Sample Midterm 2
1) Estimate the area under the curve “2 to the power of x2“ between x=-2 and x=2 using
the trapezoidal rule with 4 intervals. Give a precise theoretical bound to the error.
2) Find the area of the surface of revolution obtained by rotating f(x)=sin(x) about the xaxis from 0 to π.
3) Find the centroid of the region shown:
4) Find the slope of the tangent line to the curve r=θ at θ= π.
5) Determine if the series is convergent or not and if yes – find its sum.
* {-2, 5/2, -25/8, 125/32 …. }
* Σ∞i=1(3n8-n+1)
* Σ∞i=1(1 / n(n+1)(n+2))
6) Use the integral test to determine if the series is convergent.
* Σ∞i=1(ne-n)
* Σ∞i=2(1 / n(ln n)2)
Solutions for Sample Midterm 2
1) f(x) = ”2 to the power of x2“ = 2exp(x2), a=-2, b=2, n=4.
∆x = (b-a) / n = (2-(-2)) / 4 = 4 / 4 = 1. x0 = -2; x1 = -1; x3 = 0; x4 = 1; x5 = 2.
Trapezoidal = ∆x/2 * [ f(x0) + 2f(x1) + 2f(x3) + 2f(x4) + f(x5) ] =
= ½ * [ f(-2) + 2f(-1) + 2f(0) + 2f(1) + f(2) ] =
= ½ * [ 24 + 2*21 + 2*20 + 2*21 + 24 ] = ½ * [ 16 + 4 + 1 + 4 + 16 ] =
= 41 / 2 = 20.5
Error ≤ M2*(b-a) *∆x2/12 .
M2 = max |(f ’(x))| within (a,b). f ’(x) = 2exp(x2) * ln2 * 2x.
This naturally maximizes on the interval at x=2 (or -2). So M2 = 24 * ln2 * 4 = 72 ln2.
So the error ≤ 72 ln2 * 1 / 12 = 6 ln2 ≈ 4.159.
2) f(x)=sin(x); f ‘ (x)=cos(x). Surface of revolution S = 2πa∫bf(x)√(1+(f ‘ (x))2) dx
S = 2π0∫π sin(x)√(1+(cos(x))2)dx = [u=cos(x) du=sin(x)dx] = 2π1∫-1√(1+u2)du =
= (u/2) √(1+u2) + (1/2)ln(u + √(1+u2)) 1|-1 = (1/2) √2 + (1/2)ln(1+√2) – 0 – (1/2)ln(1) =
(√2 + ln(1+√2))/2
3) A= a∫bf(x)dx = -2∫0f(x)dx + 0∫3f(x)dx + 3∫4f(x)dx = -2∫0 3 dx + 0∫3 (3-2+1)dx + 3∫4 3 dx =
= 3x -2|0 + 2x 0|3 + 3x 3|4 = 3(0-(-2)) + 2(3-0) + 3(4-3) = 6 + 6 + 3 = 15
_
x = (1/A) a∫bxf(x)dx = (1/A) [-2∫0xf(x)dx + 0∫3xf(x)dx + 3∫4xf(x)dx] =
= (1/15) [-2∫0 3x dx + 0∫3 (3-2+1)x dx + 3∫4 3x dx] =
= (1/15) [(3/2)x2 -2|0 + (2/2)x2 0|3 + (3/2)x2 3|4 = (1/15) [(3/2)(0-4) + (9-0) + (3/2)(16-9)] =
= (1/15) [-6 + 9 + 21/2] = (1/15)(27/2)= 27/30 = 9/10
_
y = (1/2A) a∫bf2(x)dx = (1/A) [-2∫0 f2(x)dx + 0∫3 f2(x)dx + 3∫4 f2(x)dx] =
= (1/30) [-2∫0 32 dx + 0∫3 (32-22+12) dx + 3∫4 32 dx] = (1/30) [-2∫0 9dx + 0∫3 6dx + 3∫4 9dx] =
= (1/30) [9x -2|0 + 6x 0|3 + 9x 3|4 = (1/30) [ 9(0-(-2)) + 6(3-0) + 9(4-3)] =
= (1/30) [18 + 18 + 9] = (1/30)(45)= 45/30 = 1.5
4) r = θ; x = r cos(θ) = θcos(θ); y = r sin(θ)= θsin(θ);
dy/dθ = cos(θ) – θsin(θ); dx/dθ = sin(θ) + θcos(θ);
dy/dx = (dy/dθ)/(dx/dθ) = (cos(θ) – θsin(θ)) / (sin(θ) + θcos(θ)).
dy/dx (θ=π) = (cos(π) – π sin(π)) / (sin(π) + π cos(π)) = (-1 - 0)/(0 + -π) = -1/-π = 1/π.
5) For geometric series: ∑∞i=1 crn-1 converges to S = c/(1-r) for |r|<1 & diverges otherwise.
a) c = a1= -2; r = (a2 / a1) = (5/2) / -2 = -5/4; |r| = |-5/4| = 5/4 >1 ,so the series diverges.
b) ∑∞i=1(3n8-n+1) = ∑∞i=1(3n8-(n-1)) = ∑∞i=1(3n/8(n-1)) = ∑∞i=1(3*3(n-1)/8(n-1)) =
= ∑∞i=13*(3(n-1)/8(n-1)) = ∑∞i=13*(3/8)(n-1)
c= 3; r= 3/8; |r|= 3/8 < 1 so the series converges to S= 3 / (1-(3/8))= 3 / (5/8) =24/5.
c) ∑∞i=1(1 / n(n+1)(n+2)) use partial fraction to turn into telescopic series
a/n + b/(n+1) + c/(n+2) = 1/n(n+1)(n+2)
a(n+1)(n+2) + b(n)(n+2) + c(n)(n+1) = 1
a(n2+3n+2) + b(n2+2n) + c(n2+n) = 1
n2(a+b+c) + n(3a+2b+c) + 2a = 1
2a = 1 → a = ½;
subtract (a+b+c) = 0 from 3a+2b+c = 0; and get 2a + b = 0; → 1 + b = 0 → b = -1;
(a+b+c) = 0 → (½ - 1 + c) = 0 → c = ½.
So ∑∞i=1(1 / n(n+1)(n+2)) = ∑∞i=1 ½ [1/n – 2/(n+1) + 1/(n+2)]
This is a telescopic series in which the only terms that don’t cancel out are:
1 – (2/2) + 1/2, thus the total sum is ½ [1 – (2/2) + 1/2] = 1/2 * 1/2 = 1/4
to see that this is a telescopic series we will write all the elements with same
denominator in the same column:
1/n
: 1 1/2 1/3 1/4 1/5 1/6
-2/(n+1):
-2/2 -2/3 -2/4 -2/5 -2/6
1/(n+2) :
1/3 1/4 1/5 1/6
Sum
: 1 -1/2 0
0
0
0 … = 1 – 1/2 = 1/2
6) a. ∑∞i=1(ne-n) an= ne-n; f(n) = ne-n; f(x) = xe-x
Look at 1∫∞ f(x) dx if it converges so does our sum and vice versa.
∞
∞
-x
-x
-x
-x
1∫ f(x) dx = 1∫ xe dx [by parts u=x dv=e dx; du=dx v=∫e dx=-e dx]
= -xe-x1|∞ - 1∫∞-e-xdx = [-xe-x - e-x] 1|∞ = lim (a → ∞) [-xe-x - e-x] 1|a
= lim (a → ∞) [-ae-a - e-a - (-1e-1 - e-1)] = [0 – 0 - (-e-1 - e-1)] = 2e-1
*
By l’Hopital: lim (a→∞) [-ae-a] = -lim (a→∞) [a/ea] = -lim (a→∞) [1/-e-a] = 1/∞ = 0.
b. Σ∞i=2(1 / n(ln n)2) an=1 / n(ln n)2; f(n) =1 / n(ln n)2; f(x) = 1 / x(ln x)2;
Look at 2∫∞ f(x) dx if it converges so does our sum and vice versa.
∞
∞
2
2∫ f(x) dx = 2∫ 1/x(ln x) dx [substitute u=ln x du=1/x dx]
∞
2
-1
∞
= ln2∫ 1/u dx = [-u ] ln2| = lim (a → ∞) [-u-1] ln2|a = lim (a → ∞) [-a-1 – -ln2-1] =
0 + ln2-1 = ln2-1