Find the number of roots of $F(x)= int_0^x e^t(t^2-3t-5)mathrm dt , x>0$ in the interval $(0,4)$












3












$begingroup$



Let $$F(x)= int_0^x e^t(t^2-3t-5)mathrm dt , x>0$$
Find the number of roots of $F(x)=0$ in the interval $(0,4)$.




My attempt: I simply integrated it and got $F(x)=e^x(x(x-5))$ which has roots $0$ and $5$ and none is between $0$ and $4$. So the answer is $0$, which is correct.



My question: Is there any other way to do this question in comparatively less time?






integration roots







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$endgroup$

















    3












    $begingroup$



    Let $$F(x)= int_0^x e^t(t^2-3t-5)mathrm dt , x>0$$
    Find the number of roots of $F(x)=0$ in the interval $(0,4)$.




    My attempt: I simply integrated it and got $F(x)=e^x(x(x-5))$ which has roots $0$ and $5$ and none is between $0$ and $4$. So the answer is $0$, which is correct.



    My question: Is there any other way to do this question in comparatively less time?






    integration roots







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      Let $$F(x)= int_0^x e^t(t^2-3t-5)mathrm dt , x>0$$
      Find the number of roots of $F(x)=0$ in the interval $(0,4)$.




      My attempt: I simply integrated it and got $F(x)=e^x(x(x-5))$ which has roots $0$ and $5$ and none is between $0$ and $4$. So the answer is $0$, which is correct.



      My question: Is there any other way to do this question in comparatively less time?






      integration roots







      share|cite|improve this question











      $endgroup$





      Let $$F(x)= int_0^x e^t(t^2-3t-5)mathrm dt , x>0$$
      Find the number of roots of $F(x)=0$ in the interval $(0,4)$.




      My attempt: I simply integrated it and got $F(x)=e^x(x(x-5))$ which has roots $0$ and $5$ and none is between $0$ and $4$. So the answer is $0$, which is correct.



      My question: Is there any other way to do this question in comparatively less time?






      integration roots




      integration roots






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 10:51









      mrtaurho

      6,19571641




      6,19571641










      asked Jan 15 at 10:48









      user233797user233797

      162




      162






















          3 Answers
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          4












          $begingroup$

          You have $F(0)=0, F'(x)=e^x(x^2-3x-5)<0$ in $(0,4)$. This means $F(x)$ is strictly decreasing in $(0,4)$, so that $F(x)<0$ in $(0,4)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. It was really helpful.
            $endgroup$
            – user233797
            Jan 15 at 11:01










          • $begingroup$
            You are welcome
            $endgroup$
            – Shubham Johri
            Jan 15 at 11:02



















          2












          $begingroup$

          Hint: note that
          $$x^2-3x-5<0$$
          On the interval $(0, 4)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay I got it. Thank you.
            $endgroup$
            – user233797
            Jan 15 at 11:02



















          2












          $begingroup$

          Yes: it's easy to see, without computing the roots, that $t^2-3t-5$ has a negative and a positive root, and the latter is $>4$. Using the First Fundamental Theorem of Integral Calculus, $F(x)$ decreases on the interval $(0,4)$. Now $F(0)=0$, hence $F(x)<0$ for $0<x<4$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. I got it now.
            $endgroup$
            – user233797
            Jan 15 at 11:24












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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          You have $F(0)=0, F'(x)=e^x(x^2-3x-5)<0$ in $(0,4)$. This means $F(x)$ is strictly decreasing in $(0,4)$, so that $F(x)<0$ in $(0,4)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. It was really helpful.
            $endgroup$
            – user233797
            Jan 15 at 11:01










          • $begingroup$
            You are welcome
            $endgroup$
            – Shubham Johri
            Jan 15 at 11:02
















          4












          $begingroup$

          You have $F(0)=0, F'(x)=e^x(x^2-3x-5)<0$ in $(0,4)$. This means $F(x)$ is strictly decreasing in $(0,4)$, so that $F(x)<0$ in $(0,4)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. It was really helpful.
            $endgroup$
            – user233797
            Jan 15 at 11:01










          • $begingroup$
            You are welcome
            $endgroup$
            – Shubham Johri
            Jan 15 at 11:02














          4












          4








          4





          $begingroup$

          You have $F(0)=0, F'(x)=e^x(x^2-3x-5)<0$ in $(0,4)$. This means $F(x)$ is strictly decreasing in $(0,4)$, so that $F(x)<0$ in $(0,4)$.






          share|cite|improve this answer









          $endgroup$



          You have $F(0)=0, F'(x)=e^x(x^2-3x-5)<0$ in $(0,4)$. This means $F(x)$ is strictly decreasing in $(0,4)$, so that $F(x)<0$ in $(0,4)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 10:57









          Shubham JohriShubham Johri

          5,668918




          5,668918












          • $begingroup$
            Thank you. It was really helpful.
            $endgroup$
            – user233797
            Jan 15 at 11:01










          • $begingroup$
            You are welcome
            $endgroup$
            – Shubham Johri
            Jan 15 at 11:02


















          • $begingroup$
            Thank you. It was really helpful.
            $endgroup$
            – user233797
            Jan 15 at 11:01










          • $begingroup$
            You are welcome
            $endgroup$
            – Shubham Johri
            Jan 15 at 11:02
















          $begingroup$
          Thank you. It was really helpful.
          $endgroup$
          – user233797
          Jan 15 at 11:01




          $begingroup$
          Thank you. It was really helpful.
          $endgroup$
          – user233797
          Jan 15 at 11:01












          $begingroup$
          You are welcome
          $endgroup$
          – Shubham Johri
          Jan 15 at 11:02




          $begingroup$
          You are welcome
          $endgroup$
          – Shubham Johri
          Jan 15 at 11:02











          2












          $begingroup$

          Hint: note that
          $$x^2-3x-5<0$$
          On the interval $(0, 4)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay I got it. Thank you.
            $endgroup$
            – user233797
            Jan 15 at 11:02
















          2












          $begingroup$

          Hint: note that
          $$x^2-3x-5<0$$
          On the interval $(0, 4)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay I got it. Thank you.
            $endgroup$
            – user233797
            Jan 15 at 11:02














          2












          2








          2





          $begingroup$

          Hint: note that
          $$x^2-3x-5<0$$
          On the interval $(0, 4)$






          share|cite|improve this answer









          $endgroup$



          Hint: note that
          $$x^2-3x-5<0$$
          On the interval $(0, 4)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 10:57









          BotondBotond

          6,69531034




          6,69531034












          • $begingroup$
            Okay I got it. Thank you.
            $endgroup$
            – user233797
            Jan 15 at 11:02


















          • $begingroup$
            Okay I got it. Thank you.
            $endgroup$
            – user233797
            Jan 15 at 11:02
















          $begingroup$
          Okay I got it. Thank you.
          $endgroup$
          – user233797
          Jan 15 at 11:02




          $begingroup$
          Okay I got it. Thank you.
          $endgroup$
          – user233797
          Jan 15 at 11:02











          2












          $begingroup$

          Yes: it's easy to see, without computing the roots, that $t^2-3t-5$ has a negative and a positive root, and the latter is $>4$. Using the First Fundamental Theorem of Integral Calculus, $F(x)$ decreases on the interval $(0,4)$. Now $F(0)=0$, hence $F(x)<0$ for $0<x<4$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. I got it now.
            $endgroup$
            – user233797
            Jan 15 at 11:24
















          2












          $begingroup$

          Yes: it's easy to see, without computing the roots, that $t^2-3t-5$ has a negative and a positive root, and the latter is $>4$. Using the First Fundamental Theorem of Integral Calculus, $F(x)$ decreases on the interval $(0,4)$. Now $F(0)=0$, hence $F(x)<0$ for $0<x<4$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. I got it now.
            $endgroup$
            – user233797
            Jan 15 at 11:24














          2












          2








          2





          $begingroup$

          Yes: it's easy to see, without computing the roots, that $t^2-3t-5$ has a negative and a positive root, and the latter is $>4$. Using the First Fundamental Theorem of Integral Calculus, $F(x)$ decreases on the interval $(0,4)$. Now $F(0)=0$, hence $F(x)<0$ for $0<x<4$.






          share|cite|improve this answer









          $endgroup$



          Yes: it's easy to see, without computing the roots, that $t^2-3t-5$ has a negative and a positive root, and the latter is $>4$. Using the First Fundamental Theorem of Integral Calculus, $F(x)$ decreases on the interval $(0,4)$. Now $F(0)=0$, hence $F(x)<0$ for $0<x<4$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 11:04









          BernardBernard

          124k743118




          124k743118












          • $begingroup$
            Thank you. I got it now.
            $endgroup$
            – user233797
            Jan 15 at 11:24


















          • $begingroup$
            Thank you. I got it now.
            $endgroup$
            – user233797
            Jan 15 at 11:24
















          $begingroup$
          Thank you. I got it now.
          $endgroup$
          – user233797
          Jan 15 at 11:24




          $begingroup$
          Thank you. I got it now.
          $endgroup$
          – user233797
          Jan 15 at 11:24


















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