Given that $f(x)= \tan x - x^{1000}, x\ge0$ has positive roots $x_1,x_2,x_3...$ such that $x_i<x_j$ is $i<j ~\forall~i,j \in \mathbb > N$, then area of region enclosed by $f(x)$ with x axis between lines $x= 0$ and $x=\pi/2$ is:
A) $\ln|\sec x_1|-\dfrac{{x_1}^{1001}}{1001}$
B) $\dfrac{{x_1}^{1001}}{1001}$ - $\ln |\sec x_1|$
C) $\dfrac{{x_2}^{1001}}{1001}- \dfrac{2{x_1}^{1001}}{1001}+ 2\ln{\left|\dfrac{\sec x_1}{\sqrt {\sec x_2}}\right|}$
D) $\dfrac{{x_2}^{1001}}{1001}- \dfrac{{x_1}^{1001}}{1001}+ \ln {\left|\dfrac{\sec x_1}{\sqrt{\sec x_2}}\right|}$
Attempt:
$\text{Required area = } \displaystyle\int_0^{\pi/2} \tan x - x^{1000} = \left(\ln |\sec x| + \dfrac{x^{1001}}{1001} \right)\biggr|_0^{\pi/2}$
$x=0$ is the first root but we can't call it $x_1$ because $x_i$ denotes postive roots.
I am unable to understand the language of the question and what exactly I am supposed to do.
Could someone please explain that? Please do not provide hint-only answers because they won't help in developing conceptual clarity.
This is a bit difficult to answer based on pure calculus without taking aid of graphs.
If you consider the graph of $$f(x)=\tan x-x^4$$ here and the graph of $$f(x)=\tan x-x^5$$ here,
there is an additional root which occurs in $(0,{\pi}/{2}) $ for all powers of $x$ greater than $4$.
So, there are exactly 2 roots for $$f(x)=\tan x -x^{1000} \space \space \forall x\varepsilon(0,\pi /2)$$
$$\implies \text{Required area = } \displaystyle\int_0^{x_1} (\tan x - x^{1000})dx -\displaystyle\int_{x_1}^{x_2} (\tan x - x^{1000} )dx +\displaystyle\int_{x_2}^{\pi /2} (\tan x - x^{1000})dx$$
But since $x_2\rightarrow \pi/2$, you can ignore the last term.