Question

Let ${A_n}$ be the area bounded by the curve $y = {(\tan x)^n}$ & the lines $x = 0$ , $y = 0$ & $x = \dfrac{\pi }{4}$ . Prove that for $n > 2$ , ${A_n} + {A_{n - 2}} = \dfrac{1}{{(n - 1)}}$ & deduce that $\dfrac{1}{{(2n + 2)}} < {A_n} < \dfrac{1}{{(2n - 1)}}$ .

Answer 1

Hint: The concept that is going to be used in this problem is integration and mathematical induction. We will use integration to find the area bounded by the given curve and lines, and mathematical induction to prove the given statement and deduce the result.To solve the problem, we will first find the area ${A_n}$ by integrating the given function $y = {(\tan x)^n}$ with respect to x between the limits of $x = 0$ and $x = \frac{\pi}{4}$. Next, we will use mathematical induction to prove the statement ${A_n} + {A_{n - 2}} = \frac{1}{{(n - 1)}}$ for $n > 2$. And we will use this statement to deduce the result $\frac{1}{{(2n + 2)}} < {A_n} < \frac{1}{{(2n - 1)}}$

Complete step by step solution:
Firstly, we have to solve ${A_n}$ and ${A_{n - 2}}$ then do the further solution,
 ${A_n} = \int_0^{\dfrac{\pi }{4}} {{{(\tan x)}^n}dx}$.
Similarly, as for ${A_{n - 2}}$ we get that,
${A_{n - 2}} = \int_0^{\dfrac{\pi }{4}} {{{(\tan x)}^{n - 2}}dx} $
By adding both ${A_n}$ and ${A_{n - 2}}$ we get,
${A_n} + {A_{n - 2}} = \int_0^{\dfrac{\pi }{4}} {{{(\tan x)}^{n - 2}}dx} + \int_0^{\dfrac{\pi }{4}} {{{(\tan x)}^{n - 2}}dx} $
Now , putting $t = \tan x$
And, $dt = {\sec ^2}xdx$
$\dfrac{{dt}}{{(1 + {t^2})}} = dx$
As for ${A_2} + {A_{n - 2}}$ we get,
\[{A_2} + {A_{n - 2}} = \int_0^1 {\dfrac{{{t^n}}}{{(1 + {t^2})}}dt + \int_0^1 {\dfrac{{{t^{n - 2}}}}{{(1 + {t^2})}}dt} } \]
As per taking dt common,
${A_2} + {A_{n - 2}} = \int_0^1 {[\dfrac{{{t^n}}}{{(1 + {t^2})}} + \dfrac{{{t^{n}}}}{{(1 + {t^2}){t^2}}}]dt} $
By doing further solution we get,
${A_2} + {A_{n - 2}} = \int_0^1 {\dfrac{{({t^2} + 1){t^n}}}{{(1 + {t^2}){t^2}}}dt} $
By which we get,
\[ = \int_0^1 {{t^{n - 2}}dt} \]
After applying the limit we get,
\[{A_2} + {A_{n - 2}} = {\left[ {\dfrac{{{t^{n - 1}}}}{{(n - 1)}}} \right]_0}^1 = \dfrac{1}{{(n - 1)}}\]
Here, $u = \tan x$ where $0 < u < 1$
$u^{n - 2} < u^n$ and $u^n < u^{n + 2}$
$A_{n - 2} < A_n$ and $A_n < A_{n + 2}$
Adding $A_n$ on both sides of both equations:
$A_n + A_{n + 2} < 2A_n$ and $2A_n < A_n + A_{n + 2}$
Dividing both sides by 2:
$\frac{1}{n-1} < 2A_n < \frac{1}{n+1}$
And $A_n < \frac{1}{2n+2}$ and $A_n > \frac{1}{2n-2}$
As collapsing both equation we get the solution of the both equation,
$\dfrac{1}{{2n + 2}} < {A_n} < \dfrac{1}{{2n - 1}}$
Therefore, we got the answer as $\dfrac{1}{{2n + 2}} < {A_n} < \dfrac{1}{{2n - 1}}$ .
“Hence proved” the answer was matched to prove the solution given in the question.

Note: Integration is a powerful tool for solving mathematical problems and understanding physical phenomena. To use it effectively, it is important to follow the correct steps and use the appropriate formulas, and to pay close attention when entering the upper and lower limits of the interval in question..