How do you prove the integral of a positive function is also positive?












-1












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It seems straightforward, every time I draw a picture of a function that's always greater than zero, then the area under it is also always greater than zero. But then when I look at the definition of an integral, it's some convoluted summation that I don't use for anything. Is it actually true that if $f>0$ over some [a,b] then the $ int_{a}^{b}fdx>0$? Because I can't find any function where that's not true, assuming there's no discontinuities.






definite-integrals







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  • 4




    $begingroup$
    The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:11










  • $begingroup$
    Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition.
    $endgroup$
    – user608672
    Dec 5 '18 at 1:26












  • $begingroup$
    I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:42










  • $begingroup$
    Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:45










  • $begingroup$
    Riemann integral
    $endgroup$
    – user608672
    Dec 5 '18 at 2:40
















-1












$begingroup$


It seems straightforward, every time I draw a picture of a function that's always greater than zero, then the area under it is also always greater than zero. But then when I look at the definition of an integral, it's some convoluted summation that I don't use for anything. Is it actually true that if $f>0$ over some [a,b] then the $ int_{a}^{b}fdx>0$? Because I can't find any function where that's not true, assuming there's no discontinuities.






definite-integrals







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:11










  • $begingroup$
    Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition.
    $endgroup$
    – user608672
    Dec 5 '18 at 1:26












  • $begingroup$
    I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:42










  • $begingroup$
    Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:45










  • $begingroup$
    Riemann integral
    $endgroup$
    – user608672
    Dec 5 '18 at 2:40














-1












-1








-1





$begingroup$


It seems straightforward, every time I draw a picture of a function that's always greater than zero, then the area under it is also always greater than zero. But then when I look at the definition of an integral, it's some convoluted summation that I don't use for anything. Is it actually true that if $f>0$ over some [a,b] then the $ int_{a}^{b}fdx>0$? Because I can't find any function where that's not true, assuming there's no discontinuities.






definite-integrals







share|cite|improve this question









$endgroup$




It seems straightforward, every time I draw a picture of a function that's always greater than zero, then the area under it is also always greater than zero. But then when I look at the definition of an integral, it's some convoluted summation that I don't use for anything. Is it actually true that if $f>0$ over some [a,b] then the $ int_{a}^{b}fdx>0$? Because I can't find any function where that's not true, assuming there's no discontinuities.






definite-integrals




definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 0:55









user608672user608672

64




64








  • 4




    $begingroup$
    The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:11










  • $begingroup$
    Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition.
    $endgroup$
    – user608672
    Dec 5 '18 at 1:26












  • $begingroup$
    I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:42










  • $begingroup$
    Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:45










  • $begingroup$
    Riemann integral
    $endgroup$
    – user608672
    Dec 5 '18 at 2:40














  • 4




    $begingroup$
    The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:11










  • $begingroup$
    Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition.
    $endgroup$
    – user608672
    Dec 5 '18 at 1:26












  • $begingroup$
    I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof.
    $endgroup$
    – Matt Samuel
    Dec 5 '18 at 1:42










  • $begingroup$
    Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:45










  • $begingroup$
    Riemann integral
    $endgroup$
    – user608672
    Dec 5 '18 at 2:40








4




4




$begingroup$
The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities.
$endgroup$
– Matt Samuel
Dec 5 '18 at 1:11




$begingroup$
The only way to tell is by looking at the "convoluted summation." It's true even if there are discontinuities.
$endgroup$
– Matt Samuel
Dec 5 '18 at 1:11












$begingroup$
Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition.
$endgroup$
– user608672
Dec 5 '18 at 1:26






$begingroup$
Except your assertion is wrong. I looked at it and the definition is not itself a proof for an extrapolation. In fact, no definition is ever a proof, it's just something taken to be true, by definition.
$endgroup$
– user608672
Dec 5 '18 at 1:26














$begingroup$
I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof.
$endgroup$
– Matt Samuel
Dec 5 '18 at 1:42




$begingroup$
I don't believe anything in my comment implied that the definition was a proof. But you can use the definition in a proof.
$endgroup$
– Matt Samuel
Dec 5 '18 at 1:42












$begingroup$
Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct.
$endgroup$
– xbh
Dec 5 '18 at 1:45




$begingroup$
Which integral are you thinking about? Riemann, Lebesgue, or other kind? You need to specify it. E.G. for Riemann integrals, the conclusion is correct.
$endgroup$
– xbh
Dec 5 '18 at 1:45












$begingroup$
Riemann integral
$endgroup$
– user608672
Dec 5 '18 at 2:40




$begingroup$
Riemann integral
$endgroup$
– user608672
Dec 5 '18 at 2:40










3 Answers
3






active

oldest

votes


















1












$begingroup$

Since $f(x)>0$ it is clear that any Riemann sum of the function is bounded below by 0.



To bound it away from 0, since a Riemann integrable function is continuous almost everywhere, it is continuous at some point $x_*in[a,b]$.



$f(x_*)>0$ by hypothesis. Let $epsilon=f(x_*)/2>0$. By continuity of $f$, there exists $delta>0$ such that
$$begin{align*}left|x-x_*right|<deltaquad&Rightarrowquadleft|f(x)-f(x_*)right|<epsilon\
&Rightarrowquad f(x)>f(x_*)-epsilon=f(x_*)/2>0text{.}end{align*}$$



At this point, we have that within the non-empty interval $(max(x_*-delta,a),min(x_*+delta,b))$, the function $f(x)>f(x_*)/2>0$ is bounded away from $0$.



For any Riemann sum such that both $max(x_*-delta,a)$ and $min(x_*+delta,b)$ are partition points, it is clear that the Riemann sum is bounded below by $(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$.



The integral is the limit as the Riemann sums are refined, so
$$int_a^bf(x),mathrm{d}xgeq(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    A way to see this without using the definition of an integral is to use the fundamental theorem of calculus.



    We have that for any antiderivative of $f$, say $F$, that $int_a^x f(t)dt = F(x)-F(a)$. Now we see that for $x$ in $(a,b),$



    $$frac{d}{dx}int_a^x f(t)dt=F’(x)=f(x)>0.$$



    So if we believe that a positive derivative indicates that a function is strictly increasing, we have that $int_a^a f(t)dt = 0$ and that the integral is strictly increasing, so that $int_a^x f(t)dt > 0$ for any $x$ in $(a,b]$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You can use the definition, it just isn't immediately apparent to me how.
      $endgroup$
      – user608672
      Dec 5 '18 at 3:47










    • $begingroup$
      I gave another answer utilizing the definition.
      $endgroup$
      – John B
      Dec 5 '18 at 4:34



















    0












    $begingroup$

    Assume that $f$ is continuous.



    By the extreme value theorem, $f$ takes a minimum value on $[a,b]$. Let’s call this $m$. Note that $m$>0 since $f$ is strictly positive.



    Now let us break apart $[a,b]$ into $N$ equal parts: $[a,a+frac{b-a}{N}]$, $[a+frac{b-a}{N},a+2 frac{b-a}{N}]$, ..., $[b-frac{b-a}{N}, b]$ each of length $frac{b-a}{N} $. Then let $a_n$ be some value from the $n$th part. The integral is then given by $$int_a^b f(x)dx=lim_{N to infty} sum_{n=1}^N frac{b-a}{N} f(a_n)ge lim_{N to infty} sum_{n=1}^N frac{b-a}{N}m=(b-a)m>0.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
      $endgroup$
      – obscurans
      Dec 5 '18 at 5:18












    • $begingroup$
      I will add the assumption that f is continuous
      $endgroup$
      – John B
      Dec 5 '18 at 5:34










    • $begingroup$
      But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
      $endgroup$
      – obscurans
      Dec 5 '18 at 5:43










    • $begingroup$
      True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
      $endgroup$
      – John B
      Dec 5 '18 at 5:49











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Since $f(x)>0$ it is clear that any Riemann sum of the function is bounded below by 0.



    To bound it away from 0, since a Riemann integrable function is continuous almost everywhere, it is continuous at some point $x_*in[a,b]$.



    $f(x_*)>0$ by hypothesis. Let $epsilon=f(x_*)/2>0$. By continuity of $f$, there exists $delta>0$ such that
    $$begin{align*}left|x-x_*right|<deltaquad&Rightarrowquadleft|f(x)-f(x_*)right|<epsilon\
    &Rightarrowquad f(x)>f(x_*)-epsilon=f(x_*)/2>0text{.}end{align*}$$



    At this point, we have that within the non-empty interval $(max(x_*-delta,a),min(x_*+delta,b))$, the function $f(x)>f(x_*)/2>0$ is bounded away from $0$.



    For any Riemann sum such that both $max(x_*-delta,a)$ and $min(x_*+delta,b)$ are partition points, it is clear that the Riemann sum is bounded below by $(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$.



    The integral is the limit as the Riemann sums are refined, so
    $$int_a^bf(x),mathrm{d}xgeq(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Since $f(x)>0$ it is clear that any Riemann sum of the function is bounded below by 0.



      To bound it away from 0, since a Riemann integrable function is continuous almost everywhere, it is continuous at some point $x_*in[a,b]$.



      $f(x_*)>0$ by hypothesis. Let $epsilon=f(x_*)/2>0$. By continuity of $f$, there exists $delta>0$ such that
      $$begin{align*}left|x-x_*right|<deltaquad&Rightarrowquadleft|f(x)-f(x_*)right|<epsilon\
      &Rightarrowquad f(x)>f(x_*)-epsilon=f(x_*)/2>0text{.}end{align*}$$



      At this point, we have that within the non-empty interval $(max(x_*-delta,a),min(x_*+delta,b))$, the function $f(x)>f(x_*)/2>0$ is bounded away from $0$.



      For any Riemann sum such that both $max(x_*-delta,a)$ and $min(x_*+delta,b)$ are partition points, it is clear that the Riemann sum is bounded below by $(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$.



      The integral is the limit as the Riemann sums are refined, so
      $$int_a^bf(x),mathrm{d}xgeq(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Since $f(x)>0$ it is clear that any Riemann sum of the function is bounded below by 0.



        To bound it away from 0, since a Riemann integrable function is continuous almost everywhere, it is continuous at some point $x_*in[a,b]$.



        $f(x_*)>0$ by hypothesis. Let $epsilon=f(x_*)/2>0$. By continuity of $f$, there exists $delta>0$ such that
        $$begin{align*}left|x-x_*right|<deltaquad&Rightarrowquadleft|f(x)-f(x_*)right|<epsilon\
        &Rightarrowquad f(x)>f(x_*)-epsilon=f(x_*)/2>0text{.}end{align*}$$



        At this point, we have that within the non-empty interval $(max(x_*-delta,a),min(x_*+delta,b))$, the function $f(x)>f(x_*)/2>0$ is bounded away from $0$.



        For any Riemann sum such that both $max(x_*-delta,a)$ and $min(x_*+delta,b)$ are partition points, it is clear that the Riemann sum is bounded below by $(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$.



        The integral is the limit as the Riemann sums are refined, so
        $$int_a^bf(x),mathrm{d}xgeq(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$$






        share|cite|improve this answer









        $endgroup$



        Since $f(x)>0$ it is clear that any Riemann sum of the function is bounded below by 0.



        To bound it away from 0, since a Riemann integrable function is continuous almost everywhere, it is continuous at some point $x_*in[a,b]$.



        $f(x_*)>0$ by hypothesis. Let $epsilon=f(x_*)/2>0$. By continuity of $f$, there exists $delta>0$ such that
        $$begin{align*}left|x-x_*right|<deltaquad&Rightarrowquadleft|f(x)-f(x_*)right|<epsilon\
        &Rightarrowquad f(x)>f(x_*)-epsilon=f(x_*)/2>0text{.}end{align*}$$



        At this point, we have that within the non-empty interval $(max(x_*-delta,a),min(x_*+delta,b))$, the function $f(x)>f(x_*)/2>0$ is bounded away from $0$.



        For any Riemann sum such that both $max(x_*-delta,a)$ and $min(x_*+delta,b)$ are partition points, it is clear that the Riemann sum is bounded below by $(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$.



        The integral is the limit as the Riemann sums are refined, so
        $$int_a^bf(x),mathrm{d}xgeq(min(x_*+delta,b))-max(x_*-delta,a))f(x_*)/2>0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 3:04









        obscuransobscurans

        1,027311




        1,027311























            0












            $begingroup$

            A way to see this without using the definition of an integral is to use the fundamental theorem of calculus.



            We have that for any antiderivative of $f$, say $F$, that $int_a^x f(t)dt = F(x)-F(a)$. Now we see that for $x$ in $(a,b),$



            $$frac{d}{dx}int_a^x f(t)dt=F’(x)=f(x)>0.$$



            So if we believe that a positive derivative indicates that a function is strictly increasing, we have that $int_a^a f(t)dt = 0$ and that the integral is strictly increasing, so that $int_a^x f(t)dt > 0$ for any $x$ in $(a,b]$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You can use the definition, it just isn't immediately apparent to me how.
              $endgroup$
              – user608672
              Dec 5 '18 at 3:47










            • $begingroup$
              I gave another answer utilizing the definition.
              $endgroup$
              – John B
              Dec 5 '18 at 4:34
















            0












            $begingroup$

            A way to see this without using the definition of an integral is to use the fundamental theorem of calculus.



            We have that for any antiderivative of $f$, say $F$, that $int_a^x f(t)dt = F(x)-F(a)$. Now we see that for $x$ in $(a,b),$



            $$frac{d}{dx}int_a^x f(t)dt=F’(x)=f(x)>0.$$



            So if we believe that a positive derivative indicates that a function is strictly increasing, we have that $int_a^a f(t)dt = 0$ and that the integral is strictly increasing, so that $int_a^x f(t)dt > 0$ for any $x$ in $(a,b]$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You can use the definition, it just isn't immediately apparent to me how.
              $endgroup$
              – user608672
              Dec 5 '18 at 3:47










            • $begingroup$
              I gave another answer utilizing the definition.
              $endgroup$
              – John B
              Dec 5 '18 at 4:34














            0












            0








            0





            $begingroup$

            A way to see this without using the definition of an integral is to use the fundamental theorem of calculus.



            We have that for any antiderivative of $f$, say $F$, that $int_a^x f(t)dt = F(x)-F(a)$. Now we see that for $x$ in $(a,b),$



            $$frac{d}{dx}int_a^x f(t)dt=F’(x)=f(x)>0.$$



            So if we believe that a positive derivative indicates that a function is strictly increasing, we have that $int_a^a f(t)dt = 0$ and that the integral is strictly increasing, so that $int_a^x f(t)dt > 0$ for any $x$ in $(a,b]$.






            share|cite|improve this answer









            $endgroup$



            A way to see this without using the definition of an integral is to use the fundamental theorem of calculus.



            We have that for any antiderivative of $f$, say $F$, that $int_a^x f(t)dt = F(x)-F(a)$. Now we see that for $x$ in $(a,b),$



            $$frac{d}{dx}int_a^x f(t)dt=F’(x)=f(x)>0.$$



            So if we believe that a positive derivative indicates that a function is strictly increasing, we have that $int_a^a f(t)dt = 0$ and that the integral is strictly increasing, so that $int_a^x f(t)dt > 0$ for any $x$ in $(a,b]$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 3:27









            John BJohn B

            1766




            1766












            • $begingroup$
              You can use the definition, it just isn't immediately apparent to me how.
              $endgroup$
              – user608672
              Dec 5 '18 at 3:47










            • $begingroup$
              I gave another answer utilizing the definition.
              $endgroup$
              – John B
              Dec 5 '18 at 4:34


















            • $begingroup$
              You can use the definition, it just isn't immediately apparent to me how.
              $endgroup$
              – user608672
              Dec 5 '18 at 3:47










            • $begingroup$
              I gave another answer utilizing the definition.
              $endgroup$
              – John B
              Dec 5 '18 at 4:34
















            $begingroup$
            You can use the definition, it just isn't immediately apparent to me how.
            $endgroup$
            – user608672
            Dec 5 '18 at 3:47




            $begingroup$
            You can use the definition, it just isn't immediately apparent to me how.
            $endgroup$
            – user608672
            Dec 5 '18 at 3:47












            $begingroup$
            I gave another answer utilizing the definition.
            $endgroup$
            – John B
            Dec 5 '18 at 4:34




            $begingroup$
            I gave another answer utilizing the definition.
            $endgroup$
            – John B
            Dec 5 '18 at 4:34











            0












            $begingroup$

            Assume that $f$ is continuous.



            By the extreme value theorem, $f$ takes a minimum value on $[a,b]$. Let’s call this $m$. Note that $m$>0 since $f$ is strictly positive.



            Now let us break apart $[a,b]$ into $N$ equal parts: $[a,a+frac{b-a}{N}]$, $[a+frac{b-a}{N},a+2 frac{b-a}{N}]$, ..., $[b-frac{b-a}{N}, b]$ each of length $frac{b-a}{N} $. Then let $a_n$ be some value from the $n$th part. The integral is then given by $$int_a^b f(x)dx=lim_{N to infty} sum_{n=1}^N frac{b-a}{N} f(a_n)ge lim_{N to infty} sum_{n=1}^N frac{b-a}{N}m=(b-a)m>0.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:18












            • $begingroup$
              I will add the assumption that f is continuous
              $endgroup$
              – John B
              Dec 5 '18 at 5:34










            • $begingroup$
              But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:43










            • $begingroup$
              True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
              $endgroup$
              – John B
              Dec 5 '18 at 5:49
















            0












            $begingroup$

            Assume that $f$ is continuous.



            By the extreme value theorem, $f$ takes a minimum value on $[a,b]$. Let’s call this $m$. Note that $m$>0 since $f$ is strictly positive.



            Now let us break apart $[a,b]$ into $N$ equal parts: $[a,a+frac{b-a}{N}]$, $[a+frac{b-a}{N},a+2 frac{b-a}{N}]$, ..., $[b-frac{b-a}{N}, b]$ each of length $frac{b-a}{N} $. Then let $a_n$ be some value from the $n$th part. The integral is then given by $$int_a^b f(x)dx=lim_{N to infty} sum_{n=1}^N frac{b-a}{N} f(a_n)ge lim_{N to infty} sum_{n=1}^N frac{b-a}{N}m=(b-a)m>0.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:18












            • $begingroup$
              I will add the assumption that f is continuous
              $endgroup$
              – John B
              Dec 5 '18 at 5:34










            • $begingroup$
              But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:43










            • $begingroup$
              True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
              $endgroup$
              – John B
              Dec 5 '18 at 5:49














            0












            0








            0





            $begingroup$

            Assume that $f$ is continuous.



            By the extreme value theorem, $f$ takes a minimum value on $[a,b]$. Let’s call this $m$. Note that $m$>0 since $f$ is strictly positive.



            Now let us break apart $[a,b]$ into $N$ equal parts: $[a,a+frac{b-a}{N}]$, $[a+frac{b-a}{N},a+2 frac{b-a}{N}]$, ..., $[b-frac{b-a}{N}, b]$ each of length $frac{b-a}{N} $. Then let $a_n$ be some value from the $n$th part. The integral is then given by $$int_a^b f(x)dx=lim_{N to infty} sum_{n=1}^N frac{b-a}{N} f(a_n)ge lim_{N to infty} sum_{n=1}^N frac{b-a}{N}m=(b-a)m>0.$$






            share|cite|improve this answer











            $endgroup$



            Assume that $f$ is continuous.



            By the extreme value theorem, $f$ takes a minimum value on $[a,b]$. Let’s call this $m$. Note that $m$>0 since $f$ is strictly positive.



            Now let us break apart $[a,b]$ into $N$ equal parts: $[a,a+frac{b-a}{N}]$, $[a+frac{b-a}{N},a+2 frac{b-a}{N}]$, ..., $[b-frac{b-a}{N}, b]$ each of length $frac{b-a}{N} $. Then let $a_n$ be some value from the $n$th part. The integral is then given by $$int_a^b f(x)dx=lim_{N to infty} sum_{n=1}^N frac{b-a}{N} f(a_n)ge lim_{N to infty} sum_{n=1}^N frac{b-a}{N}m=(b-a)m>0.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 5:35

























            answered Dec 5 '18 at 4:32









            John BJohn B

            1766




            1766












            • $begingroup$
              $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:18












            • $begingroup$
              I will add the assumption that f is continuous
              $endgroup$
              – John B
              Dec 5 '18 at 5:34










            • $begingroup$
              But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:43










            • $begingroup$
              True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
              $endgroup$
              – John B
              Dec 5 '18 at 5:49


















            • $begingroup$
              $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:18












            • $begingroup$
              I will add the assumption that f is continuous
              $endgroup$
              – John B
              Dec 5 '18 at 5:34










            • $begingroup$
              But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
              $endgroup$
              – obscurans
              Dec 5 '18 at 5:43










            • $begingroup$
              True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
              $endgroup$
              – John B
              Dec 5 '18 at 5:49
















            $begingroup$
            $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
            $endgroup$
            – obscurans
            Dec 5 '18 at 5:18






            $begingroup$
            $m>0$ is false. Nothing stops $f$ from going arbitrarily close to $0$ with a discontinuity. Say $f(x)=x$ for $x>0$, $f(0)=1$ on the range $[0,1]$. This is clearly Riemann integrable.
            $endgroup$
            – obscurans
            Dec 5 '18 at 5:18














            $begingroup$
            I will add the assumption that f is continuous
            $endgroup$
            – John B
            Dec 5 '18 at 5:34




            $begingroup$
            I will add the assumption that f is continuous
            $endgroup$
            – John B
            Dec 5 '18 at 5:34












            $begingroup$
            But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
            $endgroup$
            – obscurans
            Dec 5 '18 at 5:43




            $begingroup$
            But then you didn't prove the statement $f>0Rightarrowint_a^bf(x)dx>0$, you added conditions.
            $endgroup$
            – obscurans
            Dec 5 '18 at 5:43












            $begingroup$
            True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
            $endgroup$
            – John B
            Dec 5 '18 at 5:49




            $begingroup$
            True. I am content with my assumption, however. The question asked ends with “assuming there’s no discontinuities.”
            $endgroup$
            – John B
            Dec 5 '18 at 5:49


















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