Question-208235

Question Number 208235 by efronzo1 last updated on 08/Jun/24

Answered by som(math1967) last updated on 08/Jun/24

h e r e f (x) = f^{- 1} (x)

\underset{2}{\int^{4}} \frac{1 - x}{1 + x} d x

= \underset{2}{\int^{4}} \frac{2 d x}{1 + x} - \underset{2}{\int^{4}} \frac{1 + x}{1 + x} d x

= {[2 l n (1 + x) - x]}_{2}^{4}

= 2 l n 5 - 4 - 2 l n 3 + 2

= 2 l n \frac{5}{3} - 2

Answered by shunmisaki007 last updated on 08/Jun/24

Let y = f^{- 1} (x)

f (y) = x = \frac{1 - y}{1 + y}

x (1 + y) = 1 - y

x + x y = 1 - y

x y + y = 1 - x

y (1 + x) = 1 - x

\Rightarrow y = f^{- 1} (x) = \frac{1 - x}{1 + x}

\underset{2}{\int^{4}} f^{- 1} (x) d x = \underset{2}{\int^{4}} \frac{1 - x}{1 + x} d x = \underset{2}{\int^{4}} \frac{1 - x + 1 - 1}{1 + x} d x

= \underset{2}{\int^{4}} \frac{2 - x - 1}{1 + x} d x = \underset{2}{\int^{4}} \frac{2 - (x + 1)}{1 + x} d x

= \underset{2}{\int^{4}} \frac{2}{1 + x} d x - \underset{2}{\int^{4}} \frac{1 + x}{1 + x} d x

= {[2 \ln (1 + x)]}_{2}^{4} - {[x]}_{2}^{4}

= (2 \ln (1 + 4) - 2 \ln (1 + 2)) - (4 - 2)

= 2 (\ln (5) - \ln (3)) - 2

∴ \underset{2}{\int^{4}} f^{- 1} (x) d x = 2 \ln (\frac{5}{3}) - 2 \approx - 0.987 ★