# Area that is bounded by functions

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 $$u\rightarrow \infty$$.



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?

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.
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 $$\lim_{u\to\infty}A(u)=e^3-e$$