Find the area bounded by the arc of the curve y = e^-x

Find the area bounded by the arc of the curve $y = e^{-x}$, its asymptote, and the tangent to this curve drawn from the origin.

We will find the tangent point, then split the enclosed area into two pieces at $x=0$.

Step 1: Find the asymptote

As $x \to \infty$: $y = e^{-x} \to 0$. The asymptote is $y = 0$ (the $x$-axis).

Step 2: Find the tangent from the origin

For a point $(x_0, e^{-x_0})$ on the curve, the tangent line has slope $m=-e^{-x_0}$, giving:

\[y - e^{-x_0} = -e^{-x_0}(x - x_0) \implies y = e^{-x_0}(1 + x_0 - x)\]

For this tangent to pass through the origin $(0,0)$:

\[0 = e^{-x_0}(1 + x_0) \implies 1 + x_0 = 0 \implies x_0 = -1\]

Point of tangency: $(-1, e)$. Tangent line: $y = e(1 + (-1) - x) = -ex$.

Step 3: Verify the tangent

At $x=0$: $y=0$ ✓. At $x=-1$: $y=e$ ✓. Slope of tangent: $-e$; slope of the curve at $x=-1$ is $-e^{-(-1)}=-e$ ✓.

Step 4: Determine the bounded region

The region splits naturally into two parts: from $x=-1$ to $x=0$, between the tangent line (below) and the curve (above); and from $x=0$ to $x=\infty$, between the asymptote (below) and the curve (above).

Step 5: Calculate Region 1

\[A_1 = \int_{-1}^{0} [e^{-x} - (-ex)] \, dx = \int_{-1}^{0} (e^{-x} + ex) \, dx = \left[-e^{-x} + \frac{ex^2}{2}\right]_{-1}^{0}\]

At $x=0$: $-1+0=-1$. At $x=-1$: $-e+\frac{e}{2}=-\frac{e}{2}$.

\[A_1 = -1 - \left(-\frac{e}{2}\right) = \frac{e-2}{2}\]

Step 6: Calculate Region 2

\[A_2 = \int_{0}^{\infty} e^{-x} \, dx = \left[-e^{-x}\right]_{0}^{\infty} = 0-(-1) = 1\]

Step 7: Total area

\[A = A_1 + A_2 = \frac{e-2}{2} + 1 = \frac{e}{2}\]

\[\boxed{A = \frac{e}{2} \text{ square units}}\]

Step 8: Numerical value

\[A = \frac{e}{2} \approx 1.359 \text{ square units}\]

Summary

\[\boxed{A = \frac{e}{2} \approx 1.359 \text{ square units}}\]

Key insight: The tangent from the origin to $y=e^{-x}$ touches at $(-1,e)$, and the total area between the curve, its horizontal asymptote, and this tangent is exactly $\frac{e}{2}$.