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The functions $$f_k(x)=\frac{x+k}{e^x}$$ are given.

Let $A(u)$ be the area that is bounded by $f_1, f_3$, the $x$-axis und the line $x=u$.

I want to check the area if $u\rightarrow \infty$.

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To calculate the area $A(u)$ do we calculate the area that is bounded by $f_1$ with endpoints the intersection point of that function with the $x$-axis and $x=u$ and the the area that is bounded by $f_2$ with endpoints the intersection point of that function with the $x$-axis and $x=u$ and then we subtract these two areas?

But in that way we haven't taken into consideration that the area has to be bounded by the $x$-axis, do we?

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Yes, if you look at the region where $-3 \le x \le -1$, we can see that the $x$-axis is a boundary of interest to us as well. Your computation have included that.

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we start with finding the zeros of $f_k$. $$\frac{x+k}{e^x}=0$$ $$\frac{x}{e^x}=-\frac{k}{e^x}$$ $$x=-k$$ so we see that $$A(u)=\int_{-3}^{u}f_3(x)dx-\int_{-1}^{u}f_1(x)dx$$ $$A(u)=\int_{-3}^{-1}f_3(x)dx+\int_{-1}^{u}f_3(x)-f_1(x)dx$$ $$A(u)=\int_{-3}^{-1}xe^{-x}dx+3\int_{-3}^{-1}e^{-x}dx+2\int_{-1}^{u}e^{-x}dx$$ $$A(u)=-2e^3+3(e^3-e)+2(e-e^{-u})$$ $$A(u)=e^3-e-2e^{-u}$$ So $$\lim_{u\to\infty}A(u)=e^3-e$$

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