# Range of Integrated function

1. May 4, 2017

### WubbaLubba Dubdub

1. The problem statement, all variables and given/known data
Let $g(x) = \int_0^xf(t) dt$ where $f$ is such that $\frac{1}{2} \leq f(t) \leq 1$ for $t \in [0,1]$ and $\frac{1}{2} \geq f(t) \geq 0$ for $t \in (1,2]$. Then $g(2)$ belongs to interval
A. $[\frac{-3}{2}, \frac{1}{2}]$
B. $[0, 2)$
C. $(\frac{3}{2}, \frac{5}{2}]$
D. $(2, 4)$

2. Relevant equations

3. The attempt at a solution
I got $g'(x) = f(x)$ and using this and the definite integral given, i have $g(0) = 0$
I didn't really know where to go from here, so I tried making a graph (sort of) using the minimum and maximum slopes of the function in the given intervals and found an area in which, I think the function will exist, with the interval for $g(2)$ being $[\frac{1}{2},\frac{3}{2}]$.
This isn't present in the options...Can someone please point out my mistakes and help me get the answer.

Last edited: May 4, 2017
2. May 4, 2017

### BvU

I mostly agree with your reasoning; generally the lower bound of an interval is given first, followed by the upper bound: $[{1\over 2},{3\over 2}]$ is your result. If it isn't in the list litterally, you might try to exclude the answers that certainly don't satisfy...

3. May 4, 2017

### WubbaLubba Dubdub

Ah silly me. Will edit it. I found the answer key too and it says B is correct. Is $g(0) =0$ correct?

4. May 4, 2017

5. May 4, 2017

### WubbaLubba Dubdub

Now I get it! Thank you!