy, v= 3^y/ (ln 3) , and we have
Int 3^y.y dy = int u dv – int v du
=y.3^y/(ln 3) – int 3^y/ (ln 3) dy
=y.3^y/(ln 3) – 3^y/(ln 3)^2 + c
3.Let the equations of the two surfaces be written in the form
H= x^2 – 2.y.z + y^3 -4=0 and K=x^2+1-(2-4a)y^2-az^2=0
Then,
Vf= 2xi+(3y^2-2z)i-2yk, Vk=2xi-2(2-4a)yj-2azk
Thus, the normals to the two surfaces at (1,-1,2) are given by
N1= 2i-j+2k
N2= 2i +2(2-4a)j-4ak
Since, N1.N2=2.2 -2(2-4a)-2.4a=4-4+8a-8a=0
It follows that N1 and N2 are perpendicular for all a and so the required result follows.
4.Sin x/2 = 4/(1+u^2)^1/2, cos x/2 =1/(1+u^2)^1/2
Then, cos x= cos^2 x/2 – sin^2 x/2= 1-u^2/ 1+u^2
Also du= ½ sec^2 x/2 dx
dx=2cos^2 x/2 du= 2du/1+u^2
thus the integral becomes int du/u^2+u =i/2 tan^-1 u/2 +c +1/2 tan^-1 (1/2 tanx/2) +c
5.Un= 1/(2n-1)(2n+1)= ½[1/(2n-1) – 1/(2n+1)], then
Sn= U1+U2+U3+u4+….+Un=1/2(1- 1/3)+1/2(1/3 – 1/5)+…+1/2(1/2n-1 – 1/2n+1) = ½ (1-1/3+1/3-1/5+1/5-…+1/(2n-1)-1/(2n+1))= ½(1 – 1/2n+1)
Since limit n approache to infinite Sn = limit n approache to infinite ½(1- 1/2n+1)=1/2, the series converges and its sum is 1/2
