# Let $f:[0,1]\to\mathbb R$ be continuous such that $f(t)\geq 0$ for all $t$ in $[0,1]$. What can be said about $g(x):=\int_0^x f(t)\,dt$?

I am studying for exams and am stuck on this problem.

Let $f:[0,1]\rightarrow \mathbb R$ be continuous such that $f(t)\geqslant 0$ for all $t$ in $[0,1]$. Define $g(x)=\int_{0}^{x}f(t)dt$.Then which of the following is correct?

(a) $g$ is monotone and bounded,

(b) $g$ is monotone but not bounded,

(c) $g$ is bounded but not monotone,

(d) $g$ is neither monotone nor bounded.

I take $f(t)=t+1$ and proceed according to the given condition. I see that g is bounded and monotone. Am I right? I want a better way to approach the problem. Please help.

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Hint: $g(x)$ is the area below the graph of $f$ from $0$ to $x$. –  Sigur Nov 20 '12 at 15:21

Since $f$ is continuous and nonnegative on the closed, bounded interval $[0,1]$, then there is some $M>0$ such that $0\leq f(t)\leq M$ for all $t$ in $[0,1]$. We can use this to show that $0\leq\int_0^xf(t)\,dt\leq M$ for all $x$ in $[0,1]$, and so $g$ is bounded.

Since $f$ is nonnegative, then for any $x,y$ in $[0,1]$ such that $x<y$, we have $\int_x^yf(t)\,dt\geq 0$, so since $g(y)=g(x)+\int_x^yf(t)\,dt$, then it follows that $g$ is monotone.

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If $x\in[0,1]$, you're ok with $(a)$, yes. Since $f$ is continuous over $[0,1]$ it is bounded there, whence the integral cannot be unbounded over $0\leq x \leq 1$. Note that $g'=f$, and $f\geq 0$. What does this tell you about how $g$ grows?

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You can split the problem into two parts:

1. Is $g$ monotone? Yes, because integrals are positive operators. That means that if $f \ge 0$, $\int_a^b f(x) dx \ge 0$. By choosing $a,b \in [0,1]$ such that $a<b$, we get $g(b) = g(a) + \int_a^b f(x) dx \ge g(a)$.

2. Is $g$ bounded? Again, the answer is yes. We use positivity again, as well as the fact that a continuous function takes its maximum on a compact set. Let $M = \max_{x \in [0,1]} f(x)$. Then $0 \le \int_{0}^x f(t) dt \le \int_{0}^x M dt = Mx$.

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To see that $g(x)$ is bounded note that $f$ is continuous so it has a maximum on $[0,1]$ say $M$. Can you see how to use $M$ to bound $g(x)$?

To show that $g$ is increasing you need to show that if $x<y$ then $g(x) \leq g(y)$ in particular we want to show

$$\int_0^y f(t)dt-\int_0^xf(t)dt=\int_x^yf(t)dt\geq 0.$$

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