If g(x) is the inverse of f(x) and f(x) has d | vedclass

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(A) Let $I = \int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy$.Since $g(y)$ is the inverse of $f(x)$,we have $y = f(x)$,which implies $x = g(y)$.When $x =

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) Let $I = \int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy$.Since $g(y)$ is the inverse of $f(x)$,we have $y = f(x)$,which implies $x = g(y)$.When $x = 1$,$y = f(1) = 2$. When $x = 5$,$y = f(5) = 10$.Using the substitution $y = f(x)$,we have $dy = f'(x) dx$.Alternatively,using the property of the integral of inverse functions: $\int_{a}^{b} f(x) dx + \int_{f(a)}^{f(b)} g(y) dy = b f(b) - a f(a)$.Here,$a = 1$ and $b = 5$.So,$\int_{1}^{5} f(x) dx + \int_{f(1)}^{f(5)} g(y) dy = 5 f(5) - 1 f(1)$.Substituting the given values $f(1) = 2$ and $f(5) = 10$:$= 5(10) - 1(2) = 50 - 2 = 48$.Thus,the value is $48$.

If $g(x)$ is the inverse of $f(x)$ and $f(x)$ has domain $x \in [1, 5]$,where $f(1) = 2$ and $f(5) = 10$,then the value of $\int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy$ is equal to:

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