Using integration, find
I =
Z 3
0
sinh 
1 −
x
3

dx


To evaluate the integral

\[
I = \int_0^3 \sinh\left(1 - \frac{x}{3}\right) dx,
\]

we can proceed step by step.

### Step 1: Use the definition of hyperbolic sine
The function \(\sinh(u)\) is defined as:

\[
\sinh(u) = \frac{e^u - e^{-u}}{2}.
\]

Thus, \(\sinh\left(1 - \frac{x}{3}\right)\) becomes:

\[
\sinh\left(1 - \frac{x}{3}\right) = \frac{e^{1 - \frac{x}{3}} - e^{-\left(1 - \frac{x}{3}\right)}}{2}.
\]

So, the integral can be rewritten as:

\[
I = \frac{1}{2} \int_0^3 \left( e^{1 - \frac{x}{3}} - e^{-\left(1 - \frac{x}{3}\right)} \right) dx.
\]

### Step 2: Separate the integrals
We now separate the integral into two parts:

\[
I = \frac{1}{2} \left( \int_0^3 e^{1 - \frac{x}{3}} dx - \int_0^3 e^{-\left(1 - \frac{x}{3}\right)} dx \right).
\]

### Step 3: Solve each integral
Now we will solve each integral separately.

#### First integral:
\[
\int_0^3 e^{1 - \frac{x}{3}} dx.
\]
Let \(u = 1 - \frac{x}{3}\), hence \(du = -\frac{1}{3} dx\), and \(dx = -3 du\).

When \(x = 0\), \(u = 1\); and when \(x = 3\), \(u = 0\). Thus, the integral becomes:

\[
\int_1^0 e^u (-3) du = 3 \int_0^1 e^u du = 3(e^1 - e^0) = 3(e - 1).
\]

#### Second integral:
\[
\int_0^3 e^{-\left(1 - \frac{x}{3}\right)} dx.
\]
Let \(v = 1 - \frac{x}{3}\), hence \(dv = -\frac{1}{3} dx\), and \(dx = -3 dv\).

When \(x = 0\), \(v = 1\); and when \(x = 3\), \(v = 0\). Thus, the integral becomes:

\[
\int_1^0 e^{-v} (-3) dv = 3 \int_0^1 e^{-v} dv = 3(-e^{-1} + e^0) = 3(1 - \frac{1}{e}).
\]

### Step 4: Combine results
Now, putting everything together:

\[
I = \frac{1}{2} \left[ 3(e - 1) - 3\left(1 - \frac{1}{e}\right) \right].
\]

Simplify:

\[
I = \frac{3}{2} \left[ (e - 1) - \left(1 - \frac{1}{e}\right) \right] = \frac{3}{2} \left( e - 1 - 1 + \frac{1}{e} \right) = \frac{3}{2} \left( e - 2 + \frac{1}{e} \right).
\]

Thus, the final answer is:

\[
I = \frac{3}{2} \left( e - 2 + \frac{1}{e} \right).
\]

---

Would you like any further details or explanations?

Here are five related questions you might find interesting:
1. How would you solve the integral if the limits were different?
2. Can you solve a similar integral involving \(\cosh(x)\)?
3. How does the result change if we integrate \(\sinh(kx)\) for a constant \(k\)?
4. What is the geometrical interpretation of integrating \(\sinh(x)\) over an interval?
5. Can the integral be solved using a substitution without separating terms?

**Tip:** Always check for substitutions in complex integrals to simplify the process.

Using integration, find I = ∫₀³ sinh(1 - x/3) dx.

Learn how to solve the definite integral I = ∫₀³ sinh(1 - x/3) dx using substitution and hyperbolic function properties. This step-by-step guide covers integration techniques and is suitable for students in advanced calculus courses.

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