MA114 Exam 2 Solution Fall 2007
This is a solution of the second exam in MA114 Sections 001 to 006.
The statements of the problems are not verbatim. Also, problem 8 parts c
and d did not actually appear on the exam (Instead the points for them were added to parts a and b of the same problem.)
- (10 points) Respond as indicated to each of the following. You do not need to show work for this problem.
- Among the choices below, circle the substitution that would be the most appropriate to compute `int sqrt(1-x^2) dx`.
- `u=sin(x)`
- `x=tan(u)`
- `x=sin(u)`
- `u = e^x`
- `x = u^4
The correct answer is (iii) `x=sin(u)`. This substitution allows `sqrt(1-x^2)` to be replaced with `cos(u)` if one chooses the domain to be
`u in [-pi/2, pi/2]`.
- Circle the correct choice. If `T(x) = 3 - 7(x - a) + t(x - a)^2` is the degree 3 Taylor polynomial forthe function f(x) at x = a, then the equation of the tangent line to the graph of `f(x)` at `x=a` is:
- `y = -7 + 5(x-a)`
- `y=3-7(x-a)`
- `y=3 - 7x`
The correct answer is (ii) `y = 3 - 7(x-a)` because the right hand side is equal to `f(a) + f'(a)(x-a)`.
- Answer "true" or "false". If `p(x)` is any polynomial, then the partial fraction decomposition of `f(x) = (p(x))/(x^3+x)` can be written in the form `f(x) = q(x) + (Ax+B)/(x^2+1) + C/x` where `A`, `B`, and `C` are constants and `q(x)` is a polynomial.
This is true.
- Answer "true" or "false". If `{a_n}` is a sequence and `lim_(n->oo) a_n = L`, then for every `epsilon > 0`, there is an `N` such that if `|L-a_n| < epsilon`, then `n > N`.
This is false.
- Answer "true" or "false". The partial fraction decomposition of `f(x) = (x^2)/((x-a)^2(x-2))` is of the form `f(x) = A/(x-1) + (Bx)/(x-1) + C/(x-2)` where `A`, `B`, `C` are constants.
This is false.
- (10 points) Use partial fractions to evaluate `int 1/((x-1)(x-2)) dx`. SHOW YOUR WORK.
The partial fraction decomposition is of the form
`1/((x-1)(x-2)) = A/(x-1) + B/(x-2)`
After clearing denominators, one has `1= A(x-2) + B(x-1)`. Substituting `x=2`
gives `B=1` and substituting `x=1` gives `A = -1`. So, one has
`int (dx)/((x-1)(x-2)) = int -1/(x-1) + 1/(x-2)\ dx = -ln(|x-1|) + ln(|x-2|) + C = ln(|(x-2)/(x-1)|) + C`
- (10 points) Calculate the partial fraction expansion of `f(x) = (x^3+x+1)/(x(x^2+1))`. DO NOT integrate the result. SHOW YOUR WORK.
Because the numerator has degree greater than that of the denominator, one starts by doing the division to get a quotient of 1 with
a remainder of 1. The decomposition would be of the form:
`f(x) = 1 + 1/(x(x^2+1)) = 1 + A/x + (Bx+C)/(x^2+1)`
where `A`, `B`, and `C` are constants. Clearing the denominators in the
remainder part gives:
`1 = A(x^2+ 1) + (Bx + C)x`
Substituting `x= 0` gives `A = 1`. Substituting `x = 1` and `x = -1` gives
`1 = 2 + B + C` and `1 = 2 + B - C`. Adding the two equations gives
`2 = 4 + 2B` or `B = -1`. Subtracting the two equations gives `C = 0`.
So the partial fraction decomposition is
`f(x) = 1 + 1/x - x/(x^2 + 1)`
- Evaluate each of the following integrals. SHOW YOUR WORK.
- (5 points) `int sin^3(x+2) cos(x + 2) dx`
Substitute `u = sin(x+2)`. One has `du = cos(x+2) dx` and so
`int sin^3(x+2) cos(x+2) dx = int u^3 du = u^4/4 = (sin^4(x+2))/4 + C`
- (5 points) `int sin(3x)cos(5x) dx`
The addition formulas for sine give:
`sin(3x + 5x) = sin(3x)cos(5x) + cos(3x)sin(5x)`
and
`sin(3x - 5x) = sin(3x)cos(5x) - cos(3x)sin(5x)`
Adding the two identities gives
`sin(8x) + sin(-2x) = 2sin(3x)cos(5x)`
So,
`int sin(3x)cos(5x)dx = int 1/2(sin(8x)-sin(2x))dx = -1/16 cos(8x) + 1/4 cos(2x) + C`
- Evaluate each of the following. SHOW YOUR WORK!
- (5 points) `int sin^2(x) dx`
Subtract `cos(2x) = cos^2(x) - sin^2(x)` from `1 = cos^2(x) + sin^2(x)` to get
`1 - cos(2x) = 2 sin^2(x)`
Using this, one can integrate:
`int sin^2(x) dx = int (1 - cos(2x))/2 dx = x/2 - (sin(2x))/4 + C`
- (5 points) `int_0^(1/3) 1/sqrt(9x^2+1) dx`
Substitute `3x = tan(theta)` for `theta in (-pi/2, pi/2)`. One has
`3dx = sec^2(theta) d theta` and the limits `x = 0, 1/3` correspond to
`theta = 0, pi/4` respectively. The substitution yields:
`int_0^(1/3) 1/sqrt(9x^2+1) dx = int_0^(pi/4) (sec^2(theta) d theta)/(3 sqrt(tan^2(theta) + 1)) = int_0^(pi/4) (sec^2(theta) d theta)/(3 |sec(theta)|)`
Since `theta in [-pi/2,pi/2]`, one has `|sec(theta)| = sec(theta)` and so
the integral simplifies to
`1/3 int_0^(pi/4) sec(theta) d theta = 1/3 int_0^(pi/4) (sec^2(theta) + sec(theta)tan(theta))/(sec(theta) + tan(theta)) d theta = 1/3{:ln(|sec(theta) + tan(theta)|)|_0^(pi/4) = 1/3 ln(sqrt(2) + 1) - ln(1) = 1/3 ln(sqrt(2)+1)`
So, the value of the integral is `1/3 ln(sqrt(2)+1) approx 0.2937911956`.
- SHOW YOUR WORK on all parts of this problem.
- (4 points) Calculate the Taylor polynomial of degree 3 at `x=0` for `f(x) = ln(1-x)`.
Start by calculating the derivatives and their values at `x=0`:
Function | Value at 0 |
`f(x) = ln(1-x)` | 0 |
`f'(x) = -1/(1-x)` | -1 |
`f''(x) = -1/((1-x)^2)` | -1 |
`f'''(x) = -2/((1-x)^3)` | -2 |
The Taylor polynomial of order 3 at `x=0` is
`T_3(x) = f(0) + f'(0)x + f''(0)(x^2)/2 + f'''(0) (x^3)/6 = 0 - x - (x^2)/2 - (x^3)/3`
- (4 points) Calculate the Taylor polynomial of degree 2 for `f(x) = 1 + 2x + 3x^2 + 5x^3` at `x = 1`.
The table of derivatives and their values at `x = 1` is:
Function | Value at 1 |
`f(x) = 1 + 2x + 3x^2 + 5x^3` | 11 |
`f'(x) = 2 + 6x + 15 x^2` | 23 |
`f''(x) = 6 + 30 x` | 36 |
The Taylor polynomial of degree 2 at `x=1` is:
`T_2(x) = f(1) + f'(1)(x-1) + f''(0)((x-1)^2)/2 = 11 + 23(x-1) + 18(x-1)^2`
- (2 points) Calculate the remainder `R_2(x)` at `x=1` when `f(x) = 1 + 2x + 3x^2 + 5x^3`. You can either give the exact value or you can provide an upper bound for its absolute value.
We already have `T_2(x)` and we can extend it to `T_3(x)`: One has `f'''(x) = 30` and so
`f(x) = T_3(x) = T_2(x) + f'''(1)((x-1)^3)/6 = T_2(x) + 5(x-1)^3`
Now, we have
`f(x) = T_2(x) + R_2(x) = T_3(x) = T_2(x) + 5(x-1)^3`
and so `R_2(x) = 5(x-1)^3`.
Alternatively, one can use `|f'''(x)| = 30 = M` to get
`|R_2(x)| <= (M |x-1|^(3))/(3!) = 5|x-1|^3`
- SHOW YOUR WORK on all parts of this problem.
- (5 points)
- Calculate the trapezoid rule estimate for `int_(-1)^3 x^2 + x dx` with two subintervals.
Let `f(x) = x^2 + x`. One has `(b-a)/n = 2` and so one needs to calculate: `f(-1) = 0`, `f(1) = 2`, and `f(3) = 12`. The trapezoid rule gives
the approximation:
`(f(-1) + 2 f(1) + f(3)) (b-a)/(2n) = (0 + 4 + 12)4/4 = 16`
- Use the appropriate error formula to calculate the maximum error in this estimate.
One has `f''(x) = 2` and so one can let `K=2` in the estimate:
`K/12 ((b-a)^3)/(n^2) = 2/12 (4^3)/4 = 8/3`
Note that the exact value of the integral is 40/3 and so this is an example
where the error is the largest possible.
- (5 points)
- Calculate Simpson's rule estimate for `int_(-1)^3 x^3 + x\ dx` for 4 subintervals.
Let `f(x) = x^3 + x`. One has `Delta = (b-a)/n = 1` and so, one needs to calculate: `f(-1) = -2`, `f(0) = 0`, `f(1) = 2`, `f(2) = 10`, and `f(3) = 30`. The Simpson's rule estimate is:
`(Delta)/3 (f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)) = 1/3(-2+0+4+40+30) = 72/3 = 24`
- Use the appropriate error formula to calculate the maximum error in this estimate.
One has `f^(4)(x) = 0` since `f(x)` is only of degree 3. So, the
error upper bound of `K/180 ((b-a)^5)/n^4` is zero because we can take
`K = 0`.
- The table below gives the values of a function `f(x)` and its first four derivatives at selected values of `x`. Use this information to answer the four parts of the question that follow. Provide a brief explanation and/or calculations that make it clear how you arrived at your answer.
x | -2 | -1 | 0 | 1 | 2 |
`f(x)` | 7/6 | 3/2 | 5/6 | 1/6 | -15/2 |
`f'(x)` | -5/6 | 0 | -5/6 | -4/3 | -39/2 |
`f^(2)(x)` | 22/3 | -2 | 2/3 | -14/3 | -38 |
`f^(3)(x)` | -22 | 0 | 2 | -16 | -54 |
`f^(4)(x)` | 32 | 12 | -8 | -28 | -48 |
- (3 points) Calculate the Taylor polynomial of degree 3 for `f(x)` at `x=0`. Do not simplify your answer.
One has
`T_3(x) = f(0) + f'(0)x + f''(0)(x^2)/2 + f'''(0)(x^3)/6 = 5/6 -5/6 x + 1/3 x^2 + 1/3 x^3`
- (3 points) Calculate the Taylor polynomial of degree 3 for `f(x)` at `x=1`. Do not simplify your answer.
One has
`T_3(x) = f(1) + f'(1)(x-1) + f''(1)((x-1)^2)/2 + f'''(1)((x-1)^3)/6 = 1/6 -4/3 (x-1) - 7/3 (x-1)^2 - 8/3 (x-1)^3`
- (2 points) Let `h(x) = 3(x-1) f(x)`. Calculate the Taylor polynomial of degree 3 for `h(x)` at `x=1`. Do not simplify your answer.
The Taylor polynomial is `3(x-1)` times the answer to the last part:
`T_3(x) = 3(x-1)T_2(x) = 3(x-1)(1/6 -4/3 (x-1) - 7/3 (x-1)^2) = (x-1)/2 - 4(x-1)^2 -7(x-1)^3`
- (2 points) Calculate the Taylor polynomial of degree 3 for `f'(x)` at `x=0`.
Do not simplify your answer.
One has:
`T_3(x) = f'(0) + f''(0)x + f'''(0)(x^2)/2 + f^(4)(0)(x^3)/6 = -5/6 + 2/3 x + x^2 - 4/3 x^3`
- (10 points) On the line after each of the following, enter one of the following: (a) the value to which the sequence converges, (b) `oo`, (c) `-oo`, or (in case none of the others apply) (d) "does not exist" or simply "divergent". In each case provide a brief expllanation and/or supporting calculations for your answer.
- `lim_(n->oo) ((n-3)n)/(2n^2+1)`
The value is `1/2` because
`lim_(n->oo)((n-3)n)/(2n^2+1) = lim_(n->oo) (1 - 3/n)/(2+1/(n^2)) = 1/2`
- `lim_(n->oo) 2^(cos(n pi))`
Since `cos(n pi)` alternates between -1 and 1, the terms alternate between `1/2` and `2`. So the limit does not exist.
- `lim_(n->oo) 3^(sin((n pi)/(2n+5)))`
One has `lim_(n->oo) (n pi)/(2n+5) = lim_(n->oo) pi/(2+5/n) = pi/2` and `sin(pi/2) = 1`. So, the limit is 3.
- `lim_(n->oo) (2 sin(n))/sqrt(n+1)`
The numerator is at most 2 in absolute value and the denominator
approaches infinity as `n->oo`. So, the limit is 0.
- `lim_(n->oo) (-1)^n e^((n+1)/n)`
One has `lim_(n->oo) (n+1)/n = lim_(n->oo) (1+1/n) = 1` and
so the subsequence of even index terms approach e whereas the subsequence of
odd index terms approach -e. These not being equal, the limit does not exist.
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