Consider the region in the first quadrant bounded above by the hyperbola \(y = 1/x,\)
below by the \(x\)-axis, left by the line \(x = m,\) and right by the line \(x = 2m.\)
Prove that the area of the region is independent of \(m.\)
SOLUTION
The integral expression for the area is
\[\int_m^{2m} \frac{1}{x} \di x \pd\]
By FTC II, we find an antiderivative of \(1/x\)—one antiderivative is \(\ln \abs x\)—and
evaluate it at \(2m\) and \(m.\)
Doing so gives
\[\ln \abs x \intEval_m^{2m} = \ln(2m) - \ln m = \ln \par{\frac{2m}{m}} = \ln 2\]
Thus, the area is fixed at \(\ln 2\)
and does not depend on \(m.\)
EXERCISE 2
A constant force \(F\) is applied to a point on the rim of a circular, uniform, solid disk of mass \(m\) and radius \(R\)
(Figure 1).
This force supplies a constant angular acceleration of
\[\alpha = \frac{2F}{mR} \pd\]
The angular velocity, \(\omega,\) describes how fast the disk spins,
while the angular displacement, \(\theta,\) represents the angle by which a point on the rim has rotated.
If the disk starts from rest with no initial rotation, then derive an expression for \(\theta(t),\)
the disk's angular displacement as a function of time.
SOLUTION
The angular displacement \(\theta,\)
angular velocity \(\omega,\)
and angular acceleration \(\alpha\) are
analogous to a particle's position, velocity, and acceleration when moving in a straight line.
Hence, \(\alpha = \textderiv{\omega}{t}\) and \(\omega = \textderiv{\theta}{t}.\)
Since \(\alpha = 2F/mR,\) we have the differential equation
\[
\deriv{\omega}{t} = \frac{2F}{mR} \pd
\]
All quantities on the right are constants, so the general solution is
\[\omega = \frac{2F}{mR} t + C \pd\]
Because the disk starts from rest, \(\orange{\omega(0) = 0} \col\)
\[\orange{\omega(0)} = 0 + C = \orange{0} \implies C = 0 \pd\]
But \(\omega = \textDeriv{\theta}{t},\) so we have
\[
\deriv{\theta}{t} = \frac{2F}{mR} t \cma
\]
or
\[\theta = \frac{F}{mR} t^2 + D \pd\]
Since the disk has no initial rotation, \(\orange{\theta(0)} = \orange{0}.\)
Substituting this initial condition shows
\[\orange{\theta(0)} = 0 + D = \orange{0} \implies D = 0 \pd\]
Hence, the angular displacement function is
\[\theta(t) = \frac{F}{mR} t^2 \cma\]
as requested.
EXERCISE 3
For any nonzero constant \(a,\) let
\[g(x) = \frac{x \sin x}{(x^2 + a^2) \ln (1 + \atan 9x)} \pd\]
For \(x \in (0, \pi),\) show that
\[0 \leq \int_0^\pi g(x) \di x \leq \int_0^\pi g(x) \csc x \di x \pd\]
SOLUTION
On \((0, \pi),\) note that \(x,\) \(\sin x,\) and \((x^2 + a^2)\) are all positive.
Additionally, \(\atan 9x \gt 0,\) so \(\ln(1 + \atan 9x) \gt \ln 1 = 0.\)
Hence, the function
\[g(x) = \frac{x \sin x}{(x^2 + a^2) \ln (1 + \atan 9x)}\]
is positive on \((0, \pi).\) In addition,
\[g(x) \csc x = \frac{x}{(x^2 + a^2) \ln (1 + \atan 9x)} \pd\]
Because \(0 \leq \sin x \leq 1\) on \((0, \pi),\) we have
\[0 \leq \frac{x \sin x}{(x^2 + a^2) \ln (1 + \atan 9x)} \leq \frac{x}{(x^2 + a^2) \ln (1 + \atan 9x)} \pd\]
Inequalities are preserved with definite integrals, so
\[0 \leq \int_0^\pi \frac{x \sin x}{(x^2 + a^2) \ln (1 + \atan 9x)} \di x \leq \int_0^\pi \frac{x}{(x^2 + a^2) \ln (1 + \atan 9x)} \di x \pd\]
This analysis shows that
\[0 \leq \int_0^\pi g(x) \di x \leq \int_0^\pi g(x) \csc x \di x \pd\]
EXERCISE 4
Antidifferentiate the following function:
\[f(x) = (\csc x) \sin \par{e^x} + x e^x (\csc x) \cos \par{e^x} - x (\cos x) (\csc^2 x) \sin \par{e^x} \pd \]
SOLUTION
We rewrite the third term of \(f,\) attaining
\[f(x) = (\csc x) \sin \par{e^x} + x e^x (\csc x) \cos \par{e^x} + x (-\csc x \cot x) \sin \par{e^x} \pd \]
The right side is a Product Rule expansion—namely,
\[f(x) = \deriv{}{x} \parbr{x \csc x \sin \par{e^x}} \pd \]
Thus, the antiderivative is
\[F(x) = \boxed{x (\csc x) \sin \par{e^x} + C}\]
EXERCISE 5
The left side of a uniform plane wall is maintained at \(40 \celcius,\)
while the right side is maintained at \(20 \celcius.\)
The wall is \(0.1\) meter thick and neither generates nor stores heat.
Let the \(x\)-axis be positioned such that \(x = 0\) is the left side and \(x = 0.1\)
is the right side.
(See Figure 2.)
By the heat conduction equation,
the wall's temperature \(T\) as a function of \(x\) satisfies the differential equation
\[\derivOrder{T}{x}{2} = 0 \pd\]
For \(0 \leq x \leq 0.1,\)
find \(T(x),\) the temperature distribution through the wall.
What is the temperature at the midpoint of the wall thickness—that is,
at \(x = 0.05 \ques\)
If the wall is made of a different material, such as one with a greater insulation,
then does the answer in part (b) change?
SOLUTION
Integrating the differential equation \(\textderivOrder{T}{x}{2} = 0\) twice gives
\[
\ba
\deriv{T}{x} &= C \nl
T(x) &= Cx + D \cma
\ea
\]
where \(C\) and \(D\) are constants.
Note the initial conditions \(T(0) = 40\) and \(T(0.1) = 20.\)
Substituting \(\orange{T(0) = 40}\) shows
\[\orange{T(0)} = 0 + D = \orange{40} \implies D = 40 \pd\]
Then substituting \(\orange{T(0.1) = 20}\) yields
\[\orange{T(0.1)} = 0.1 C + 40 = \orange{20} \implies C = -200 \pd\]
Hence,
\[T(x) = \boxed{-200x + 40}\]
The temperature at the center of the wall is
\[T(0.05) = -200(0.05) + 40 = \boxed{30 \celcius}\]
No.
The temperature profile \(T(x) = -200x + 40\)
is true for any wall material in the given context because
the differential equation \(\textderivOrder{T}{x}{2} = 0\)
shows no reference to any material properties, such as conductivity.
Trust in the math!
EXERCISE 6
Evaluate the following limit if it exists:
\[\lim_{x \to 0} \frac{2}{x^2} \int_0^{x} \sin u \di u \pd \]
SOLUTION
We treat this expression as a fraction whose numerator is \(2 \int_0^x \sin u \di u\)
and whose denominator is \(x^2.\)
Each function approaches \(0\) as \(x \to 0,\)
so this limit is of the indeterminate form \(\indZero.\)
Because both functions are differentiable,
we apply L'Hopital's Rule:
FTC I gives the derivative of the numerator to be \(2 \sin x,\)
and differentiating the denominator gives \(2x.\)
So our limit is
\[\lim_{x \to 0} \frac{2 \sin x}{2x} = \lim_{x \to 0} \frac{\sin x}{x} = \boxed 1\]
You may remember \(\lim_{x \to 0} (\sin x)/x = 1;\)
if not, then you can apply L'Hopital's Rule again to attain this result.
EXERCISE 7
Let \(S\) be the region in the first quadrant bounded by the graph of \(y = \sqrt x\)
and the line \(x = m.\)
The value of \(m\) is increasing at a constant rate of \(3\) units per minute.
How quickly is the area of \(S\) increasing with time when \(m = 16 \ques\)
SOLUTION
The expression for the area of \(S\) is
\[A = \int_0^m \sqrt x \di x \pd\]
Let \(t\) be time; then we require \(\textDeriv{A}{t}\)
when \(m = 16.\)
But we need a way to relate this rate to the given rates.
(See Section 2.7
to review the procedure of working with related rates.)
Note that \(\textDeriv{m}{t} = 3.\)
By the Chain Rule,
\[\deriv{A}{t} = \deriv{A}{m} \deriv{m}{t} = 3 \deriv{A}{m} \pd\]
Then FTC I gives
\[\deriv{A}{m} = \sqrt m \pd\]
So when \(m = 16,\) we have
\[\deriv{A}{t} \intEval_{m = 16} = 3 \sqrt{16} = \boxed{12}\]
The area is increasing at a rate of \(12\) square units per minute.
EXERCISE 8
Let \(Q\) be a quantity that changes with time \(t\)
and whose value at \(t = 0\) is \(Q_0.\)
Suppose that \(Q(t)\) increases at a rate of \(f(t)\) and decreases at a rate of \(g(t),\)
where \(f\) and \(g\) are differentiable functions.
Prove that \(Q\) has a relative maximum at the critical number \(t = c\) when \(f'(c) \lt g'(c).\)
SOLUTION
By the Second-Derivative Test (see Section 3.3),
\(Q\) has a relative maximum at a critical number \(c\) if \(Q''(c) \lt 0.\)
We therefore aim to show this.
The net rate of change of \(Q\) is \(f - g,\)
so by the Net Change Theorem
\[Q(t) = Q_0 + \int_0^t \parbr{f(u) - g(u)} \di u \pd\]
Note that \(u\) is a "dummy variable"; \(t\) is already used for the upper bound of the integral.
By FTC I we differentiate to obtain
\[Q'(t) = f(t) - g(t) \pd\]
Because \(c\) is a critical number, we have \(Q'(c) = 0 \col\)
\[Q'(c) = f(c) - g(c) = 0 \implies f(c) = g(c) \pd\]
Also, \(f\) and \(g\) are differentiable, so
\[Q''(t) = f'(t) - g'(t) \pd\]
Thus, \(Q''(c) \lt 0\) is satisfied if \(f'(c) \lt g'(c).\)
EXERCISE 9
If \(\int_0^1 x \sin \par{x^3} \di x = K,\)
then calculate \(\int_0^1 x^5 \sin \par{x^9} \di x\) in terms of \(K.\)
SOLUTION
To simplify \(\int_0^1 x^5 \sin \par{x^9} \di x,\)
we substitute \(u = x^3.\)
Then \(\dd u = 3x^2 \di x,\)
from which \(\dd x = \dd u/(3x^2).\)
When \(x = 0,\) \(u = 0;\)
when \(x = 1,\) \(u = 1.\)
Thus,
\[
\ba
\int_0^1 x^5 \sin \par{x^9} \di x &= \tfrac{1}{3} \int_0^1 u \sin \par{u^3} \di u \nl
&= \boxed{\frac{K}{3}}
\ea
\]
EXERCISE 10
Let \(k\) and \(m\) be positive numbers.
Show that with \(n\) subintervals, an endpoint approximation
estimates
\(\int_m^{2m} \dd x/x^k\)
with an error bound given by
\[
\abs{E_n} \leq \frac{k}{2n m^{k - 1}} \pd
\]
SOLUTION
If \(f(x) = 1/x^k,\) then we have
\[f\,'(x) = -kx^{-k - 1} = -\frac{k}{x^{k + 1}} \pd\]
The maximum value of \(\abs{f\,'(x)} = k/x^{k + 1}\)
on \([m, 2m]\) is
\[\abs{f\,'(m)} = \frac{k}{m^{k + 1}} \pd\]
[Note that \(\abs{f\,'(m)}\) is large when the denominator is small,
so \(x = m\) minimizes the denominator over our interval.]
So the error bound in the endpoint approximation is
\[
\abs{E_n} \leq \frac{\ds \par{\frac{k}{m^{k + 1}}} (2m - m)^2}{2n}
= \frac{\ds \par{\frac{k}{m^{k + 1}}} m^2}{2n}
= \frac{k}{2n m^{k - 1}} \cma
\]
as requested.
EXERCISE 11
Prove that Simpson's Rule always provides exact approximations
to definite integrals of cubic functions.
SOLUTION
A cubic function is of the form
\[f(x) = Ax^3 + Bx^2 + Cx + D \pd\]
We want to prove that \(\int_a^b f(x) \di x\)
can be calculated exactly using Simpson's Rule with any even \(n.\)
The error bound to Simpson's Rule is
\[\abs{E_S} \leq \frac{M_4 (b - a)^5}{180 n^4} \cma\]
where \(\abs{f^{(4)}(x)} \leq M_4\) on \([a, b].\)
Simpson's Rule is exact if and only if \(E_S = 0\) for all even \(n.\)
Taking derivatives of \(f,\) we see
\[
\ba
f\,'(x) &= 3Ax^2 + 2Bx + C \cma \nl
f\,''(x) &= 6Ax + 2B \cma \nl
f\,'''(x) &= 6A \cma \nl
f^{(4)}(x) &= 0 \pd
\ea
\]
Since the maximum value of \(\abs{f^{(4)}(x)}\) is always \(0,\)
we have \(M_4 = 0.\)
Hence, \(E_S = 0\) and so the error in Simpson's Rule is \(0\)
for all even \(n.\)
EXERCISE 12
This exercise examines the Weierstrass substitution,
famously called the world's sneakiest substitution.
Prove the trigonometric identities
\[
\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}
\and \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \pd
\]
(Hint: Start with the double-angle identities for sine and cosine;
then use the Pythagorean identity \(1 + \tan^2 \theta\) \(= 1/\cos^2 \theta.\))
If \(t = \tan(x/2),\) then show that
\[\dd x = \frac{2}{t^2 + 1} \di t \pd\]
Use the substitution \(t = \tan(x/2)\) to evaluate
\[\int_0^{\pi/2} \frac{1}{1 + \sin x + \cos x} \di x \pd\]
SOLUTION
By the double-angle identity for sine,
\[
\ba
\sin x &= 2 \sin \tfrac{x}{2} \cos \tfrac{x}{2} \nl
&= \frac{\dfrac{2 \sin \tfrac{x}{2} \cos \tfrac{x}{2} }{\cos^2 \tfrac{x}{2}}}{\dfrac{1}{\cos^2 \tfrac{x}{2}}} \nl
&= \frac{2 \tan \tfrac{x}{2}}{1 + \tan^2 \tfrac{x}{2}} \cma
\ea
\]
as requested.
Likewise, by the double-angle identity for cosine,
\[
\ba
\cos x &= \cos^2 \tfrac{x}{2} - \sin^2 \tfrac{x}{2} \nl
&= \frac{\dfrac{\cos^2 \tfrac{x}{2} - \sin^2 \tfrac{x}{2}}{\cos^2 \tfrac{x}{2}}}{\dfrac{1}{\cos^2 \tfrac{x}{2}}} \nl
&= \frac{1 - \tan^2 \tfrac{x}{2}}{1 + \tan^2 \tfrac{x}{2}} \cma
\ea
\]
as requested.
If \(t = \tan(x/2),\) then \(x = 2 \atan t\) and so
\[
\dd x = \frac{2}{t^2 + 1} \di t \cma
\]
as requested.
Note that
\[
\sin x = \frac{2t}{1 + t^2}
\and \cos x = \frac{1 - t^2}{1 + t^2} \pd
\]
When \(x = 0,\) \(t = \tan 0 = 0.\)
When \(x = \pi/2,\) \(t = \tan(\pi/4) = 1.\)
So the integral becomes
\[
\ba
\int_0^1 \frac{1}{\ds 1 + \frac{2t}{t^2 + 1} + \frac{1 - t^2}{t^2 + 1}} \par{\frac{2}{t^2 + 1}} \di t
&= \int_0^1 \frac{2}{t^2 + 1 + 2t + 1 - t^2} \di t \nl
&= \int_0^1 \frac{2}{2t + 2} \di t \nl
&= \int_0^1 \frac{1}{t + 1} \di t \nl
&= \ln \abs{t + 1} \intEval_0^1 \nl
&= \boxed{\ln 2}
\ea
\]
