Title: Integraalifunktio ja määrätty integraali

Määrää sellainen funktion \[f(x)=[value]\] integraalifunktio, jonka kuvaaja kulkee origon \((0,0)\) kautta. Luonnollinen logaritmi merkitään ln, esim. ln(2). Arcustangentti merkitään atan, esim. atan(1). Itseisarvo merkitään abs, esim. abs(x).

Find the integral function of the function \[f(x)=[value]\] the graph of which is passing through the origin \((0,0)\). The natural logarithm of a is written as ln(a), e.g. ln(3). The arcus tangent of b is written as atan(b), e.g. atan(1). The absolute value of c is written as abs(c), e.g. abs(-2).

Integraalifunktio onThe integral function is .
Feedback: Malliratkaisu: The model solution: \[ F(x) = \int f(x)\,\mathrm{d}x = \int [value]\,\mathrm{d}x \] Osamurtokehitelmä: Partial fraction expansion: \[ \begin{align*} f(x) &=[value] = [value]+[value] \\ &= [value] \\ &=[value] . \end{align*} \] Osoittajia vertaamalla saadaan yhtälöryhmä The system of equations follow by comparing the numerators: \[ \left\{\begin{array}{rrrllll} A&+B&&=0&|\,\cdot1&&\\ &[value]B&+C&=[value]&&&\\ A&&+[value]C&=0.&|\,\cdot(-1)&&\\ \end{array}\right. \] Eliminoidaan \(A\) kertomalla puolittain sopivasti ja laskemalla yhteen. Eliminate \(A\) by multiplying both sides of the equation appropriately and adding together. \[ \left\{\begin{array}{rrll} B&-[value]C&=0&|\,\cdot [value]\\ [value]B&+C&=[value]&|\,\cdot(-1)\\ \end{array}\right. \] Eliminoidaan \(B\) kertomalla puolittain sopivasti ja laskemalla yhteen. Eliminate \(B\) by multiplying both sides of the equation appropriately and adding together. \[ ([value])C=-[value], \] josta \(C=[value]\). Edelleen sijoittamalla saadaan \(A=-[value]\) ja \(B=[value]\). Tällöin which gives \(C=[value]\). By substituting it follows \(A=-[value]\) ja \(B=[value]\). Hence \[ f(x) =[value]+[value]. \] \[ \begin{align*} F(x) &= \int f(x)\,\mathrm{d}x = \int ([value]+[value]+[value])\,\mathrm{d}x\\ &=[value]+[value]+[value]+C. \end{align*} \] Integraalifunktion kuvaaja kulkee origon \((0,0)\) kautta, joten The graph of the integral function is passing through the origin \((0,0)\) and consequently \[ \begin{align*} F(0) =[value]+[value]+[value]+C &=0\\ C &=[value]. \end{align*} \] Kysytty integraalifunktio on The integral function is \[ F(x) = [value]+[value]+[value]. \]

Determine a function \(y\) such that for all \(x\) it holds thatMääritä sellainen funktio \(y\), että kaikilla \(x\) pätee

\[y(x)=[value]+x^{[value]}\int_{0}^{[value]} y(t) \, \textrm{d}t. \]

Hint: Based on the equation conclude that \(y\) is of the form \(y(x)=[value]+Ax^{[value]}\), where \(A\) is a constant. Substitute an expression of this form to the given equation and determine a suitable value for \(A\).Vihje: Päättele, että yhtälön perusteella \(y\) on muotoa \(y(x)=[value]+Ax^{[value]}\), missä \(A\) on vakio. Sijoita tällainen lauseke yhtälöön ja tutki, miten \(A\) täytyy valita.

\(y(x)=~\)
Feedback: Model solution: Mallivastaus: \(y\colon [0,[value]]\to \mathbb{R} \ \ \ \ x\in [0,[value]]\) \[y(x)=[value]+x^{[value]}\underbrace{\int_{0}^{[value]}y(t) \, \textrm{d}t}_{A} \ \ \Vert y(x)=[value]+Ax^{[value]} \] \begin{eqnarray*} y(x) & = & [value]+x^{[value]}\int_{0}^{[value]}([value]+At^{[value]}) \, \textrm{d}t \\ & = & [value]+x^{[value]}\left( \int_{0}^{[value]}[value] \, \textrm{d}t + \int_{0}^{[value]}At^{[value]} \, \textrm{d}t\right) \\ & = & [value]+x^{[value]}\left( \bigg/_{\!\!\!0}^{\,[value]}[value]t+\bigg/_{\!\!\!0}^{\,[value]}\frac{1}{[value]}At^{[value]}\right) \\ & = & [value]+x^{[value]}(([value]-0)+(\frac{1}{[value]}A\cdot [value]^{[value]}-\frac{1}{[value]}A\cdot 0^[value])) \\ & = & [value]+x^{[value]}([value]+[value]A) \end{eqnarray*} \begin{eqnarray*} \Rightarrow A & = & [value]+[value]A \\ [value]A & = & [value] \\ A & = & [value] \end{eqnarray*} \[\Rightarrow y(x)=[value] \]

Compute the improper integralLaske epäoleellinen integraaliLaske epäoleellinen integraali

\[ \int_{[value]}^{[value]} \textrm{e}^{-x} [value] \, \textrm{d}x. \]

Give your answer in exact form! Anna vastaus tarkkana arvona!

\(\int_{[value]}^{[value]} \textrm{e}^{-x} [value] \, \textrm{d}x=~\)
Feedback: Model solution: Mallivastaus: \[\int_{0}^{\infty} \textrm{e}^{-x} {[value]} \, \textrm{d}x. \] Integration by parts: Osittaisintegrointi: \(\displaystyle\int_a^{b} u(x)v'(x) \, \textrm{d}x=\bigg/_{\!\!\!\!a}^{\,b} u(x)v(x)-\int_{a}^{b} u'(x)v(x) \, \textrm{d}x\) \(u(x)=\textrm{e}^{-x} \Rightarrow u'(x)=-\textrm{e}^{-x}\) \(v'(x)= [value] \Rightarrow v(x)= -\frac{1}{[value]} \cos({[value]}x)\) \begin{eqnarray*} \int_{[value]}^{\infty} \textrm{e}^{-x} {[value]} \, \textrm{d}x & = & \bigg/_{\!\!\!\!0}^{\,\infty} -\textrm{e}^{-x} \cdot \frac{1}{[value]} \cdot \cos([value]x)+\int_{0}^{\infty} \textrm{e}^{-x} \cdot \frac{1}{[value]}\cos([value]x) \, \textrm{d}x \\ & = & \frac{1}{[value]}+\frac{1}{[value]} \int_{0}^{\infty} \textrm{e}^{-x} \cos([value]x) \, \textrm{d}x \end{eqnarray*} Applying integration by parts again, we getKäytetään osittaisintegrointia uudestaan. Nyt \(u(x)=\textrm{e}^{-x} \Rightarrow u'(x)=-\textrm{e}^{-x} \) \(v'(x)= \cos([value]x) \Rightarrow v(x)=\frac{1}{[value]} \sin([value]x)\) \begin{eqnarray*} \frac{1}{[value]}+\frac{1}{[value]} \int_{0}^{\infty} \textrm{e}^{-x} \cos([value]x) \, \textrm{d}x & = & \frac{1}{[value]}+\frac{1}{[value]} \bigg(\bigg/_{\!\!\!\!0}^{\,\infty} \frac{1}{[value]} \textrm{e}^{-x} \sin([value]x)+\int_{0}^{\infty} -\textrm{e}^{-x}\cdot \frac{1}{[value]} \sin([value]x) \, \textrm{d}x \bigg) \\ & = & \frac{1}{[value]}+\frac{1}{[value]} \bigg( 0+ \frac{1}{[value]} \int_{0}^{\infty} -\textrm{e}^{-x} \sin({[value]}x) \, \textrm{d}x \bigg) \\ & = & \frac{1}{[value]}-\frac{1}{[value]}\int_{0}^{\infty} \textrm{e}^{-x} \sin([value]x) \, \textrm{d}x. \end{eqnarray*} Whoa! The original integrand appeared also on the right hand side of the equation! Let us move it to the left side.Ohoh! Alkuperäinen integroitava ilmestyi myös yhtälön toiselle puolelle! Siirretään se vasemmalle puolelle. \begin{eqnarray*} \int_{0}^{\infty} \textrm{e}^{-x}\sin([value]x) \, \textrm{d}x+\frac{1}{[value]} \int_{0}^{\infty} \textrm{e}^{-x}\sin([value]x) \, \textrm{d}x & = & \frac{1}{[value]} \\ [value] \int_{0}^{\infty} \textrm{e}^{-x} \sin([value]x) \, \textrm{d}x & = & \frac{1}{[value]} \\ \int_{0}^{\infty} \textrm{e}^{-x} \sin([value]x) \, \textrm{d}x & = & [value]. \end{eqnarray*}

Compute the following integrals: Laske seuraavat integraalit:

(a)\[\int_{[value]}^{[value]}([value]x-[value]\sqrt{x}+[value]) \, \textrm{d}x \] (b)\[\int_{0}^{[value]} \cos(\frac{x}{2}) \, \textrm{d}x \] (c)\[\int_{[value]}^{[value]} [value] \, \textrm{d}x \]

Give your answers in exact form! The base of natural logarithm should be typed as e, e.g. 2*e. For square root, use command sqrt(a), e.g. sqrt(2). Anna vastaukset tarkkoina arvoina! Neliöjuuri a merkitään sqrt(a), esim. sqrt(2). Luonnollisen logaritmin kantaluku e merkitään e, esim. 2*e.

(a) | \(\int_{[value]}^{[value]}([value]x-[value]\sqrt{x}+[value]) \, \textrm{d}x=~\)
(b) | \(\int_{0}^{[value]} \cos(\frac{x}{2}) \, \textrm{d}x=~\)
(c) | \(\int_{[value]}^{[value]} [value] \, \textrm{d}x=~\)
Feedback: Model solution: Mallivastaus: (a) \begin{eqnarray*} \int_{[value]}^{[value]}([value]x-[value]\sqrt{x}+[value]) \, \textrm{d}x & = & \bigg/_{\!\!\!\![value]}^{\,[value]}([value]x^2-[value]\cdot \frac{2}{3}x^{3/2}+[value]x) \\ & = & [value]\cdot [value]^2-[value]\cdot [value]^{3/2}+[value]\cdot [value]-[value]\cdot [value]^2+[value]\cdot [value]^{3/2}-[value]\cdot [value] \\ & = & [value] \end{eqnarray*} (b) \[\int_{0}^{[value]} \cos(\frac{x}{2}) \, \textrm{d}x=\bigg/_{\!\!\!\!0}^{\,[value]}2\sin(\frac{x}{2})=2\sin([value])-2\sin(0)=2\sin([value])=[value]. \] (c) \[\int_{[value]}^{[value]} [value] \, \textrm{d}x=\bigg/_{\!\!\!\![value]}^{\,[value]} [value]=[value]. \]

Calculate the integrals Laske integraalit

\[\text{(a) } \int_{0}^[value] [value] \, \mathrm{d}x\] \[\text{(b) } \int_{0}^[value] [value] \, \mathrm{d}x\]

(Hint: What does \(D(\ln(\cos(x))\) simplify to?)(Vihje: Mitä \(D(\ln(\cos(x))\) on?)

Provide your answer in exact form. For natural logarithm, use command ln(a), e.g. ln(2). The base of natural logarithm should be typed as e, e.g. 2*e. For square root, use command sqrt(a), e.g. sqrt(2).Anna vastauksesi tarkkoina arvoina. Luonnollinen logaritmi a:sta merkitään ln(a), esim. ln(2). Luonnollisen logaritmin kantalukua merkitään e, esim. 2*e. Neliöjuuri a:sta merkitään sqrt(a), esim. sqrt(2).

Answers: Vastaukset:

(a) |  |
(b) |  |
Feedback: Model solution: Mallivastaus: (a) \begin{eqnarray*} \int_{0}^{[value]} |[value]-x| \, \mathrm{d}x & = & \int_{0}^{[value]} ([value]-x) \, \mathrm{d}x+\int_{[value]}^{[value]} (x-[value]) \, \mathrm{d}x \\ & = & \bigg/_{\!\!\!\!0}^{\,[value]} ([value]x-\frac{1}{2}x^2)+\bigg/_{\!\!\!\![value]}^{\,[value]} (\frac{1}{2}x^2-[value]x) \\ & = & [value]\cdot [value]-\frac{1}{2}\cdot [value]^2+\frac{1}{2}\cdot [value]^2-[value]\cdot [value]-\frac{1}{2}\cdot [value]^2+[value]^2 \\ & = & [value]. \end{eqnarray*} (b) \[D\left(\ln\left(\cos(x)\right)\right)=\frac{1}{\cos(x)}(-\sin(x))=\frac{-\sin(x)}{\cos(x)}=-\tan(x).\] \[\int_{0}^{[value]} \tan(x) \, \mathrm{d}x = \bigg/_{\!\!\!\!0}^{\,[value]} -\ln\left(\cos(x)\right)=-\ln\left(\cos\left([value]\right)\right)-\left[-\ln\left(\cos(0)\right)\right] = [value].\]

Calculate the following integrals. Give your answers in exact form. For square root, use command sqrt(a), e.g. sqrt(2).

Määritä seuraavat integraalit. Anna vastauksesi tarkkana arvona. Neliöjuuri merkitään sqrt(a), esim. sqrt(2).

(a) | \(\displaystyle \int_{[value]}^{[value]} [value] \, \mathrm{d}x =~\) |
(b) | \(\displaystyle \int_{[value]}^{[value]} [value] \, \mathrm{d}x =~\) |
Feedback: Model solution: Mallivastaus: (a) \[\begin{array}{rll} \int_[value]^[value] \left([value]\right) \mathrm{d}x & = & \Bigg/_{\!\!\!\![value]}^{\,[value]} \left( [value]\right) \\ & = & [value] - \left([value]\right) \\ & = & [value] \end{array}\] (b) \[\begin{array}{rll} \int_[value]^[value] [value] \mathrm{d}x & = & \Bigg/_{\!\!\!\![value]}^{\,[value]} \left( [value] \right)\\ & = & -[value]\cdot \cos([value]) - \left(-[value]\cdot \cos([value])\right) \\ & = & [value] \end{array}\]