Definite integral of T z T_z from 1 1 to π \pi

Calculus Level 3

For T z = 2 z − 1 2 z + 1 T_z=\frac{2z-1}{2z+1} , ∫ 1 π T z d z = a \large{\int_1^\pi \mathrm{T_z}\,\mathrm{d}z=a} . What is ⌊ 100 a ⌋ \lfloor 100a \rfloor ? (Hint: the answer is a perfect cube.)


The answer is 125.

Zeeshan Ali by Zeeshan Ali , 25, Pakistan

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1 solution

Vincent Moroney
Jul 9, 2018

∫ 1 π 2 x − 1 2 x + 1 d x = ∫ 1 π 2 x + 1 − 2 2 x + 1 d x = ∫ 1 π ( 1 − 2 2 x + 1 ) d x = ( π − 1 ) + ln ⁡ ( 3 2 π + 1 ) ≈ 1.254 ⌊ 125.4 ⌋ = 125 \begin{aligned} \int_1^{\pi} \frac{2x-1}{2x+1} \,dx = & \int_1^{\pi} \frac{2x+1-2}{2x+1}\,dx \\ = & \int_1^{\pi} \Big(1- \frac{2}{2x+1} \Big)\,dx \\ = & (\pi-1) + \ln\Big(\frac{3}{2\pi +1} \Big) \approx 1.254 \\ & \left \lfloor{125.4}\right \rfloor = \boxed{125} \end{aligned}

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