Show that if $f : [a, b] \to \mathbb{R}$ is non-negative and continuous on $[a, b]$ and $$\int_{a}^{b} f(x)dx = 0$$ then $f(x)=0$, for all $x ∈ [a, b]$.
I'm having difficulties in proving things to do with integrals, how do I start this?
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3 Answers
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Suppose that $f(z)> 0$ for some $z\in[a,b]$. Then, since $f$ is continuous, there exists a neighbourhood $(c,d)$ around $z$ such that $f(x)>\frac{f(z)}{2}>0$ for all $x\in(c,d)$. But then $\int_c^df(x)dx>0$. You should be able to complete the proof from there.
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It's worth getting some intuition for problems of this kind.
The fact that the function is nonnegative means if you plot the function then it doesn't dip below the x-axis.
If it is continuous, then very roughly speaking, you can plot the function without taking your hand off the page.
Finally, if the definite integral is zero then the area under the curve is zero.
To prove this result, you can think about what it would mean if the function was positive at some point. In particular, think about what it would mean for such a function to be continuous in the region where it is positive. Finally you can think about what the value of the integral is over the region where the function is positive.
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Suppose that there exists $x_0 $ s.t $f(x_0) > 0 $.
f is continuous so there exists $\delta >0 $ s.t for each x if $x\in[x_0-\delta,x_0+\delta] $ than $f(x) > \dfrac {f(x_0)}{2} > 0 $ (for $\epsilon = (\dfrac {f(x_0)}{2} ) $ now split the integral of [a,b] to $[a,x_0-\delta],[x_0-\delta,x_0 +\delta] ,[x_0+\delta, b] $ and get to a contradiction.