Question

Let $f\left(x\right)={\int }_0^x{e}^{t-\left[t\right]}dt(x>0)$, where $\left[x\right]$ denotes greatest integer less than or equal to $x$, is

• A. Continuous and differentiable $\forall xϵ(0,3]$
• B. Continuous but not differentiable $\forall xϵ(0,3]$
• C. $f\left(1\right)=e$
• D. $f\left(2\right)=2(e-1)$

Answer 1

The correct option is A Continuous and differentiable $\forall xϵ(0,3]$

Let $\left(x\right)$ denote the fractional part of $x$ such that $\left(x\right)+\left[x\right]=x$

Using the property of definite integrals and the fact that $0\le \left(t\right)<1$, we can write the given function as:

$f\left(x\right)={\int }_0^x{e}^{\left(t\right)}dt={\int }_0^1{e}^{t}dt+{\int }_1^2{e}^{t}dt+...+{\int }_{\left[x\right]-1}^{\left[x\right]}{e}^{t}dt+{\int }_{\left[x\right]}^{\left[x\right]+\left(x\right)}{e}^{t}dt$

$\Rightarrow f\left(x\right)=(e-1)+(e^2-e)+...+(e^{\left[x\right]}-e^{\left[x\right]-1})+(e^x-e^{\left[x\right]})=e^x-1$

Thus the function is continuous and differentiable.

Hence, option A is the correct answer.