Title: Numeerinen integrointi ja geometrisia sovellutuksia

The graphs of the functions \(f,g\), together with the \(x\)-axis and vertical lines \(x=c\) define sets in the plane. Determine the areas of these sets, whenFunktioiden \(f,g\) kuvaajat rajaavat yhdessä \(x\)-akselin ja pystysuorien \(x=c\) kanssa tasojoukkoja. Määritä niiden pinta-alat, kun

(a) \[f(x)=1/x^{[value]},~~~0 < x \leq 1 \] (b) \[g(x)=1/x^{[value]}, \ \ \ x\geq 1. \]

Give your answers in exact form. For \(\infty\), type \({\tt inf}\).Anna vastaukset tarkkoina arvoina. \(\infty\) merkitään \({\tt inf}\).

AnswerVastaus (a):

AnswerVastaus (b):
Feedback: Model solution: Mallivastaus: (a) \[\begin{align*} A =&\lim_{\varepsilon \to 0^{+}} \int_{\varepsilon}^{1} \frac{1}{x^[value]} \, dx \\ =&\lim_{\varepsilon \to 0^{+}} \int_{\varepsilon}^{1} x^{[value]} \, dx \\ \\ =&\lim_{\varepsilon \to 0^{+}} \bigg/_{\!\!\!\varepsilon}^{\,1} -\frac{x^{[value]}}{[value]} \\ \\ =&\lim_{\varepsilon \to 0^{+}} \bigg/_{\!\!\!\varepsilon}^{\,1} [value] \\ \\ =&\lim_{\varepsilon \to 0^{+}} \left( -\frac{1}[value]+\underbrace{\frac{1}{[value] \varepsilon}}_{\infty} \right) \\ \\ \\ =&\underline{\underline{\infty}} \\ \end{align*}\] (b) \[\begin{align*} A =&\int_{1}^{\infty} \frac{1}{x^{[value]}} \\ \\ =&\bigg/_{\!\!\!1}^{\,\infty} [value] \\ \\ \\ =&-\underbrace{\frac{1}{\infty}}_{0}+\frac{1}{[value]} \\ =&\underline{\underline{[value]}}. \\ \end{align*}\]

Määritä käyrän \(y=f(x)=\frac{[value]x^[value]}{x^[value]+1}\), \(x\)-akselin sekä suorien \(x=-1\) ja \([value]\) rajoittaman äärellisen alueen pinta-ala.

Anna vastauksesi tarkkana arvona. Luonnollinen logaritmi a:sta merkitään ln(a), esim. ln(2).

Calculate the finite area restricted by the curve \(y=f(x)=\frac{[value]x^[value]}{x^[value]+1}\), \(x\) -axis and the lines \(x=-1\), \([value]\).

Give your answer as exact value.The natural logarithm of a is written as ln(a), e.g. ln(3).

Pinta-ala onThe area is .
Feedback: Malliratkaisu: Koska \(y=f(x)=\frac{[value]x^[value]}{x^[value]+1}\le0\), kun \(-1\le x\le 0\), ja \(y=f(x)=\frac{[value]x^[value]}{x^[value]+1}>0\), kun \( 0 < x \le [value] \), niin The model solution: Because \(y=f(x)=\frac{[value]x^[value]}{x^[value]+1}\le0\) when \(-1\le x\le 0\), and \(y=f(x)=\frac{[value]x^[value]}{x^[value]+1}>0\) when \( 0 < x \le [value] \), it follows that \[ \begin{align*} A&=\int\limits_{-1}^{[value]}\left|\frac{[value]x^[value]}{x^[value]+1}\right|\,dx =\int\limits_{-1}^{0}(-\frac{[value]x^[value]}{x^[value]+1})\,dx+\int\limits_{0}^{[value]}\frac{[value]x^[value]}{x^[value]+1}\,dx\\ &={\Bigl/}_{\!\!-1}^{\,0}(-[value])+{\Bigl/}_{\!\!0}^{\,[value]}[value]\\ &=-\ln(1)+\ln(2)+\ln([value])-\ln(1)=\ln([value]). \end{align*} \] Kuva: Illustration: [value]

The curve \(y=[value]\), where \(0\leq x \leq [value]\) revolves around the \(x\)-axis. Determine the volume of this solid. Käyrä \(y=[value]\), missä \(0\leq x \leq [value]\), pyörähtää \(x\)-akselin ympäri. Määritä näin saadun kappaleen tilavuus.

The symbol \(\pi\) is written as \({\tt pi}\).Symboli \(\pi\) merkitään \({\tt pi}\).

Answer: Vastaus:
Feedback: Model solution: Mallivastaus: \(y={[value]}\). Method 1Tapa 1 \begin{eqnarray*} V & = & [value]\pi \int_{0}^{[value]} (\sin(x))^2 \, \textrm{d}x \\ & = & [value]\pi \int_{0}^{[value]} \frac{1}{2}(1-\cos(2x)) \, \textrm{d}x \\ & = & [value]\cdot \frac{1}{2}\pi ([value]-0)-[value]\cdot \frac{1}{4}\pi (\sin([value])-\sin(0)) \\ & = & [value]. \end{eqnarray*} Method 2Tapa 2 Applying integration by parts (\(\int_a^{b} u(x)v'(x) \, \textrm{d}x= |_{a}^{b} u(x)v(x)-\int_{a}^{b} u'(x)v(x) \, \textrm{d}x\))Osittaisintegroinnilla (\(\int_a^{b} u(x)v'(x) \, \textrm{d}x= |_{a}^{b} u(x)v(x)-\int_{a}^{b} u'(x)v(x) \, \textrm{d}x\)) \(u(x)=\sin(x) \ \ \ u'(x)=\cos(x)\) \(v'(x)=\sin(x) \ \ \ v(x)=-\cos(x)\). \begin{eqnarray*} \int_{0}^{[value]} \sin(x) \sin(x) \, \textrm{d}x & = & \bigg|_{0}^{[value]} (-\cos(x))\sin(x)-\int_{0}^{[value]} (-\cos(x))\cos(x) \, \textrm{d}x \\ \int_{0}^{[value]} \sin^2(x) \, \textrm{d}x & = & \bigg|_{0}^{[value]} (-\cos(x)) \sin(x)+\int_{0}^{[value]} (1-\sin^2(x)) \, \textrm{d}x \\ 2 \int_{0}^{[value]} \sin^2(x) \, \textrm{d}x & = & -\bigg|_{0}^{[value]} \cos(x)\sin(x)+\int_{0}^{[value]} 1 \, \textrm{d}x \\ \int_{0}^{[value]} \sin^2(x) \, \textrm{d}x & = & \frac{1}{2}([value]+[value])=[value] \\ \Rightarrow [value]\pi \int_{0}^{[value]} \sin^2(x) \, \textrm{d}x & = & [value]. \end{eqnarray*}

The arc of the parabola \(y=[value]x^2, \ \ \ 0 \leq x \leq [value]\) revolves around the \(y\)-axis. Find the volume of this parabolic cup. Hint: Integrate the cross section of the cup in the form \[V= \int_{0}^{[value]} A(y) \, \textrm{d}y. \] Give your answer in exact form. The symbol \(\pi\) is written as \({\tt pi}\). Paraabelin kaari \(y=[value]x^2, \ \ \ 0 \leq x \leq [value]\) pyörähtää \(y\)-akselin ympäri. Laske näin saadun paraboloidisen maljan tilavuus. Vihje: Integroi maljan poikkileikkauksen pinta-alaa muodossa \[V= \int_{0}^{[value]} A(y) \, \textrm{d}y. \] Anna vastaus tarkkana arvona. Symbolia \(\pi\) merkitään \({\tt pi}\).

Answer: Vastaus:
Feedback: Model solution: Mallivastaus: \[y=[value]x^2, \ \ \ 0 \leq x \leq [value]. \] \begin{eqnarray*} x^2 & = & \frac{1}{[value]} y \ \ \Arrowvert \sqrt{} \\ x & = & \sqrt{\frac{1}{[value]}y} \ \ \ \text{inverse functionkäänteisfunktio} \end{eqnarray*} WhenKun \(x=0, \ \ \ y=0,\) Whenja kun \(x=[value], \ \ \ y=[value]\). \begin{eqnarray*} A & = & \int_{0}^{[value]} \pi \frac{1}{[value]} y \, \textrm{d}y \\ & = & \pi \frac{1}{[value]} \bigg|_{0}^{[value]} \frac{y^2}{2} \\ & = & \pi \frac{1}{[value]} \left( \frac{[value]^2}{2}-\frac{0^2}{2} \right) \\ & = & \pi \cdot \frac{1}{[value]}\cdot [value] \\ & = & [value] \end{eqnarray*}