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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 $f(x)$

2) Also, the question said to find where $f$ is non-differentiable and to justify why?

Please help me out. Thanks

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    In general, start by computing a few values of your function. Say, $f(0.1), f(0.5), f(0.9), f(1.1), f(100)$. Then compute $g(0.1), g(0.5), g(0.9), g(1.1), g(100)$ – 2017-02-16
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    What is $f$ equal to at the points $t=0,1$? – 2017-02-16
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    @avs can't we just prove that f(t) is continuous at [0,1] and then put value of f(t) = t and then solve the integration to get g(x) = x^2/2. Is this correct? Also how to find where f is non-differentiable? – 2017-02-16
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    @joeb u mean to f(t) ?? actually f(t) is equal to t when t is greater than and equal to 0 and less that and equal to 1. – 2017-02-16
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    Yes that's what I meant. You should update your post and state $f(t)=t$ for $0 \leqslant t \leqslant 1$. And you were right before by writing $g(x) = \int_0^x f(t) dt$ instead of $f(x) = \int_0^x f(t)dt$ - the latter is a circular definition and doesn't make sense. Moreover, $g$ has a chance to differentiable at $t = 0$, but it will depend on what $f(0)$ is equal to, so again, you need to alter your post a bit – 2017-02-16
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    @joeb thanx for correcting me..hope now the question is clear. But still i don't get how to find g(x) and how to justify the point where it is non-differentiable. – 2017-02-16
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    One fact that might help is that a function is continuous anywhere it is differentiable. (However, note that continuity at a point does not imply diferentiability there.) Will this help you narrow your search? – 2017-02-16

1 Answers 1

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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 $x = 1$.

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    @ joeb thanxx..by applying fundamental theorem to the integral g(x) you mean to solve the integral..?? I.e, I would get g(x) = x^2/2, is this correct? And after finding g(x) when t>1 what am I supposed to do? Do i have to add all the g(x) which I'm getting from different conditions? – 2017-02-16
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    When $x > 1$, we can write $g(x) = \int_0^1 t dt + \int_1^x 4 dt = \left[ \frac{t^2}{2} \right]_{t=0}^1 + \left[ 4t \right]_{t=1}^x =\left( \frac{1^2}{2} - \frac{0^2}{2}\right) + \left(4x - 4\right)$ – 2017-02-16
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    So yeah, I meant solve the integral (or integrals if there were two) – 2017-02-16
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    so would the answer be g(x) = x^2/2+1/2+(4x-4) ?? Is this correct form? – 2017-02-16
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    I updated my answer so that there shouldn't be any confusion – 2017-02-16