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I was given the PDF $f_T(t)=Cte^{-t/2}I_{[0,\infty)}(t)$ "with some redundancy" as stated in the problem, what exactly does that imply? I need to find the CDF to solve the question. I know that I need to take the integral of the PDF to find the CDF. But the indicator function, specifically $[0,\infty)$, makes me unsure about the interval of integration. Would the PDF be integrated from $0$ to $t$? Or otherwise?

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    When is $f_T(t)\ne 0$ assuming $C\ne 0$? – 2017-01-24
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    This is not a PDF. – 2017-01-24
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    @Did Does that change the way I should approach the problem? In the question it says that this is a PDF, I'm not sure what to do anymore – 2017-01-24
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    Check the expression of $f_T$ in your book or whatever. – 2017-01-24
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    @Did I totally misread the exponent and thought the $t$ was $1$, it's supposed to be $e^{-t/2}$ – 2017-01-24

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The indicator function indicates that the expression is $C t e^{-1/2}$ whenever $t\in[0;\infty)$ and zero elsewhere.

$$\Bbb I_{[0;\infty)}(t) = \begin{cases}1&:& t\in[0;\infty) \\ 0 &:& \text{elsewhere}\end{cases}$$

As such the CDF should be $~F_T(t)=\int_0^t f_T(\tau)\operatorname d \tau ~\cdot~\mathbb I_{[0;\infty)}(t)$.

However, $\int_0^\infty C \tau e^{-1/2}\operatorname d \tau \neq 1$, so the function you have presented is not a probability density function, and so will not give a valid cumulative distribution function.

Check that what you have typed is what is in the book.   If it is, check that it is what the book meant.

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    So I did read the exponent wrong and e is supposed to be raised to the power of $-t/2$ – 2017-01-24
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    Yes, *that* is a pdf. You can find the CDF from that. – 2017-01-24
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I´m almost sure that $f_T(t)=C\cdot t\cdot e^{-1/2\cdot\color{red}t}I_{[0,\infty)}$.

The indicator variable shows where $f_T(t)=0$. The density can be also written as

$$f_T(t)=\begin{cases}{} C\cdot t\cdot e^{-1/2\cdot t}, \ \text{if} \ x\geq 0 \\ 0 , \ \text{if} \ x<0 \end{cases}$$

One property of a pdf is

$$\int_{-\infty}^{\infty} f(x) \, dx=1$$

In your case

$\int_{-\infty}^{0} f_T(t) \, dt + \int_{0}^{\infty} f_T(t) \, dt=1$

$\int_{-\infty}^{0} 0 \, dt + \int_{0}^{\infty} C\cdot t\cdot e^{-1/2\cdot t} \, dt=1$

$\int_{0}^{\infty} C\cdot t\cdot e^{-1/2\cdot t} \, dt=1$

To calculate the LHS you can use partial integration. Define $u(t)=t\Rightarrow u'(t)=1$ and $v'(t)=e^{-1/2t}\Rightarrow v(t)=-2e^{-1/2t}$

$\int u(t)\cdot v'(t)\, dt=u(t)\cdot v(t)-\int u'(t)\cdot v(t)\, dt$

$\int t\cdot e^{-1/2t} \, dt=-2t\cdot e^{-1/2t}-\int 1\cdot (-2e^{-1/2t})\, dt$

$=-2t\cdot e^{-1/2t}-4\cdot e^{-1/2t}$

Therefore $C\cdot [t\cdot e^{-1/2t}]_0^{\infty}-4\cdot C\cdot [e^{-1/2t}]_0^{\infty}=1$

$0-4\cdot C\cdot (0-1)=1$

$4C=1\Rightarrow C=\frac14$