Let '$f$' be defined by $f(t)$ { $0$ when $t<0$ ; $t$ when $0\le t\le1$; $4$ when $t>1$. Determine the function $g(x)$ = $\int_0^x f(t) \ dt$ 1) I used the fundamental theorem of calculus to solve this question but I don't know how to apply that theorem in this question to find the function $g(x)$
2) Also, the question said to find where $f$ is non-differentiable and to justify why?
Please help me out.
Thanks
 A: Call $g(x) = \int_0^xf(t)dt$. We find a piecewise expression for $g$ in terms of $x$ as follows...
$\mathbf{Case 1}(x< 0)$: In this case we have 
$g(x) = \int_0^xf(t)dt = - \int_x^0 f(t) dt = - \int_x^0 0 dt = 0$.
$\mathbf{Case 2}( 0 \leqslant x \leqslant 1)$: In this case we have
$g(x) = \int_0^x f(t) dt = \int_0^x t \,dt = \left[ \frac{t^2}{2} \right]_{t=0}^x = \frac{(x)^2}{2} - \frac{(0)^2}{2} = \frac{x^2}{2}$.
$\mathbf{Case 3}(x > 1)$: In this case we have 
$g(x) = \int_0^x f(t) dt = \int_0^1 f(t)dt + \int_1^x f(t) dt $
$= \int_0^1 t \, dt + \int_1^x 4 dt = \left[ \frac{t^2}{2} \right]_{t=0}^1 +\left[ 4t \right]_{t=1}^x$
$= \left( \frac{(1)^2}{2} - \frac{(0)^2}{2} \right) + \left(4(x) - 4(1)\right) = 4x - \frac{7}{8}$. 
(Note that $f(t)=4$ for all $t \in (1,x]$ which is enough to allow us to write $\int_1^x f(t)dt = \int_1^x 4\,dt$ despite the fact that $f(1) \neq 4$).
$\mathbf{Conclusion}$: The piecewise expression for $g$ is
$g(x) = \begin{cases} 0, & x<0 \\
\frac{x^2}{2}, & 0 \leqslant x \leqslant 1
\\ 4x - \frac{7}{8}, & x >1 \end{cases}$
As for differentiability, a reformulated version of the fundamental theorem says that $g$ will be differentiable precisely where $f$ is continuous, which is everywhere except at the point $x = 0$. You can also look at the piecewise expression we just found for $g$ to confirm that $g$ is continuous everywhere except at $x = 1$.
