\\\\(
\nonumber
\newcommand{\bevisslut}{$\blacksquare$}
\newenvironment{matr}[1]{\hspace{-.8mm}\begin{bmatrix}\hspace{-1mm}\begin{array}{#1}}{\end{array}\hspace{-1mm}\end{bmatrix}\hspace{-.8mm}}
\newcommand{\transp}{\hspace{-.6mm}^{\top}}
\newcommand{\maengde}[2]{\left\lbrace \hspace{-1mm} \begin{array}{c|c} #1 & #2 \end{array} \hspace{-1mm} \right\rbrace}
\newenvironment{eqnalign}[1]{\begin{equation}\begin{array}{#1}}{\end{array}\end{equation}}
\newcommand{\eqnl}{}
\newcommand{\matind}[3]{{_\mathrm{#1}\mathbf{#2}_\mathrm{#3}}}
\newcommand{\vekind}[2]{{_\mathrm{#1}\mathbf{#2}}}
\newcommand{\jac}[2]{{\mathrm{Jacobi}_\mathbf{#1} (#2)}}
\newcommand{\diver}[2]{{\mathrm{div}\mathbf{#1} (#2)}}
\newcommand{\rot}[1]{{\mathbf{rot}\mathbf{(#1)}}}
\newcommand{\am}{\mathrm{am}}
\newcommand{\gm}{\mathrm{gm}}
\newcommand{\E}{\mathrm{E}}
\newcommand{\Span}{\mathrm{span}}
\newcommand{\mU}{\mathbf{U}}
\newcommand{\mA}{\mathbf{A}}
\newcommand{\mB}{\mathbf{B}}
\newcommand{\mC}{\mathbf{C}}
\newcommand{\mD}{\mathbf{D}}
\newcommand{\mE}{\mathbf{E}}
\newcommand{\mF}{\mathbf{F}}
\newcommand{\mK}{\mathbf{K}}
\newcommand{\mI}{\mathbf{I}}
\newcommand{\mM}{\mathbf{M}}
\newcommand{\mN}{\mathbf{N}}
\newcommand{\mQ}{\mathbf{Q}}
\newcommand{\mT}{\mathbf{T}}
\newcommand{\mV}{\mathbf{V}}
\newcommand{\mW}{\mathbf{W}}
\newcommand{\mX}{\mathbf{X}}
\newcommand{\ma}{\mathbf{a}}
\newcommand{\mb}{\mathbf{b}}
\newcommand{\mc}{\mathbf{c}}
\newcommand{\md}{\mathbf{d}}
\newcommand{\me}{\mathbf{e}}
\newcommand{\mn}{\mathbf{n}}
\newcommand{\mr}{\mathbf{r}}
\newcommand{\mv}{\mathbf{v}}
\newcommand{\mw}{\mathbf{w}}
\newcommand{\mx}{\mathbf{x}}
\newcommand{\mxb}{\mathbf{x_{bet}}}
\newcommand{\my}{\mathbf{y}}
\newcommand{\mz}{\mathbf{z}}
\newcommand{\reel}{\mathbb{R}}
\newcommand{\mL}{\bm{\Lambda}}
\newcommand{\mnul}{\mathbf{0}}
\newcommand{\trap}[1]{\mathrm{trap}(#1)}
\newcommand{\Det}{\operatorname{Det}}
\newcommand{\adj}{\operatorname{adj}}
\newcommand{\Ar}{\operatorname{Areal}}
\newcommand{\Vol}{\operatorname{Vol}}
\newcommand{\Rum}{\operatorname{Rum}}
\newcommand{\diag}{\operatorname{\bf{diag}}}
\newcommand{\bidiag}{\operatorname{\bf{bidiag}}}
\newcommand{\spanVec}[1]{\mathrm{span}{#1}}
\newcommand{\Div}{\operatorname{Div}}
\newcommand{\Rot}{\operatorname{\mathbf{Rot}}}
\newcommand{\Jac}{\operatorname{Jacobi}}
\newcommand{\Tan}{\operatorname{Tan}}
\newcommand{\Ort}{\operatorname{Ort}}
\newcommand{\Flux}{\operatorname{Flux}}
\newcommand{\Cmass}{\operatorname{Cm}}
\newcommand{\Imom}{\operatorname{Im}}
\newcommand{\Pmom}{\operatorname{Pm}}
\newcommand{\IS}{\operatorname{I}}
\newcommand{\IIS}{\operatorname{II}}
\newcommand{\IIIS}{\operatorname{III}}
\newcommand{\Le}{\operatorname{L}}
\newcommand{\app}{\operatorname{app}}
\newcommand{\M}{\operatorname{M}}
\newcommand{\re}{\mathrm{Re}}
\newcommand{\im}{\mathrm{Im}}
\newcommand{\compl}{\mathbb{C}}
\newcommand{\e}{\mathrm{e}}
\\\\)
Exercise 1: Definite Integrals
A
Determine an indefinite integral to each of the functions
$$x^3\,,\,\,\frac{1}{x^3}\,\,\,\,\mathrm{and}\,\,\,\,\sin(3x-\frac{\pi}{2})\,.$$
Show hint
Note that the first example was considered in Exercise 1 this Long Day, and the next two are covered in Exercise 2 on the Long Day.
Show hint
Concerning the second example: Use the rewriting $\displaystyle{\frac{1}{x^3}=x^{-3}}\,.$
Show answer
$$\frac{1}{4}\,x^4\,,\,\,-\frac{1}{2x^2}\,\,\,\,\mathrm{and}\,\,\,\,-\frac{1}{3}\,\cos(3x-\frac{\pi}{2})\,.$$
B
Compute the following definite integrals
$$\int_0^{1}x^3\,\mathrm{d}x\,,\,\,\int_1^{2}\frac{1}{x^3}\,\mathrm{d}x\,\,\,\,\mathrm{and}\,\,\,\,\int_{-\frac{\pi}{2}}^{0}\sin(3x-\frac{\pi}{2})\,\mathrm{d}x\,.$$
Show answer
$$\frac{1}{4}\,,\,\,\frac{3}{8}\,\,\,\,\mathrm{and}\,\,\,\,\frac{1}{3} \,.$$
Exercise 2: Integration by Parts. By Hand
A
Determine an indefinite integral for the function $\,x\cos(x)\,,$ and check whether this is correct.
Show hint
Use Theorem 23.7 by setting $f(x)=\cos(x)$ and $g(x)=x\,.$
Show answer
$\cos(x)+x\sin(x)\,.$
Since $(\cos(x)+x\sin(x))’=-\sin(x)+\sin(x)+x\cos(x)=x\cos(x)\,$ we have shown that the indefinite integral is ok.
B
Determine the indefinite integral $\displaystyle{\,\int{t\e^t \mathrm dt}\,},$ and check that it is correct.
Show hint
Use Theorem 23.7 by setting $f(t)=\e^t$ and $g(t)=t\,.$
Show answer
$t\e^t-\e^t+\mathrm{konst}\,.$ Check by differentiation.
C
Determine an indefinite integral to the function $\,x^2\ln(x)\,,\,\,x>0\,.$
Show hint
Use Theorem 23.7 by setting $f(x)=x^2$ and $g(x)=\ln(x)\,.$
Show answer
$\displaystyle{\frac 13\, x^3\,\ln(x)-\frac 19\, x^3}\,.$
D
A first order linear differential equation is given by
$\,x’(t)-2x(t)=3t\,.$ Solve it using the general solution formula .
Show hint
You may need to be able to compute the indefinite integral
$$\displaystyle{\int e^{-2t}3t\mathrm dt}\,.$$
Show answer
$\displaystyle{x(t)= -\frac 32\,t-\frac 34+c\cdot\e^{2t}\,,\,\,c\in \reel\,}.$
Exercise 3: Integration by Substitution
To the questions in this Exercise, use the substitution formula
$$\int{f(g(x))g'(x)\mathrm dx}=\int{f(t)\,\mathrm dt}\,\,\,\mathrm{where}\,\,\,t=g(x)\,.$$
A
Determine an indefinite integral to $\,\displaystyle{x\e^{x^2}}\,.$
Show hint
Can you identify an inner function $g(x)$ and an outer function $f(x)\,?$
Show hint
If we put $\,g(x)=x^2\,$ and $\,f(x)=\e^x\,,$ we have
$$f(g(x))\cdot g'(x)=\e^{x^2}\cdot 2x=2\cdot(x\e^{x^2})\,.$$
Show answer
$\displaystyle{\int{x\e^{x^2}\,\mathrm dx}=\frac 12\,\,\int{f(t)\,\mathrm dt}=\frac 12\,\e^t=\frac 12\,\e^{x^2}\,.}$
B
Find the indefinite integral $\displaystyle{\int \frac{x}{x^2+1} \, \mathrm dx\,.}$
Show hint
Can you identify an inner function $g(x)$ and an outer function $f(x)\,?$
Show hint
If we put $\,g(x)=x^2+1\,$ and $\,f(x)=\displaystyle{\frac{1}{x}}\,,$ we have
$$f(g(x))g'(x)=2\,\frac{x}{x^2+1}\,.$$
Show answer
$\displaystyle{\int \frac{x}{x^2+1} \, \mathrm dx=\frac 12\,\,\,\int{f(t)\,\mathrm dt}=\frac 12\,\,\,\int{\frac{1}{t}}\,\mathrm dt=\ln(t)+k=\frac 12\ln(x^2+1)+k}\,.$
C
Find and indefinite integral to $\,\displaystyle{\frac{\sin (x)}{3 -\cos(x)}}\,$ and then determine $\displaystyle{\int_0^{\pi} \frac{\sin (x)}{3 -\cos(x)} \mathrm dx\,.}$
Show hint
Put $g(x)$ equal to the denominator, etc.
Show answer
$\displaystyle{\int_0^{\pi} \frac{\sin (x)}{3 -\cos(x)} dx = \ln(4)-\ln(2)=\ln(2)\,.}$
Exercise 4: Parametrization and a Curvilinear Integral. By Hand
Intro: If a curve in the $(x,y)$ -plane is given as the graph for a function
$$y=f(x)\,,\,\,x\in \left[a,b\right]$$
it is easy to state a parametric representation for the curve:
$$\begin{matr}{c}x\\\\y\end{matr}=\mathbf r(u)=\begin{matr}{c}u\\\\f(u)\end{matr}\,,\,\,u\in \left[a,b\right]\,.$$
A curve $K$ is given as a segment of the graph for the function $\ln(x)\,:$
$$K=\left\lbrace (x,y)\in\reel^2\,\vert\, y=\ln(x)\,,\,\,x\in\left[ 1\,,\,2\sqrt 2\right] \right\rbrace .$$
A
State a parametric representation for the curve and determine the Jacobi-function that belongs to the parametric representation
Show answer
The parametric representation: $\mr(u)=(\,u,\ln (u)\,)\,\,\mathrm{hvor}\,\ u\in\left[ 1\,,\,2\sqrt 2\right]\,.$
Jacobi$\displaystyle{(u)=|\,\mr’(u)\,|=\sqrt{1+\frac{1}{u^2}}=\frac{1}{u}\sqrt{u^2+1}}\,.$
B
Compute the curvilinear integral $\displaystyle{\int_Kx^2\,d\mu\,.}$
Show hint
$x$ must be substituted by the first coordinate $u\,$ of the parametric representation. By this (when we remember to multiply with the Jacobi-function) the integrand $\displaystyle{u\sqrt{u^2+1}}$ . Find an indefinite integral to the integrand by the method of substitution, by considering $g(x)=x^2+1$ as an inner function and $f(x)=\sqrt x$ as an outer function.
Show answer
$$\int_Kx^2ds=\int_1^{2\sqrt 2}u\sqrt {u^2+1}\,\mathrm dt=9-\frac{2\sqrt 2}{3}.$$
Exercise 5: Area and Volume. Advanced
Please enjoy the architect Norman Forster’s glass-skyscraber The Gherkin in London in the Figure below!
We have in the $\,(x,z)$ -plane in the space delimited a field $\,A\,$ by the coordinate axes and the graph for the function
$$x=f(z)=\frac 12\,\sqrt{-z^2+2z+3}\,,\,\,\,\,z\in\left[0,3\right]\,.$$
A
A solid model of The Gherkin appears when we rotate the field $\,A\,$ about the $\,z$ -axis by the angle $\,2\pi\,.$ Determine the volume of the model.
Show hint
There is probably a formula for such a volume $\ldots\,$
Show hint
The volume formula: $\displaystyle{\pi\int_{}^{}{f(z)^2}\,\mathrm dz }\,.$
B
Determine the area of the field $\,A\,.$
Show hint
A lot of acrobatics is needed in order to find an indefinite integral for $\,f\,!$ Set the inner function to $\,\displaystyle{g(z)=\frac 12 z -\frac 12\,}$ and the outer to $\,\displaystyle{h(z)=2\sqrt{1-z^2}}\,$ and determine $\,h(g(z))g’(z)\,.$
Show hint
Then you get $\,\displaystyle{\int f(z)\,\mathrm dz=2\int{\sqrt{1-t^2}}\mathrm dt}\,.$
Show hint
Now you may use this rewriting:
$$\sqrt{1-t^2}=\frac{1}{\sqrt{1-t^2}}+t\cdot \frac{-t}{\sqrt{1-t^2}}\,.$$
Remember or find also the derivative of $\,\mathrm{Arcsin}(t)\,.$
Show hint
$$\int \sqrt{1-t^2}\,\mathrm dt=\frac 12 t\sqrt{1-t^2}+\frac 12 \mathrm{Arcsin}(t)\,.$$
Show answer
The area: $\,\displaystyle{\frac 23\,\pi+\frac{\sqrt{3}}{4}}\,.$