Finding the Range of an Integrated Function with Given Constraints

[WubbaLubba Dubdub]

Homework Statement

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. $(2, 4)$

Homework Equations

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.

 

[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...

 

[WubbaLubba Dubdub]

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...

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

 

[BvU]

Ah silly me. Will edit it.
better leave as is or the thread becomes unintellegible...

I found the answer key too and it says B is correct.
yes. If it belongs to [1/2, 3/2] it certainly belongs to [0,2). The others all miss something

Is $g(0) =0$ correct?
yes.

 

[WubbaLubba Dubdub]

Now I get it! Thank you!