Question

If f(x) can be written as $f\left(x\right)={\int }_1^{x-1}{e}^{y^2}dy$ then which of the following is true about f(x)?

• A. f(x) is monotonically increasing in the interval [0,1]
• B. f(x) is strictly increasing in the interval [0,1]
• C. f(x) is monotonically increasing for all real values of x.
• D. f(x) is strictly increasing for all real values of x.

Answer 1

The correct option is D f(x) is strictly increasing for all real values of x.

Given function f(x) is written in terms of integral. And if you have observed you must have realised that the given integrand in the integral is one of those non integrable function about which we read. We can not integrate these functions so we don’t know how that function will look like but without knowing we can differentiate. THANKS to Leibnitz Bro.
So, we’ll apply Leibnitz formula to find out the monotonicity of f(x).
$f\left(x\right)={\int }_1^{(x-1)}{e}^{y{}^2}dy{f}^{\prime }\left(x\right)=e^{(x-1{)}^2}.1-e.0\left[using\frac{d}{dx}{\int }_{\left(a\right(x\left)\right)}^{\left(b\right(x\left)\right)}f\right(t)dt=f(b\left(x\right)).b^{\prime }(x)-f(a\left(x\right)).a^{\prime }(x\left)\right]f^{\prime }\left(x\right)=e^{(x-1{)}^2}$
We have found f’(x). Now we have to check where the functions is monotonous. Remember how we do it? We’ll find out the intervals where f’(x) is positive or negative.
$Since,(x-1{)}^2\ge 0\forall xϵRSo,e^{(x-1{)}^2}>0\forall xϵR$
And therefore we can say that function is strictly increasing for all real values of x.