Find the interval in which $int_{-1}^{x} (e^{t} -1)(2-t) dt , (x>-1)$.












1












$begingroup$



Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.




By solving integral
$$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
= [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
and this gives me
$$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
now, i don't know how to proceed.
I don't how to find roots so that i can get the interval.



Is there some easy way to solve the problem?






calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.




    By solving integral
    $$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
    = [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
    and this gives me
    $$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
    now, i don't know how to proceed.
    I don't how to find roots so that i can get the interval.



    Is there some easy way to solve the problem?






    calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.




      By solving integral
      $$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
      = [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
      and this gives me
      $$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
      now, i don't know how to proceed.
      I don't how to find roots so that i can get the interval.



      Is there some easy way to solve the problem?






      calculus







      share|cite|improve this question











      $endgroup$





      Find the interval in which $F(x) = int_{-1}^{x} (e^{t}-1)(2-t) dt$ , $(x>-1)$ is increasing.




      By solving integral
      $$int_{-1}^{x} ((2-t)e^{t} -2 +t)dt
      = [(2-t)e^{t} + e^{t} -2t + t^{2}/2]_{-1}^{x}$$
      and this gives me
      $$3e^{x} -xe^{x} -2x + x^{2}/2 - 4/e - 5/2$$
      now, i don't know how to proceed.
      I don't how to find roots so that i can get the interval.



      Is there some easy way to solve the problem?






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 11:49









      Robert Z

      94.2k1061132




      94.2k1061132










      asked Jan 1 at 11:39









      MathsaddictMathsaddict

      2908




      2908






















          1 Answer
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          $begingroup$

          Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
          $$f(x)=(e^x-1)(2-x),$$
          so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
            $endgroup$
            – Mathsaddict
            Jan 1 at 11:51











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          1 Answer
          1






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          active

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          3












          $begingroup$

          Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
          $$f(x)=(e^x-1)(2-x),$$
          so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
            $endgroup$
            – Mathsaddict
            Jan 1 at 11:51
















          3












          $begingroup$

          Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
          $$f(x)=(e^x-1)(2-x),$$
          so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
            $endgroup$
            – Mathsaddict
            Jan 1 at 11:51














          3












          3








          3





          $begingroup$

          Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
          $$f(x)=(e^x-1)(2-x),$$
          so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.






          share|cite|improve this answer









          $endgroup$



          Hint. Note that by the Fundamental Theorem of Calculus the derivative of $F$ is the integrand function
          $$f(x)=(e^x-1)(2-x),$$
          so in order to find the interval where $F$ is increasing, it suffices to discuss the sign of $f$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 1 at 11:45









          Robert ZRobert Z

          94.2k1061132




          94.2k1061132








          • 1




            $begingroup$
            Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
            $endgroup$
            – Mathsaddict
            Jan 1 at 11:51














          • 1




            $begingroup$
            Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
            $endgroup$
            – Mathsaddict
            Jan 1 at 11:51








          1




          1




          $begingroup$
          Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
          $endgroup$
          – Mathsaddict
          Jan 1 at 11:51




          $begingroup$
          Oh yeah, by applying leibnitz rule, I am getting this.The roots of derivative are $0$ and $2$. And the interval is $[0,2]$. Thanks.
          $endgroup$
          – Mathsaddict
          Jan 1 at 11:51


















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