^Integral calculus ^Integrální počet
Robert Mařík Mendel University Brno
16. května 2005
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•  Write appropriate functions or numbers into blank fields and press Enter.
•  Use functions and mathematical notation as explained in the file instrukce.pdf.
•  The green boundary indicates correct answer, the red boundary indicates wrong answer.
•  If you cannot solve the problem, click Ans to see the correct answer. If there are more fields to be filled, click repeatedly.
•  Vepište do políček co tam patří a stiskněte Enter.
•  Zápis funkcí provádějte tak, jak je vysvětleno v nápovědě v souboru instrukce.pdf.
•  Zelený okraj obélníku znamená správnou odpověď, červený špatnou.
•  Kliknutím na Ans se zobrazí správný výsledek - s případě že problém nejste schopni vyřešit. Je-li v otázce více políček, klikněte na Ans opakovaně.
1.    Testi
^Indefinite integrals by formulas
Quiz
1.   iexdx =
2.   ie2xdx =
3.  / rl + 3é>~x) dx =
4.     (ex + l) dx = 5./i(^ + ^)dx = ,,,I±^)dx = 7. / -^dx =
g-2x
8- / TT^dx =
JUžití vzorců
C
hC
C
c
c
c
c
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9. 10.
11. 12. 13. 14. 15. 16. 17. 18. 19.
3 • 2xdx =
x2 + x + 4
x x+1
dx =
2x
dx =
x + 4) dx
/ \/x(l — Vx")dx =
/	(x + l)(x-l)
	7                  C x1
/	X        A -r,-----dx = x2+ 6
/	~ô----jdx = x2+ 6
/	x2 + 2J -r.—-dx = x2 + l
/	x + 5 , -r.—-dx = x2+ 4
í	1 — cos2 x . ------^-----dx =
dx =
cos2 x
C
C
C
C
C
C C C
C
C
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„„   f sin x n
20.  / ------dx =
J  cos x
21.   / 2 sin x cos xdx =
22.   / sin í x — — J dx =
23.   / sin(/r — x)dx =
24.   /Vxdx =
25.   íe3x+1dx =
26.   Í2ex-2dx =
27.  [e5-3xdx = 1
28.
29.
30.
3 +x2 1
\/3H
-4 cos2(2x)
dx = =dx =
dx =
C
C
C C C
C
C
+ C C
c c
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31.
32.
x I dx =
x------F= I dx =
x,
33. í(x + l)2dx =
34.
3x + 5
-dx =
C
C
C
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35. SSWrite correct numbers inside the    IWVepište správná čísla do malých
small colored rectangles and then    podbarvenych políček a potom na-
write the primitive function (white     lezněte primitivní funkci (bílé po-
field).                                                      líčko).
(a)
(b)
(c)
(d)
x3 + l
3x
x2+ 4
x2-l
x2 + l
dx =
dx =
dx =
2x + l
2x + l
dx =
(x3 + iy
x3 + l
dx
C
(x2+ 4)'
dx
x2 + l
C dx + C
X +
2x + 2 '■ + 2x + 1
2x + l
2x + l
-dx
C
x + 5 x2+ 4
1
dx =
x2 + 2x + 5
x2 - 3x + 4
Vx2 + x + 1
dx =
dx =
dx =
x + 1
x2 + 4x + 6
dx =
2x
x2 + 4             x2 + 4
+ C 1
dx
C
dx
-dx
C 1
+ C 2x + 4 x2 + 4x + 6
dx
-dx
C
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(j) / sinxcosxdx =  /
sin
x I dx
C
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2.    Test2
^Integration by parts
iWIntegrace per-partés
^tiWhen integrating by parts we use      IWPro integraci per-partés používáme the formula                                              následující vzorec
u(x)v' (x)dx = u{x)v(x) — / u(x)v' (x)dx
(Eq:l)
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Quiz
$äWe use the integration by parts      IWTypicky používáme integraci per-
especially for integrals of the type
partes pro integrály typu
p (x) f (ax + b)dx,
(Eq:2)
§B where p (x) is a polynomial and
IWkde p (x) je polynom a
f (x) E {ex,sinx,cosx,atanx,mrax}
§8Here atan(x) is the usual arctangent functions.
^Question: Are the following integrals like (Eq:2)? Are the integral convenient for integration by parts?
IWZde atan(x) je obvyklá funkce ar-kustangens.
hdOtázka: Jsou následující integrály typu (Eq:2)? Je vhodné je integrovat metodou per-partés?
1.
dx
2.   / xex dx
3.   / x2exdx
4. í(3x + l)e-x+1dx
Yes	No
Yes	No
Yes	No
Yes	No
5.   / (x+ 4) atan-dx
6.   / xsinx2dx
7.   / x2 In xdx
8.   / atan xdx
9.   / xlnxcosxdx
10.   / x cos3 xdx
11.   /(2 + x)cos(2x)dx
12.   í(x3 - 1) sin í y - x") dx
Yes	No
Yes	No
Yes	No
Yes	No
Yes	No
Yes	No
Yes	No
Yes	No
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Quiz ES Integrate
1=1 (x2+ x — 2)sinxdx.
u = 1.
v' =
2.1 =
v =
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Jlntegrujte
dx
_
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I = -{x2 + 2x + l)
cos x
§SNow we have an expression which can be written as above (check it yourself). We integrate by parts in
I (x + l) cosxdx.
2     (x + l)cosxdx.
IWNyni máme něco, co se dá přepsat do výše uvedeného tvaru (zkontrolujte si) do tvaru. Integrujeme výraz I (x + l) cos xdx. Použijeme opět metodu per-partés.
u =
u =
I = — (x2 + 2x + 1) cos x + 2Í
dx
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I = —(x2 + 2x + l)cosx + 2Í (x + 1) sinx = — (x2+ 2x +l)cosx+ 2( (x +1) sinx
sinx
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sinxdx
cos x + C
I
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Quiz Ě8 Integrate I =     atanxdx.
u = 1.
v1 =
2.1 =
v =
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dx
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3. íSäNow we have an expression which can be written in the form (check it yourself). Find out the number which has to be in the first colored field. When you find out this number, the integration is easy.
I = x atan x — = x atan x — ŕ = x atan x —
IWNyní máme něco, co se dá přepsat (zkontrolujte si) do tvaru. Zjistíte-li, jaké číslo je potřeba zapsat do prvního podbarveného obdélníčku, je integrace snadná.
X        A
x^Tídx
)Jx^Tídx
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4. fläThe result is
atan xdx = x atan x — - ln( 1 + x2
^Výsledek je
C
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Quiz ^Integrate
I = i(x2 - l)exdx
Jmtegrujte
1. §8We integrate by parts with u (x) =    NIntegrujeme per-partés při volbě (x2 — 1). With this notation we have    u(x) = x2 — 1 (use zero constant of integration in responses)
u = x v' =
u  =
V =
2. ^Integration by parts gives
I =
hUPo použití vzorce pro integraci per-partés máme ...
dx
3.  ÜÖWe integrate once more by parts        hdBudeme  integrovat ještě jednou
per-partés
u = 2x                       u' =
v' =                             v =
4.  SSThe second integration by parts gi-    ^Opětovné   použití   vzorce   per-ves ...                                                    partes dává ...
I =
dx
5. fSThe result after the last integration    IWPo poslední integraci a po snadné and simplifications is ...                        úpravě obdržíme ...
1=                                                                         +C
Quiz ^Integrate                                           IWIntegrujte
I =     xln(x + l)dx
1.  lääWe integrate by parts with u(x) =     h»JBudeme integrovat per-partés při ln(x + l).                                               volbě u (x) = ln(x + 1).
u = ln(x + 1)             u' =
v' =                             v =
2.  ^Integration by parts gives ...              hJAplice vzorce per-partés dává ...
1=                             -f                             dx
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3.  ÜÖThe expression denoted by A is a rational function which is not proper.
I Divide the numerator by the denominator and write this functionas a sum of polynomial and proper function. Write the polynomial into the first field and the proper function into the second one.
A=                      +
N-------------„-------------'      V-------------«------------
polynomial                      remainder
4.  sJSThe integration and simplification    h*JFinální integrací a úpravou získá-give ...                                                   váme ...
1 =                                                                         +C
hdVýraz označený jako A je racionální funkce a je nutno ji intgerovat tak, že nejprve vydělíme čitatel jmenovatelem. Napište do prvního políčka podíl a do druhého zbytek po dělení.
Quiz Find the following integral: I =  / (x + l)e xdx
1. We integrate by parts with u (x) = (x + 1). With this notation we have (use zero constant of integration in responses):
u'(x) =
v'{x) = v(x) =
2. Integration by parts gives
1=                             - f                             dx
3. Integration gives the indefinite integral
1=             "'                                   "                                +C
Quiz Find the following integral: I =  / (x2 — 1) sinxdx
1. We integrate by parts with u(x) = (x2 — 1). With this notation we have (use zero constant of integration in responses):
u'(x) =
v'(x) = v(x) =
2. Integration by parts gives
I =                         - í                        dx
Go to the next page.
3. Now you have I = — (x2 — 1) cos(x) + 2 / xcos(x)dx. We integrate by parts
with u (x) = x. With this notation we have (use zero constant of integration in responses):
u'(x) =
v'{x) = v(x) =
4. Integration by parts gives
I = — (x2 — l)cosx +
+ 2
5. Integration gives the indefinite integral I =
dx
C
Quiz Find the following integral: I =  / In xdx
1. We integrate by parts with u(x) = lnx. With this notation we have (use zero constant of integration in responses):
u'(x) =
v'(x) = v(x) =
2. Integration by parts gives
1=                     - í                     dx
3. Integration gives the indefinite integral
1=                                                  °                                 +C
Quiz Find the following integral: I =  / x2 atan xdx (we use "atan(x)" for the usual arctangent function).
1. We integrate by parts with u{x) = atanx. With this notation we have (use zero constant of integration in responses):
u'(x) =
v'(x) = v(x) =
2. Integration by parts gives
1=                     - í                     dx
3. Integration gives the indefinite integral
/=                                           '                           +C
Quiz Find the following integral: I =  / (x + 3)e2xdx
1. We integrate by parts with u (x) = (x + 3). With this notation we have (use zero constant of integration in responses):
u'(x) =
v'{x) = v(x) =
2. Integration by parts gives
1=                             - f                             dx
3. Integration gives the indefinite integral
1=             "'                                   "                                +C
3.    Test3
^Integration by substitution                 IWIntegrace substitucí
§8 When integrating by substitution we use the formula
í f{(p{x))<p'{x)dx =  í f(t)dt                         (Eq:3)
(i.e. we substitute <p(x) = t and (p'(x)dx = dt)or
Jf(x)dx = Jf(cp(t))<P'(t)dt                       (Eq:4)
(i.e. we substitute x = <p(t) and dx = (p'(t)dt). Iw Pro integraci pomocí substituce používáme výše uvedené vzorce.
/x + v/x — 4 -----------         dx (x + l)\/x — 4
1. We use the substitution x — 4 =
2. With this substitution we have
dx =                          dŕ                                               x
t =
3. Susbtitution gives
1= f                                                                     dŕ
4. We have to divide the numerator by the denominator. This gives a sum of polynomial and proper rational fraction (which is also a partial fraction). Write this polynomial into the first and the partial fraction into the second field.
1= f            +                                                                     dŕ
5. Integration in ŕ gives
I =
6. The back substitution gives the result in the variable x
/ =                         "'                                 + C
^     .      t..       i     r        r    in          •          •                   it             ľ  SÍn(x) COS (j)    ,
Quiz Find the following integral: I =  / -----^-,—^^dx
&       &             J     sin(x) + l
1. We use the substitution t =
2. With this substitution we have dř =                          dx
3. Susbtitution gives
1= f                                                                   dt
4. We have to divide the numerator by the denominator. This gives a sum of polynomial and proper rational fraction (which is also a partial fraction in our particular example). Write this polynomial into the first and the partial fraction into the second field. 1=1+                   at
5. Integration in t gives I =
6. The back substitution gives the result in the variable x
/ =                          '                         + C
Quiz Convert the following integral by substitution into an integral of rational
function: I =     x\-------dx
J     V x-1
x +1
1. We use the substitution t2 =------- With this substitution we have
x — 1
X =
dx =                                     dx
2. Susbtitution and simpification give
1=1                                                   dt
Quiz Find the following integral: I =     x2e x dx
1. With substitution —x3 = t we have
• dx = dŕ
2. Substitution gives
' = /                       d(
3. Integration in t gives the indefinite integral I =
4. In the original variable x we have I =
C C
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Quiz Find the following integral: I =     sin5 xdx
1. With substitution cos x = t we have
• dx = dŕ
2. Substitution gives
' = /                       d(
3. Integration in t gives the indefinite integral I =
4. In the original variable x we have I =
C C
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~ .   ~ ,    .                       .    ,    .           , T       f  sin x — cos x   . Quiz Substitute tan x = t in the integral í = / —~----------5—dx
J  sin x + cos3 x
1. With substitution tan x = t we have (write expression in t)
x =
2. Differentiating we get dx =                           • dr
3. From the right triangle with angle x, opposite side t, adjacent side 1 and hypotenuse v 1 + t2 (draw such an triangle) we have the following relations between sin(x), cos(x) and new variable t:
sin(x) =                              (write expression in t)
cos(x) =                              (write expression in t)
4. Substitution gives
I=f                                                                                   dŕ
5. Now we stop. However, you can evaluate this integral using partial fractions.
Quiz Evaluate integral I =--------dx by substitution.
B           J  l + ex        y
1. Differentiating ex = t we get
• dx = dř
2. From ex = ř we have (write x as a function of t)
Differentiating this relation we have dx =                           • dř
3. After substitution we have
1=  í                                                                  dř
4. Decomposition into partial fraction and integration give the integral in the variable ř:
/=                                                              +C
5. We return to the original variable x. We have
1=                                                                         +C
4.    Test4
^Definite integral in geometry
IWAplikace v geometrii
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Quiz áSThe function on the picture is the hd Na obrázku je funkce y = ex pře-function y = ex reflected about the y- vrácená okolo osy y a posunutá o jed-axis and moved by one above. (In no- nicku nahoru. (V notaci tohoto doku-tation of this document the function mentu je možno funkci ex zapsat jako ex can be written as exp (x), or exp (x), nebo e" (x).) Označený re-e ~ (x).) The green region correponds gion odpovídá intervalu x G [0,2]. to the interval x E [0,2].
A
2
1. §8Write an analytical formula for the    l*JNapište analytický tvar funkce, function.
y =
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2. ^Express the area of the green region    fcWVyjádřete obsah vybarveného regi-as the definite integral.                           onu jako určitý integrál.
S =
dx
3. ^Complete the following formula.    ^Doplňte vzorec, který potom pou-This formula may be used later for    žijte pro integraci, integration.
e xdx =
C.
4. ^Integrerate and use the Newton-    kJIntegrujte a použijte Newtonovu-Leibniz formula.                                    Leibnizovu formuli.
S =
5. ^Substitute the limits and evaluate    b^Dosaďte meze a dopočíteje inte-the integral.                                            grál-
S =
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6. áSWrite the volume of the of the solid of revolution formed by revolving the green region about the x-axis as a definite integreal.
^Vyjádřete jako určitý integrál objem tělesa, které vznikne rotací tohoto obrazce okolo osy x.
V = n
■^Simplifying and integrating we get (use zero constant of integration) ...
dx.
NPo umocnění integrandu a po integraci (volte nulovou integrační konstantu) máme pro objem vztah ...
V  = rc
8. SSThe volume is ...
V =
Ti.
^Výsledný objem je
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I
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Quiz $IThe functions on the picture are hJNa obrázku jsou funkce y = ex y = ex and y = e~x (In notation a y = e~x (V notaci tohoto doku-of this document we can write the mentu je možno funkci ex zapsat jako function ex as exp(x) or e" (x) exp (x), nebo e" (x) a funkci e~x and the function e~x as exp(-x) jako exp (-x) , nebo e" (-x).) Ozna-ore'(-x).) The green region corre- čený region odpovídá intervalu x G ponds to x G [0,1].                               [0,1].
1.  ÜÜThe black curve is
y =
2.  šUThe red curve is
y =
JČerná funkce ie
JČervená funkce je
3. SÜiArea of the green region can be eva-    IWObsah   vybarveného   regionu   je luated as a definite integral...               možno vyjádřit jako určitý integrál
S = 4. ^Integration gives
dx
JPo integraci dostaneme
S =
5. ^Substituting limits and simplifying    iNJPo dosazení mezí a výpočtu dostá-we obtain                                               váme
S =
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ÜÖThe volume of the solid of revolution which can be obtained by revolving the green region about the x-axis can be evaluated as the definite integral ...
V = 71
^Algebraic simplifications and integration give (use a zero constant of integration) ...
NdObjem tělesa, které vznikne rotací tohoto obrazce okolo osy x je možno vyjádřit jako určitý integrál...
dx.
l*JPo umocnění integrandu a po integraci (volte nulovou integrační konstantu) máme pro objem vztah ...
V  = rc
8. ÜThe volume is ...
V =
Ti.
IWVýsledný objem je
Quiz ÜÜThe functions on the picture are hdNa obrázku jsou funkce y
2
x   a
y   =   x   and y   =
2 (In the y =
2 (v notaci tohoto doku-
notation of this document you can mentu lze tyto funkce zapsat např write   something   like   y=x"2   and jako y=x"2 a y=x"2/2+2). y=x"2/2 + 2).
1.  SäThe black curve is:
y =
2.  šSThe red curve is:
y =
3.  §SFind the intercepts of both curves.
a =           , b =
JČerná křivka je grafem funkce:
^Červená křivka je grafem funkce:
bJNajděte průsečíky křivek.
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4. ÜÖExpress the area of the shaded region as an definite integral.
IWVyjádřete obsah vyšrafované plochy pomocí určitého integráluje.
S =
dx.
5. ÜäöThe function inside integral is a polynomial. Find the coefficinets of this polynomial.
NJIntegrand lze zapsat jako polynom. Doplňte koeficienty tohoto polynomu.
S =
dx.
6. ^Integrate  and use  the  Newton-Leibniz formula.
^Integrujte a použijte Newtonovu-Leibnizovu formuli
S =
7.  áSWrite the integral which express the volume of the solid obtained by a revolution of the shaded region about the x-axis.
v = */
8.  stlThe function in the integral can be expressed as a polynomial. Complete the coefficients of the polynomial.
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IWRotuje-li vyšrafovaná plocha okolo osy x, získáme rotační těleso, jehož objem je možno zapsat ve tvaru určitého integrálu. Napište tento intgerál.
dx.
NIntegrand lze vyjádřit jako polynom (doplňte čísla)
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V = 7T
9. disintegrate  and use  the  Newton-Leibniz formula.
V = 71
10. ^Substitute the limits and evaluate the integral.
V =                   71.
P rev. Page
dx.
IhJIntegrujte a použijte Newtonovu-Leibnizovu formuli.
NDosaďte horní a dolní mez a vypočtěte integrál.
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Quiz Na obrázku je funkce y = y/x posunutá o jedničku nahoru a o jedničku doprava. (V notaci tohoto dokumentu je možno funkci y/x zapsat jako sqrt (x), nebo x" (1/2).)
1   2
1, Analytický tvar funkce je y =
2. Obsah vybarveného regionu je možno vyjádřit jako určitý integrál
S =
dx
3. Pro integraci lze použít vzorec /^dx = /xidx =
4. Po aplikaci tohoto vzorečku dostáváme
S =
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5. Po dosazení mezí a výpočtu dostáváme S =
6. Objem tělesa, které vzikne rotací tohoto obrazce je možno vyjádřit jako určitý integrál
V  = n                                                    dx.
7. Po umocnění integrandu a po integraci (volte nulovou integrační konstantu) máme pro objem vztah
V  = 71
8. Výsledný objem je V =                n.
^That's all. The user is kindly asked to send his comments to these quizzes to my E-mail address.
IWToťvše. Prosím uživatele, aby své případné komentáře a náměty zasílali na moji E-mailovou adresu.
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