What is the integral from 1 to 2 of (f(t)+f^{-1}(t)) ?

The integral from 1 to 2 of (f(t)+f^{-1}(t)) is 3/(2f)+(3f)/2

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integral from 1 to 2 of (f(t)+f^{-1}(t))

\int_{1}^{2} (f (t) + f^{- 1} (t)) d t

\frac{3}{2 f} + \frac{3 f}{2}

Solution steps

\int_{1}^{2} f t + f^{- 1} t d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} f^{- 1} t d t + \int_{1}^{2} f t d t

\int_{1}^{2} f^{- 1} t d t = \frac{3}{2 f}

\int_{1}^{2} f t d t = \frac{3 f}{2}

= \frac{3}{2 f} + \frac{3 f}{2}

What is the integral from 1 to 2 of (f(t)+f^{-1}(t)) ?
The integral from 1 to 2 of (f(t)+f^{-1}(t)) is 3/(2f)+(3f)/2