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paulika911
Pe: 2010-02-19T17:07:51+02:00 In: In:

integrala

functia F(x):R->R,F(x)=sinx,este primitiva a functiei f:R->R.Cat este f(pi/3)?[/code]

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10 raspunsuri

  1. Anonymous

    F'(x)=f(x) => f(x)=cos x
    f(PI/3)=cos PI/3 =1/2

  2. paulika911

    ms mult…mi-am batut capu 2 ore,si nu mi-am dat seama ca era simplu.multumesc mult.

  4. Anonymous

    Cu placere.

  5. emavali

    Am nevoie de rezolvarea unor exercitii, dar nu am instalat programul pentru scrierea in termeni matematici, astfel am sa incerc sa le scriu prin litere, cum pot, dupa cum am mai observat pe-aici.
    Va multumesc anticipat !
    1) Fie f;[0,1] apartinand lui R, f(x)=x3/x+1, Sa se calculeze integrala(x+1).f(x)dx ( am folosit pentru inmultire punctul normal)
    2)Se considera functiile f,g;(o,+ infinit)apartinand lui R,
    f(x)=ex( e la puterea x ) si g(x)=1/x
    a) sa se calculeze primitivele functiei f+g
    b)sa se arate ca integrala de 2 sus si 1 jos(f2( f la patrat)+g2( g la a 2a)(x))dx=e4(e la a 4a)-e2-1/2
    / reprezinta fractia.
    3)Se considera functia f;[1, +infinit]apartinand lui R,
    f(x)=1/x(1+lnx)
    a)sa se calculeze integrala de „e” sus, 1 jos f'(x)dx
    4)Sa se calculeze;
    a) integrala de 2 sus, 1 jos(x2-2x+3)dx=
    b) integrala de 1 sus, 0 jos(x+5/x)d=
    c)integrala de 1 sus, 0 jos(x-3ex( e la puterea x))dx=
    d)integrala de 2 sus, 1 jos(1+radical din x)2(la puterea a 2a)dx=
    Va rog ajutati-ma si, daca se poate, mai intai scrieti in termeni matematici exercitiile( m-am chinuit ceva sa le scriu astfel), apoi rezolvarea. Va multumesc !

  6. emavali

    …precizez ca la …e4-e2-1/2, fractia se aplica la intreaga formula prezentata aici , nu doar la 1.

  8. BlackRing
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    fara -> dupa x+1(nu stiu de ce apare)

  9. BlackRing

    $\begin{array}{l} f,g:\left( {0, + \infty } \right) \to R,f\left( x \right) = {{\rm{e}}^x},g\left( x \right) = \frac{1}{x} \\ a) \smallint \nolimits^ \left( {f + g} \right)\left( x \right){\rm{d}}x = \smallint \nolimits^ f\left( x \right){\rm{d}}x + \smallint \nolimits^ g\left( x \right){\rm{d}}x = \smallint \nolimits^ {{\rm{e}}^x}{\rm{d}}x + \smallint \nolimits^ \frac{1}{x}{\rm{d}}x = \\ {{\rm{e}}^x} + \ln \left| x \right| + C \\ b) \smallint \limits_1^2 {f^2}\left( x \right){\rm{d}}x + \smallint \limits_1^2 {g^2}\left( x \right){\rm{d}}x = \smallint \limits_1^2 {{\rm{e}}^{2x}}{\rm{d}}x + \smallint \limits_1^2 \frac{1}{{{x^2}}}{\rm{d}}x = \frac{1}{2} \cdot \left. {{{\rm{e}}^{2x}}} \right|_1^2 + \left. {\frac{{{x^{ - 1}}}}{{ - 1}}} \right|_1^2 = \frac{{{{\rm{e}}^4} - {{\rm{e}}^2}}}{2} - \frac{1}{2} + 1 = ... \\ \end{array}$

  10. BlackRing


    $\begin{array}{l} f:\left[ {1,\infty } \right) \to R,f\left( x \right) = \frac{1}{{x\left( {1 + \ln x} \right)}}{\rm{ }}sau{\rm{ }}\frac{{1 + \ln x}}{x}? \\ {\rm{In cazul }}\frac{{1 + \ln x}}{x},{f^\prime }\left( x \right) = \frac{{\frac{1}{x} \cdot x - 1 - \ln x}}{{{x^2}}} = \frac{{ - \ln x}}{{{x^2}}} \\ - \smallint \limits_1^e \frac{{\ln x}}{{{x^2}}}{\rm{d}}x = - \smallint \limits_1^e {\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)^\prime } \cdot \ln x{\rm{d}}x = - \frac{{{x^{ - 1}}}}{{ - 1}} \cdot \ln \left. x \right|_1^e - \smallint \limits_1^e \left( {\frac{1}{x} \cdot \frac{1}{x}} \right){\rm{d}}x = \\ - \frac{{{x^{ - 1}}}}{{ - 1}} \cdot \ln \left. x \right|_1^e - \left. {\frac{{{x^{ - 1}}}}{{ - 1}}} \right|_1^e = ... \\ \end{array}$

    Calculele le faci tu ca sunt obosit …

  12. BlackRing

    $\begin{array}{l} a) \smallint \limits_1^2 \left( {{x^2} - 2x + 3} \right){\rm{d}}x = \left. {\left( {\frac{{{x^3}}}{3} - \frac{{2{x^2}}}{2} + 3x} \right)} \right|_1^2 = ... \\ b) \smallint \limits_0^1 \left( {x + \frac{5}{x}} \right){\rm{d}}x = \left. {\left( {\frac{{{x^2}}}{2} + 5\ln \left| x \right|} \right)} \right|_0^1 = ... \\ c) \smallint \limits_0^1 \left( {x - 3{{\rm{e}}^x}} \right){\rm{d}}x = \left. {\left( {\frac{{{x^2}}}{2} - 3{{\rm{e}}^x}} \right)} \right|_0^1 = ... \\ d) \smallint \limits_1^2 {\left( {1 + \sqrt x } \right)^2}{\rm{d}}x = \smallint \limits_1^2 \left( {1 + 2\sqrt x + x} \right){\rm{d}}x = \smallint \limits_1^2 {\rm{d}}x + 2 \smallint \limits_1^2 {x^{\frac{1}{2}}}{\rm{d}}x + \smallint \limits_1^2 x \cdot {\rm{d}}x = \left. x \right|_1^2 + 2\left. {\frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}}} \right|_1^2 + \left. {\frac{{{x^2}}}{2}} \right|_1^2 = 2 - 1 + 2 \cdot \frac{2}{3} \cdot \left. {{x^{\frac{3}{2}}}} \right|_1^2 + 2 - \frac{1}{2} = ... \\ \end{array}$

  13. emavali

    Multumesc, multumesc foarte mult, BlackRing !

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