Calculate the area bounded by the curve y = a/2 (e^(x/a)

Calculate the area bounded by the curve $y = \frac{a}{2}\left(e^{x/a} + e^{-x/a}\right)$ and the lines $x = 0$, $y = 0$, $x = b$.

We will integrate the exponential form of the catenary directly.

Step 1: Recognize the curve

The curve is:

\[y = \frac{a}{2}\left(e^{x/a} + e^{-x/a}\right) = a\cosh\frac{x}{a}\]

This is a catenary, the shape of a hanging chain. Note that $y(0) = \frac{a}{2}(1 + 1) = a > 0$, so the curve is above the $x$-axis.

Step 2: Set up the integral

The area bounded by the curve from $x = 0$ to $x = b$ and the $x$-axis is:

\[A = \int_0^b y \, dx = \int_0^b \frac{a}{2}\left(e^{x/a} + e^{-x/a}\right) dx\]

Step 3: Evaluate the integral

\[A = \frac{a}{2}\int_0^b \left(e^{x/a} + e^{-x/a}\right) dx\]

\[= \frac{a}{2}\left[\int_0^b e^{x/a} \, dx + \int_0^b e^{-x/a} \, dx\right]\]

First integral:

\[\int_0^b e^{x/a} \, dx = \left[a e^{x/a}\right]_0^b = a(e^{b/a} - e^0) = a(e^{b/a} - 1)\]

Second integral:

\[\int_0^b e^{-x/a} \, dx = \left[-a e^{-x/a}\right]_0^b = -a(e^{-b/a} - 1) = a(1 - e^{-b/a})\]

Step 4: Combine the results

\[A = \frac{a}{2}\left[a(e^{b/a} - 1) + a(1 - e^{-b/a})\right]\]

\[= \frac{a^2}{2}\left[e^{b/a} - 1 + 1 - e^{-b/a}\right]\]

\[= \frac{a^2}{2}\left(e^{b/a} - e^{-b/a}\right)\]

Using the definition of hyperbolic sine: $\sinh t = \frac{e^t - e^{-t}}{2}$

\[A = a^2 \cdot \frac{e^{b/a} - e^{-b/a}}{2} = a^2 \sinh\frac{b}{a}\]

\[\boxed{A = a^2 \sinh\frac{b}{a} = \frac{a^2}{2}\left(e^{b/a} - e^{-b/a}\right) \text{ square units}}\]

Step 5: Verification for special cases

Case 1: $b = 0$

\[A = a^2 \sinh 0 = 0\]

This makes sense since there's no region when $b = 0$. ✓

Case 2: $b = a$

\[A = a^2 \sinh 1 = a^2 \cdot \frac{e - e^{-1}}{2} = \frac{a^2(e - 1/e)}{2} \approx 1.175 a^2\]

Case 3: $a = 1, b = 1$

\[A = \sinh 1 = \frac{e - 1/e}{2} \approx \frac{2.718 - 0.368}{2} \approx 1.175\]

Step 6: Geometric interpretation

The curve $y = a\cosh\frac{x}{a}$ is the catenary with:

• Minimum at $(0, a)$
• The shape of a hanging chain under uniform gravity
• Parameter $a$ controls the "width" or "flatness" of the curve

The area under this curve from $x = 0$ to $x = b$ grows as $\sinh\frac{b}{a}$, which means:

• For small $b/a$: approximately linear growth $A \approx ab$
• For large $b/a$: exponential growth dominated by $e^{b/a}$

Summary

The area bounded by the catenary $y = \frac{a}{2}\left(e^{x/a} + e^{-x/a}\right)$, the lines $x = 0$, $x = b$, and the $x$-axis is:

\[\boxed{A = a^2 \sinh\frac{b}{a} = \frac{a^2}{2}\left(e^{b/a} - e^{-b/a}\right) \text{ square units}}\]

Key steps:

1. Recognize catenary: $y = a\cosh\frac{x}{a}$
2. Set up integral: $A = \int_0^b \frac{a}{2}\left(e^{x/a} + e^{-x/a}\right) dx$
3. Integrate exponentials: $\frac{a^2}{2}\left[(e^{b/a} - 1) + (1 - e^{-b/a})\right]$
4. Simplify: $\frac{a^2}{2}(e^{b/a} - e^{-b/a}) = a^2\sinh\frac{b}{a}$

Key insight:

The catenary curve naturally integrates to give a hyperbolic sine function, reflecting the deep connection between hyperbolic functions and exponentials. The parameter $a$ scales both the height and the area quadratically.