Title: Integroimistekniikoita

Determine the given integrals using the method of integration by parts. Note that arctan(\(x\)) = atan(\(x\)) = tan\(^{-1}(x)\). On this exercise, use the notation atan(\(x\)).Laske annetut integraalit käyttäen osittaisintegrointia. Huomaa, että arctan(\(x\)) = atan(\(x\)) = tan\(^{-1}(x)\). Käytä tässä tehtävässä merkintää atan(\(x\)). Ja muista integroimisvakio.

(a) | \(\displaystyle \int [value] \mathrm{d}x =~\) |
(b) | \(\displaystyle \int [value] \mathrm{d}x =~\) |
Feedback: Model solution: Mallivastaus: The formula for integration by parts is of the formOsittaisintegrointikaava on muotoa \[ \int f(x) g'(x)\mathrm{d}x = f(x)g(x) - \int f'(x) g(x) \mathrm{d}x. \] To apply the formula, we have to choose a function \(f\) to differentiate and a function \(g'\) to integrate.Osittaisintegrointikaavan käyttämiseksi tulee valita integroitavasta funktiosta \(f\), jota derivoidaan ja \(g'\), jota integroidaan. (a)The integrand is of the form \(x \cdot \cos(x)\). We choose \(f(x) = x\) and \(g'(x) = \cos(x)\). Now \(f'(x) = 1\) and \(g(x) = \sin(x)\). With these selections, we can integrate the given function.Integroitava funktio on muotoa \(x \cdot \cos(x)\). Valitaan \(f(x) = x\) ja \(g'(x) = \cos(x)\). Tällöin \(f'(x) = 1\) ja \(g(x) = \sin(x)\). Tällä valinnalla saadaan funktio integroitua. \[ \int [value] \mathrm{d}x = x \cdot \sin(x) - \int \sin(x) \mathrm{d}x = x \cdot \sin(x) + \cos(x) + C \] (b) Now the integrand is \(1 \cdot \tan^{-1}(x)\). We choose \(f(x) = \tan^{-1}(x)\) and \(g'(x) = 1\). Now \(f'(x) = \frac{1}{1+x^2}\) and \(g(x) = x\). Choosing the constant 1 to represent the other function is quite common and often used trick.Nyt integroitava funktio on \(1 \cdot \tan^{-1}(x)\). Valitaan \(f(x) = \tan^{-1}(x)\) ja \(g'(x) = 1\). Tällöin \(f'(x) = \frac{1}{1+x^2}\) ja \(g(x) = x\). Etukertoimen 1 valitseminen toiseksi funktioksi on varsin yleinen ja toimiva temppu. \[ \int [value] \mathrm{d}x = x \cdot \tan^{-1}(x) - \int \frac{x}{1+x^2} \mathrm{d}x \] The remaining integral can be calculated after we introduce the derivative of the inner function \(1+x^2\), i.e. the term \(2x\), to the numerator.Jäljelle jäävä integraali voidaan laskea, kun saadaan osoittajaan sisäfunktion \(1+x^2\) derivaatta \(2x\). \[ \int \frac{x}{1+x^2} \mathrm{d}x = \frac{1}{2} \int \frac{2x}{1+x^2} \mathrm{d}x = \frac{1}{2} \text{ln} |1+x^2| \] The absolute value bars can be removed as \(1+x^2\) is always positive. The answer isItseisarvomerkit voidaan poistaa, sillä \(1+x^2\) on aina positiivinen. Vastaukseksi saadaan \[ \int [value] \mathrm{d}x = x \cdot \tan^{-1}(x) - \frac{1}{2} \text{ln} (1+x^2) + C. \]

Tarkastellaan määrätyn integraalin \[ \int\limits_{0}^{1} \frac{x}{(x+[value])(x+[value])^2}\,\mathrm{d}x \] tarkan arvon laskemista.

\(\textrm{a) }\) Kirjoita muuttujasta \(x\) riippuva integraalifunktio (integrointivakio \(C=0\)) ennen ylä- ja alarajasijoitusta. Luonnollinen logaritmi merkitään ln, esim. ln(2). Itseisarvo merkitään abs, esim. abs(-2).

\(\textrm{b) }\) Mikä on määrätyn integraalin tarkka arvo?

Consider the evaluation of the exact value for the definite integral \[ \int\limits_{0}^{1} \frac{x}{(x+[value])(x+[value])^2}\,\mathrm{d}x. \]

\(\textrm{a) }\) Write the expression for the integral function (the constant of integration C=0) depending on the variable \(x\) before substituting lower and upper limits of integration. The natural logarithm is written as ln, e.g. ln(3). The absolute value is written as abs, e.g. abs(-2).

\(\textrm{b) }\) What is the exact value of the definite integral?

\(\textrm{a) }\) Muuttujasta \(x\) riippuva integraalifunktio on \(\textrm{a) }\) The integral function depending on the variable \(x\) is .

\(\textrm{b) }\) Määrätyn integraalin tarkka arvo on \(\textrm{b) }\) The exact value of the definite integral is .
Feedback: Malliratkaisu: \[ \int\limits_{0}^{1} \frac{x}{(x+[value])(x+[value])^2}\,\mathrm{d}x \] Osamurtokehitelmä: \[ \begin{align*} \frac{x}{(x+[value])(x+[value])^2} &= [value]+[value] +[value]\\ &= \frac{[value]A+[value]B+([value])C}{(x+[value])(x+[value])^2} \\ &= \frac{Ax^2+[value]Ax+[value]A+Bx^2+[value]Bx+[value]B+Cx+[value]C}{(x+[value])(x+[value])^2} \\ &=\frac{(A+B)x^2+([value]A+[value]B+C)x+[value]A+[value]B+[value]C}{(x+[value])(x+[value])^2}. \end{align*} \] Osoittajia vertaamalla saadaan yhtälöryhmä \[ \left\{\begin{array}{rrrllll} A&+B&&=0&|\,\cdot ([value])&&|\,\cdot ([value])\\ [value]A&+[value]B&+C&=1&|\,\cdot 1&&\\ [value]A&+[value]B&+[value]C&=0.&&&|\,\cdot 1\ \end{array}\right. \] Eliminoidaan \(A\) kertomalla puolittain sopivasti ja laskemalla yhteen. \[ \left\{\begin{array}{rrll} B&+C&=1&|\,\cdot ([value]) \\ [value]B&+[value]C&=0&|\,\cdot 1\\ \end{array}\right. \] Eliminoidaan \(B\) kertomalla puolittain sopivasti ja laskemalla yhteen. \[ C=[value], \] josta edelleen sijoittamalla saadaan \(B=[value]\) ja \(A=[value]\). Tällöin \[ \begin{align*} \int\limits_{0}^{1} \frac{x}{(x+[value])(x+[value])^2}\,\mathrm{d}x &=\int\limits_{0}^{1}([value]+[value] -[value])\,\mathrm{d}x\\ &={\Bigl/}_{\!\!0}^{1\,}([value]\ln\left|x+[value]\right|+[value]\ln\left|x+[value]\right|+\frac{[value]}{x+[value]})\\ &=[value]\ln([value])+[value]\ln([value])+\frac{[value]}{[value]}-([value]\ln([value])+[value]\ln([value])+1)\\ &=[value]\ln([value])-\frac{1}{[value]}. \end{align*} \] Model solution: \[ \int\limits_{0}^{1} \frac{x}{(x+[value])(x+[value])^2}\,\mathrm{d}x \] Partial fraction expansion: \[ \begin{align*} \frac{x}{(x+[value])(x+[value])^2} &= [value]+[value] +[value]\\ &= \frac{[value]A+[value]B+([value])C}{(x+[value])(x+[value])^2} \\ &= \frac{Ax^2+[value]Ax+[value]A+Bx^2+[value]Bx+[value]B+Cx+[value]C}{(x+[value])(x+[value])^2} \\ &=\frac{(A+B)x^2+([value]A+[value]B+C)x+[value]A+[value]B+[value]C}{(x+[value])(x+[value])^2}. \end{align*} \] The system of equations follow by comparing the numerators: \[ \left\{\begin{array}{rrrllll} A&+B&&=0&|\,\cdot ([value])&&|\,\cdot ([value])\\ [value]A&+[value]B&+C&=1&|\,\cdot 1&&\\ [value]A&+[value]B&+[value]C&=0.&&&|\,\cdot 1\ \end{array}\right. \] Eliminate \(A\) by multiplying both sides of the equation appropriately and adding together. \[ \left\{\begin{array}{rrll} B&+C&=1&|\,\cdot ([value]) \\ [value]B&+[value]C&=0&|\,\cdot 1\\ \end{array}\right. \] Eliminate \(B\) by multiplying both sides of the equation appropriately and adding together. \[ C=[value], \] from which it follows by substituting \(B=[value]\) and \(A=[value]\). Hence \[ \begin{align*} \int\limits_{0}^{1} \frac{x}{(x+[value])(x+[value])^2}\,\mathrm{d}x &=\int\limits_{0}^{1}([value]+[value] -[value])\,\mathrm{d}x\\ &={\Bigl/}_{\!\!0}^{1\,}([value]\ln\left|x+[value]\right|+[value]\ln\left|x+[value]\right|+\frac{[value]}{x+[value]})\\ &=[value]\ln([value])+[value]\ln([value])+\frac{[value]}{[value]}-([value]\ln([value])+[value]\ln([value])+1)\\ &=[value]\ln([value])-\frac{1}{[value]}. \end{align*} \]

Määritä funktion \( f(x)=[value] \) integraalifunktio. Ilmoita integraalifunktio ilman integroimisvakiota.

\( \int f(x) \, \textrm{d}x= \)
Feedback: Mallivastaus: Funktion \( (ax+b)^n \) integraalifunktio on \( \frac{1}{n+1}\cdot \frac{1}{a}\cdot(ax+b)^{n+1} \). Siis \( \begin{eqnarray*} \int [value] \, dx & = & \frac{1}{[value]+1}\cdot \frac{1}{1} \cdot \left( [value] \right)^{[value]+1} \\ & = & [value] \\ & = & [value]. \end{eqnarray*} \)

Determine the following integral functions. 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.Määrää seuraavat integraalifunktiot. Luvun \(a\) luonnollinen logaritmi merkitään muodossa ln(a), esim. ln(2). Luonnollisen logaritmi kantaluku \(\mathrm{e}\) merkitään muuttujalla e, esim. 2*e.

(a) | \(\displaystyle\int [value] \, \mathrm{d}x=~\) |
(b) | \(\displaystyle\int [value] \, \mathrm{d}x=~\) |
(c) | \(\displaystyle\int [value] \, \mathrm{d}x=~\) |
Feedback: Model solution: Mallivastaus: To calculate the integrals, we apply the integration rule for composite functions: Funktiot saadaan integroitua soveltamalla yhdistetyn funktion integrointikaavaa: \[ \int g'(x)\cdot f'(g(x)) \mathrm{d}x = f(g(x)) + C\] (a) \[ \int \cos([value]+[value]) \mathrm{d}x =[value] \int \underbrace{[value]}_{g'(x)} \underbrace{\cos(\overbrace{[value]+[value]}^{g(x)})}_{f'(g(x))} \mathrm{d}x = [value]\] (b) \[ \int \mathrm{e}^{[value]} \mathrm{d}x = [value] \int \underbrace{[value]}_{g'(x)} \underbrace{\mathrm{e}^{\overbrace{[value]}^{g(x)}}}_{f'(g(x))} \mathrm{d}x = [value]\] (c) \[ \int \frac[value]{[value]} \mathrm{d}x = [value] \int \underbrace[value]_{g'(x)} \overbrace{(\underbrace{[value]}_{g(x)})^{-1}}^{f'(g(x))} \mathrm{d}x = [value]\]

Calculate the following integrals using the given substitutions: Laske seuraavat integraalit käyttämällä annettuja sijoituksia:

(a)\[ \int_{0}^{1/2} [value] \, \mathrm{d}x \ \ \ \ \ \text{subst.sij.} \ \ u=[value]. \] (b)\[ \int_{0}^[value] \sqrt{[value]-x^2} \, \mathrm{d}x \ \ \ \ \ \text{subst.sij.} \ \ x=[value]u. \] Hint: At the end of part b) you may apply the formula \[ \int \sqrt{1-x^2} \, \mathrm{d}x= \frac{x\sqrt{1-x^2}+\arcsin(x)}{2}+ \ \text{constant} \]Vihje: b)-kohdassa voi lopuksi käyttää kaavaa \[ \int \sqrt{1-x^2} \, \mathrm{d}x= \frac{x\sqrt{1-x^2}+\arcsin(x)}{2}+ \ \text{vakio} \]

(a) | \(\displaystyle\quad \int_{0}^{1/2} [value] \, \mathrm{d}x=~\) |
(b) | \(\displaystyle\quad \int_{0}^[value] \sqrt{[value]-x^2} \, \mathrm{d}x=~\) |
Feedback: Model solution: Mallivastaus: (a) \[ \int_{0}^{1/2} [value] \] Using the given substitutionSijoitetaan \(u=[value] \Rightarrow x=[value]\) \(\mathrm{d}x={[value]} \, \mathrm{d}u\) \(x=0 \Rightarrow u=0\) \(x=1/2 \Rightarrow u={[value]}\). \[\begin{array}{rll} \int_{0}^{1/2} [value] \, \mathrm{d}x & = & \int_{0}^{[value]} \sin(u)\cdot [value] \, \mathrm{d}u \cos(u)\cdot {[value]} \, \mathrm{d}u \\ & = & [value] \bigg/_{\!\!\!0}^{\,[value]} -\cos(u) \sin(u) \\ & = & [value] ( -\cos([value])+\cos(0) )=[value] ([value]+1)=[value] \sin([value])-\sin(0) ) = [value]\cdot [value]=[value] . \end{array}\] (b) \[ \int_{0}^{[value]} \sqrt{[value]-x^2} \, \mathrm{d}x \] Using the given substitutionSijoitetaan \(x=[value]u \Rightarrow u=\frac{x}{[value]}\) \(\mathrm{d}x = [value] \, \mathrm{d}u\) \(x=0 \Rightarrow u=0\) \(x=[value] \Rightarrow u=1\). \[\begin{array}{rll} \int_{0}^{[value]} \sqrt{[value]-x^2} \, \mathrm{d}x & = & \int_{0}^{1} \sqrt{[value]-([value]u)^2} \cdot [value] \, \mathrm{d}u \\ & = & \int_{0}^{1} \sqrt{[value](1-u^2)} \cdot [value] \, \mathrm{d}u \\ & = & [value] \int_{0}^{1} \sqrt{1-u^2} \, \mathrm{d}u \\ & = & [value] \bigg/_{\!\!\!0}^{\,1} \frac{u\sqrt{1-u^2}+\arcsin(u)}{2} \\ & = & [value]\cdot \frac{\arcsin(1)-\arcsin(0)}{2}=\frac{[value]}{2}\cdot \frac{\pi}{2}=[value]. \end{array}\]

Using a suitable substitution, calculate the integral Määritä sopivan sijoituksen avulla

\[\int_{0}^{1} \frac{1}[value] \, \textrm{d}x \]

Note: arctan should be typed as atan, e.g. atan(1).Huom. arctan merkitään atan, esim. atan(1).

\(\displaystyle\int_{0}^{1} \frac{1}[value] \, \textrm{d}x=\)
Feedback: Model solution: Mallivastaus: \[\int_{0}^{1} \frac{1}[value] \, \textrm{d}x = \int_{0}^{1} \frac{1}{(x+[value])^2+1} \, \textrm{d}x \] Using the substitutionSijoitetaan \(u=x+[value] \Rightarrow x=u-[value]\) \(\textrm{d}x=\textrm{d}u\) \(x=0 \Rightarrow u=[value]\) \(x=1 \Rightarrow u=[value]\). \begin{eqnarray*} \int_{0}^{1} \frac{1}{(x+[value])^2+1} \, \textrm{d}x & = & \int_{[value]}^{[value]} \frac{1}{1+u^2} \, \textrm{d}u \\ & = & \bigg/_{\!\!\![value]}^{\,[value]} \arctan(u) \\ & = & \arctan([value])-\arctan([value]). \end{eqnarray*}

Calculate the integralLaske integraali

\[ \int_{0}^{[value]} x^2 [value] \, \textrm{d}x \] Give your answer in exact form! Anna vastaus tarkkana arvona!

\(\pi\) should be written as pi, e.g. 2*pi. For square root, use command sqrt(a), e.g. sqrt(2).\(\pi\) merkitään pi, esim. 2*pi. Neliöjuuri a:sta merkitään sqrt(a), esim. sqrt(2).

\(\displaystyle\int_{0}^{[value]} x^2 [value] \, \textrm{d}x =\)
Feedback: Model solution: Mallivastaus: Integration by parts: Osittaisintegrointi: \[\int f'(x)g(x) \, \textrm{d}x=f(x)g(x)-\int f(x)g'(x) \, \textrm{d}x. \] \[ \int_{0}^{[value]} x^2 [value] \, \textrm{d}x \] First we determine the integral \(\int x^2 [value] \, \textrm{d}x\).Lasketaan ensin integraali \(\int x^2 [value] \, \textrm{d}x\). ChoosingValitaan \(f'=[value] \Rightarrow f=[value] \ \ \ \ g=x^2 \Rightarrow g'=2x\). \[ \int x^2 [value] \, dx=[value]-\int ([value]) \, \textrm{d}x. \] Now we apply the integration by parts again. ChoosingNyt käytetään osittaisintegrointia uudestaan. Valitaan \(f'= \cos(x) \Rightarrow f= \sin(x) \ \ \ g=-2x \Rightarrow g'=-2\). \begin{eqnarray*} [value]-\int ([value]) \, \textrm{d}x & = & [value]-([value])-\int ([value]) \, \textrm{d}x) \\ & = & [value]+C \\ & = & ( [value]) \cos(x) +2x\sin(x) + C \end{eqnarray*} We getNyt \begin{eqnarray*} \int_{0}^{[value]} x^2 [value] \, \textrm{d}x & = & \bigg/_{\!\!\!0}^{\,[value]} [([value]) \cos(x) +2x\sin(x) +C] \\ & = & (2-([value])^2)\cdot \cos([value])+2\cdot [value] \cdot \sin([value])+0-[(2-0^2)\cdot \cos(0)] \\ & = & [value]. \end{eqnarray*}

Using the method of integration by parts, determineLaske osittaisintegroinnin avulla

(a) | \(\displaystyle\quad \int x\cos([value]x) \, \mathrm{d}x=~\)\(~+C\) |
(b) | \(\displaystyle\quad \int (x+[value])e^{[value]x} \, \mathrm{d}x=~\)\(~+C\) |
Feedback: Model solution: Mallivastaus: Integration by parts: Osittaisintegrointikaava: \[\int f'(x)g(x) \, \mathrm{d}x=f(x)g(x)-\int f(x)g'(x) \, \mathrm{d}x \] (a) \[ \int x\cos([value]x) \, \mathrm{d}x\] ChoosingValitaan \(f'=\cos([value]x) \Rightarrow f=\frac{1}{[value]}\sin([value]x) \ \ \ g=x \Rightarrow g'=1\) \begin{eqnarray*} \int x\cos([value]x) \, \mathrm{d}x & = & \frac{1}{[value]} x\sin([value]x)-\int 1\cdot \frac{1}{[value]}\sin([value]x) \, \mathrm{d}x \\ & = & \frac{1}{[value]}x\sin([value]x)-\frac{1}{[value]}\cdot \frac{1}{[value]}\int [value]\sin([value]x) \, \mathrm{d}x \\ & = & \frac{1}{[value]}x\sin([value]x)-\frac{1}[value] (-\cos([value]x)+C) \\ & = & \frac{1}{[value]}x\sin([value]x)+\frac{1}[value]\cos([value]x)+C'. \end{eqnarray*} (a) \[\int (x+[value])e^{[value]x} \, \mathrm{d}x\] ChoosingValitaan \(f'=e^{[value]x} \Rightarrow f=\frac{1}{[value]}e^{[value]x} \ \ \ g=x+[value] \Rightarrow g'=1\). \begin{eqnarray*} \int (x+[value])e^{[value]x} \, \mathrm{d}x & = & \frac{1}{[value]}(x+[value])e^{[value]x}-\frac{1}{[value]} \int e^{[value]x} \, \mathrm{d}x \\ & = & \frac{1}{[value]}(x+[value])e^{[value]x}-\frac{1}[value]e^{[value]x}+C. \end{eqnarray*}

Laske osittaisintegroimalla. Muista käyttää kaikissa tuloissa kertolaskuoperaattorina tähteä ( *). Eksponenttifunktion \( e^{ax} \) voit kirjoittaa muodossa exp(a*x) tai e^(a*x).

\( \Large{\int} \normalsize [value] \, \textrm{d}x = \) + \( C \),
missä \( C \) on integrointivakio.

Osittaisintegroinnin kaava on \( \Large{\int} \normalsize f'(x)g(x) \textrm{d}x = f(x)g(x) - \Large{\int} \normalsize f(x)g'(x) \textrm{d}x \).

Miten pitää tässä tapauksessa funktiot \( f(x), f'(x), g(x) \text{ ja } g'(x) \) valita, jotta integraali saadaan laskettua mahdollisimman helposti?

\( f(x) = \) ja \( f'(x) = \)

\( g(x) = \) ja \( g'(x) = \)
Feedback: Mallivastaus: Osittaisintegroinnin kaava on \[ \Large{\int} \normalsize f'(x)g(x) \textrm{d}x = f(x)g(x) - \Large{\int} \normalsize f(x)g'(x) \textrm{d}x \]. Valitaan \( f'(x) = [value] \) ja \( g(x) = x \) (jos valittaisiin \( f'(x) = x \) ja \( g(x) = [value] \), tulisi integraalista \( \Large{\int} \normalsize f(x)g'(x) \textrm{d}x \) vaikeampi kuin alkuperäisestä integraalista ja se pitäisi osittaisintegroida kahteen kertaan). Siten \( f(x) = [value], \quad f'(x) = [value], \quad g(x) = [value] \quad \text{ja } g'(x) = [value]\) ja näin saadaan \[ \begin{align} \Large{\int} \normalsize [value] & = [value] \cdot [value] - \large{\int} \normalsize [value] \cdot [value] \textrm{d}x \\ & = \frac{x[value]}{[value]} - \frac{1}{[value]} \large \int \normalsize \frac{1}{[value]} \cdot [value] [value] \textrm{d}x \\ & = \frac{x[value]}{[value]} - \frac{[value]}{[value]} + C \\ & = [value] + C. \end{align} \]