Limits & Integrals

Calculus Level 2

f ( x ) = lim n x 2 1 + ( tan 1 x ) n \large f(x) = \lim _{n\to \infty } \frac {x^2}{1+(\tan ^{-1} x)^n}

For f ( x ) f(x) as defined above, find 0 f ( x ) d x \displaystyle \int ^{\infty }_{0} f(x)\ dx .


The answer is 1.259.

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1 solution

Blan Morrison
Dec 6, 2018

Keep in mind lim p b p = { 0 b < 1 1 b = 1 b > 1 \displaystyle\lim_{p\rightarrow \infty}b^p = \begin{cases} 0 & b<1 \\ 1 & b=1 \\ \infty & b>1 \end{cases} Therefore, f ( x ) = { x 2 1 + 0 arctan x < 1 x 2 1 + 1 arctan x = 1 x 2 1 + arctan x > 1 f(x) = \begin{cases} \frac{x^2}{1+0} & \arctan{x}<1 \\ \frac{x^2}{1+1} & \arctan{x}=1 \\ \frac{x^2}{1+\infty} & \arctan{x}>1 \end{cases} Or, in simpler terms: f ( x ) = { x 2 x < tan ( 1 ) x 2 2 x = tan ( 1 ) 0 x > tan ( 1 ) f(x) = \begin{cases} x^2 & x<\tan(1) \\ \frac{x^2}{2} & x=\tan(1) \\ 0 & x>\tan(1) \end{cases}

Because it is piecewise, we can integrate f f in 3 separate parts:

0 tan ( 1 ) x 2 d x + tan ( 1 ) tan ( 1 ) x 2 2 d x + tan ( 1 ) 0 d x \displaystyle\int_{0}^{\tan(1)}x^2 dx+\displaystyle\int_{\tan(1)}^{\tan(1)}\frac{x^2}{2}dx+\displaystyle\int_{\tan(1)}^{\infty} 0~dx

= tan 3 ( 1 ) 3 0 + 0 + 0 1.259 =\frac{\tan^3(1)}{3}-0+0+0\approx 1.259 β \beta_{\lceil \mid \rceil}

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