175 – Midterm 2

Here is the second midterm.

Question 1: Find the volume of the solid obtained by rotating about the {y}-axis the region between the {x}-axis and the curve {y=\sec^2x} for {0\le x\le\pi/4}.

Here is a graph of {y=\sec^2x} in the specified region. To compute the volume, we use the shell method, and see that it is given by

\displaystyle  V=\int_0^{\pi/4} 2\pi x\cdot\sec^2x\,dx.

We evaluate this integral using integration by parts.

We take: \displaystyle u=2\pi x,\mbox{ so }du=2\pi\,dx, and dv=\sec^2x\,dx, so v=\tan x; so \displaystyle V=\left.2\pi x\tan x\right|_0^{\pi/4}-\int_0^{\pi/4}2\pi\tan x\,dx

\displaystyle=\left.2\pi x\tan x+2\pi\ln(\cos x)\right|_0^{\pi/4}

\displaystyle=\left(2\pi\cdot\frac{\pi}4\cdot1+2\pi\ln\left(\frac1{\sqrt2}\right)\right)-(0+2\pi\ln(1))

\displaystyle=\frac{\pi^2}2-\pi\ln2.

Question 2: Solve the initial value problem {\displaystyle (t^2+3t+3)\frac{dx}{dt}=1}, {x(3)=0}.

We use the technique of separation of variables, and rewrite the given differential equation as:

\displaystyle  dx=\frac{dt}{t^2+3t+3},

so

\displaystyle  x=\int \frac{dt}{t^2+3t+3}.

To solve this integral, we begin by completing the square in the denominator of the given fraction:

\displaystyle  t^2+3t+3=\left(t+\frac32\right)^2+\left(3-\frac94\right)=\left(t+\frac32\right)^2+\frac34.

This indicates that to solve the integral we want to use the trigonometric substitution

\displaystyle  t+\frac32=\frac{\sqrt3}2\tan\theta,

or

\displaystyle  \theta=\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right).

This gives: \displaystyle \left(t+\frac32\right)^2+\frac34=\frac34\sec^2\theta,

\displaystyle dt=\frac{\sqrt3}2\sec^2\theta\,d\theta,\mbox{ and}

\displaystyle \int \frac{dt}{t^2+3t+3}=\int\frac{(\sqrt3/2)\,d\theta}{3/4}

\displaystyle=\frac2{\sqrt3}\theta+C

\displaystyle=\frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right)+C. 

We conclude by finding the value of {C}, for which we use that {x(3)=0.} This gives us

\displaystyle  0=x(3)= \frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}3+\sqrt3\right)+C,

or

\displaystyle  C=-\frac2{\sqrt3}\tan^{-1}(3\sqrt3).

Putting this together, we have

\displaystyle  x(t)= \frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right)-\frac2{\sqrt3}\tan^{-1}(3\sqrt3).

Question 3: Find {\displaystyle \int\frac{x^3+5}{x^4-x^2}\,dx}.

To solve this integral, we use the technique of partial fractions decomposition, and begin by factoring

\displaystyle  x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1).

This means we need to find constants {A,B,C,} and {D} such that

\displaystyle  \frac{x^3+5}{x^4-x^2}=\frac Ax+\frac B{x^2}+\frac C{x-1}+\frac D{x+1}.

As seen in class, there are several ways of doing this. For example, we can begin by adding the fractions on the right hand side, to obtain

\displaystyle  \frac{Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)}{x^4-x^2}.

This means that

\displaystyle  x^3+5=Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)

for all values of {x}. When {x=0,} this gives

\displaystyle  5=-B,\mbox{ or }B=-5.

When {x=1}, we have

\displaystyle  6=2C,\mbox{ or }C=3.

When {x=-1}, we have

\displaystyle  4=-2D,\mbox{ or }D=-2.

Finally, when {x=2} we have

\displaystyle  13=6A-5\cdot3+3\cdot12-2\cdot4,

or {A=0.}

Putting this together, we have

\displaystyle  \frac{x^3+5}{x^4-x^2}=-\frac 5{x^2}+\frac 3{x-1}-\frac 2{x+1},

and

\displaystyle  \int\frac{x^3+5}{x^4-x^2}\,dx=\frac5x+3\ln|x-1|-2\ln|x+1|+C.

[Actually, although this was not addressed in lecture, we can use different values of {C} in each of the intervals {(-\infty,-1),} {(-1,0),} {(0,1),} and {(1,\infty)}.]

Question 4: We have that {\displaystyle\int_0^1\frac4{1+t^2}\,dt=\pi}. Suppose we do not know the value of {\pi}. We can use the trapezoidal rule to approximate it. Find a value of {n} such that the theoretical error for the trapezoidal rule using {n} subdivisions is strictly smaller than {1/10}. Find the approximation given by the trapezoidal rule using this {n}.

The error {E_T} using the trapezoidal rule with {n} subintervals is bounded above by

\displaystyle  \frac{M_2(b-a)^3}{12n^2},

where {a=0,} {b=1,} and {M_2=\max_{0\le t\le 1}|f''(t)|} where {\displaystyle f(t)=\frac4{1+t^2}.}

To compute a bound on {M_2,} we first find {f''(t):} \displaystyle f'(t)=\frac{-8t}{(1+t^2)^2},

\displaystyle f''(t)=\frac{32t^2-8(1+t^2)}{(1+t^2)^3}

\displaystyle =\frac{24t^2-8}{(1+t^2)^3}.

An easy way of bounding this expression is to note that {24t^2-8} is increasing for {0\le t\le 1,} with value {-8} at {t=0,} and value {16} for {t=1,} so the maximum of its absolute value is {16.} Similarly, {\displaystyle\frac1{(1+t^2)^3}} is decreasing, so it maximum occurs at {t=0}, where it is 1.

Hence {M_2} is bounded above by {\displaystyle \frac{16}1=16,} and

\displaystyle  E_T\le \frac{16}{12n^2}=\frac4{3n^2}.

If we want {E_t<1/10,} it suffices that

\displaystyle  \frac4{3n^2}<\frac1{10},

or

\displaystyle  n^2>40/3=13.33\dots,

so {n\ge4.}

The approximation {T_4} is given by

\displaystyle  \frac {1/4}2\left(4+\frac8{1+\frac1{16}}+\frac8{1+\frac4{16}}+\frac8{1+\frac9{16}}+2\right)\approx3.131

which is indeed an accurate approximation to {\pi} within {1/10.}

[Here is a graph of {f''(t)} for {0\le t\le 1.} A more careful computation of {M_2} would have revealed that {M_2=8,} which gives {n\ge3.} (Note that {\max_{0\le t\le 1} f''(t)=2,} but {M_2} looks at the maximum of {|f''(t)|,} not just {f''(t)}.)

The approximation {T_3} gives the value {3.123\dots}. A less careful computation, replacing {24t^2-8} with {24t^2+8} (so we do not have to worry about both positive and negative numbers) gives {n\ge6}, with {T_6\approx3.137.} Actually, {T_2=3.1} works as well but the theoretical error does not predict this.]

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