If f(x) = x + e^x,then the area bounded by f^ | vedclass

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(B) Let $y = f(x) = x + e^x$. We want to find the area bounded by $f^{-1}(x)$ from $x=1$ to $x=1+e$ with the $x$-axis.This is equivalent to the area b

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$\frac{3}{2}$

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(B) Let $y = f(x) = x + e^x$. We want to find the area bounded by $f^{-1}(x)$ from $x=1$ to $x=1+e$ with the $x$-axis.This is equivalent to the area bounded by $f(y)$ from $y=1$ to $y=1+e$ with the $y$-axis,which is $\int_{1}^{1+e} f^{-1}(x) dx$.Using the property of inverse functions,the area is given by $\int_{f^{-1}(1)}^{f^{-1}(1+e)} x f'(x) dx$.For $f(x) = x + e^x$,$f(0) = 0 + e^0 = 1$ and $f(1) = 1 + e^1 = 1+e$.Thus,the area is $\int_{0}^{1} x (1 + e^x) dx$.$= \int_{0}^{1} x dx + \int_{0}^{1} x e^x dx$.$= \left[ \frac{x^2}{2} \right]_{0}^{1} + \left[ x e^x - e^x \right]_{0}^{1}$.$= (\frac{1}{2} - 0) + ((1 \cdot e^1 - e^1) - (0 \cdot e^0 - e^0))$.$= \frac{1}{2} + (0 - (-1)) = \frac{1}{2} + 1 = \frac{3}{2}$ $sq. units$.

If $f(x) = x + e^x$,then the area bounded by $f^{-1}(x)$,the ordinates $x = 1$ and $x = 1 + e$,and the $x$-axis is (in $sq. units$):

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