5

Suppose that $f$ and $g$ are real valued continuous functions on $[0,1]$ such that $f(t)<g(t)$ for all $t\in[0,1]$. Is it true that $$\int_{0}^{1}f(t) \ dt<\int_{0}^{1}g(t) \ dt \ ?$$ I know that this result is true when we replace "$<$" by "$\leq$".

Calculix
  • 3,354
  • 1
    Even if $f(t)\le g(t)$ for all $t$ and $f(t)<g(t)$ for at least one point in $[0,1]$ then also $\int_{0}^{1}f(t) \ dt<\int_{0}^{1}g(t) \ dt $ holds. –  Oct 16 '19 at 15:54
  • 1
    I assume this is only true when $f$ and $g$ are continuous (as in my case)? Since continuity implies strict inequality in a neighbourhood of the point where strict inequality holds. – Calculix Oct 16 '19 at 16:07
  • 1
    At the point of strict inequality, continuity is sufficient. –  Oct 16 '19 at 16:12
  • 1
    Why the downvote? This is a legitimate question, I have seen much worse. – Giuseppe Negro Oct 16 '19 at 20:20

2 Answers2

9

Yes, of course, because the integral of a positive continuous function is positive: $$\int_0^1g(x)dx-\int_0^1f(x)dx=\int_0^1(g(x)-f(x))dx>0.$$

Michael Rozenberg
  • 194,933
1

lets suppose that: $$f(t)<g(t)\Rightarrow f(t)-g(t)<0$$ intuitively this would give: $$\int_0^1\left[f(t)-g(t)\right]dt<0$$ which is the same as: $$\int_0^1f(t)dt<\int_0^1g(t)dt$$ since integrals follow the property: $$\int\left[\alpha(x)+\beta(x)\right]dx=\int\alpha(x)dx+\int\beta(x)dx$$

Henry Lee
  • 12,215