Show that if $\int_{a}^{b}g(t)dt=0$ then $g(t)=0$ for all $t\in[a,b]$

Question

Suppose $g$ is a continuous function on $[a,b]$ and that $g(t)\ge0$ for all $t\in[a,b]$. Show that if $\int_{a}^{b}g(t)dt=0$ then $g(t)=0$ for all $t\in[a,b]$

I know that there exists some function $G(x)= \int_{a}^{b}g(t)dt$ where $G(x)$ is differentialable at any $c$ in the interval where $G'(c)=g(c)$. Now since $g(t)$ is always positive it should follow that $G'(x)$ is always positive therefore $G(x)$ would be increasing.

Now given $G(x)= \int_{a}^{b}g(t)dt =0$ would mean that $G(b)-G(a)=0$ thus $G(b)=G(a)$ now since $G(x)$ is increasing that would mean that $g(t)=0$ for all $t$ in the interval.

if $g(x_0)>0$ then $g$ is nonzero in a neighbourhood of $x_0$. — Angina Seng, May 08 '18 at 06:39
i'm not familiar with the neighbourhood ideas. Could I still solve the problem by trying to use the fundamental theorem of calculus? — HighSchool15, May 08 '18 at 06:44

score 2 · Accepted Answer · answered May 08 '18 at 06:55

Your answer is very close to the right one.

Let $G$ be a primitive of $g$ on $[a,b]$, Since $g\geqslant 0$, $G$ is increasing and $a<b$ implies that $G(a)\leqslant G(b)$. Thus, $\int_a^b g(x) dx \geqslant 0$.

If $\int_a^b g(x) dx = 0$ then $G(a)=G(b)$ and since $G$ is increasing, it is constant over $[a,b]$ therefore its derivative $g$ is zero.

The reciprocal is fairly obvious.

Notice how you wrote $G(x) =\int_a^b g(t) dt$ but that expression does not depend on $x$.

score 0 · Answer 2 · answered May 08 '18 at 06:54

0

We have

$G(x)= \int_{a}^{x}g(t)dt$ is differentiable on $[a,b]$
$G'(x) = g(x) \geq 0 \Rightarrow G$ is non-decreasing
$0 = G(b) \geq G(x) \geq G(a) = 0 \Rightarrow G(x) = 0 \Rightarrow g(x) = G'(x) = 0 $ on $[a,b]$

answered May 08 '18 at 06:54

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Show that if $\int_{a}^{b}g(t)dt=0$ then $g(t)=0$ for all $t\in[a,b]$

2 Answers2