Apply the Newton-Leibniz formula to calculate

Apply the Newton-Leibniz formula to calculate the following integral: $\int_0^1 \sinh x \, dx$.

We will use the Newton-Leibniz formula:

\[\int_a^b f(x) \, dx = F(b) - F(a)\]

where $F(x)$ is an antiderivative of $f(x)$.

Step 1: Recall the definition of hyperbolic sine

The hyperbolic sine function is defined as:

\[\sinh x = \frac{e^x - e^{-x}}{2}\]

Step 2: Find the antiderivative

We need to find $F(x)$ such that $F'(x) = \sinh x$.

We know that:

\[\frac{d}{dx}[\cosh x] = \sinh x\]

where $\cosh x = \frac{e^x + e^{-x}}{2}$ is the hyperbolic cosine function.

We can verify this:

\[\frac{d}{dx}\left[\frac{e^x + e^{-x}}{2}\right] = \frac{e^x - e^{-x}}{2} = \sinh x\]

Therefore, the antiderivative is:

\[F(x) = \cosh x + C\]

Step 3: Apply the Newton-Leibniz formula

\[\int_0^1 \sinh x \, dx = \left[\cosh x\right]_0^1\]

\[= \cosh(1) - \cosh(0)\]

Step 4: Evaluate the hyperbolic cosine values

At $x = 1$:

\[\cosh(1) = \frac{e^1 + e^{-1}}{2} = \frac{e + \frac{1}{e}}{2} = \frac{e^2 + 1}{2e}\]

At $x = 0$:

\[\cosh(0) = \frac{e^0 + e^{0}}{2} = \frac{1 + 1}{2} = 1\]

Step 5: Calculate the result

\[\int_0^1 \sinh x \, dx = \cosh(1) - 1 = \frac{e + e^{-1}}{2} - 1\]

\[= \frac{e + \frac{1}{e}}{2} - 1 = \frac{e^2 + 1}{2e} - 1\]

\[= \frac{e^2 + 1 - 2e}{2e} = \frac{e^2 - 2e + 1}{2e}\]

\[= \frac{(e - 1)^2}{2e}\]

Numerical value:

\[\cosh(1) \approx 1.5431, \quad \text{so} \quad \int_0^1 \sinh x \, dx \approx 0.5431\]

Geometric interpretation:

The integral $\int_0^1 \sinh x \, dx$ represents the area under the curve $y = \sinh x$ from $x = 0$ to $x = 1$.

Key properties of $y = \sinh x$:

• At $x = 0$: $\sinh(0) = 0$ (the curve passes through the origin)
• At $x = 1$: $\sinh(1) = \frac{e - e^{-1}}{2} \approx 1.175$
• The function $\sinh x$ is an odd function: $\sinh(-x) = -\sinh(x)$
• The function is always increasing (since $\sinh'(x) = \cosh(x) > 0$ for all $x$)
• The function is concave down for $x < 0$ and concave up for $x > 0$
• Unlike trigonometric sine, hyperbolic sine grows exponentially for large $|x|$

Final answer:

\[\boxed{\int_0^1 \sinh x \, dx = \cosh(1) - 1 = \frac{e + e^{-1}}{2} - 1 = \frac{(e-1)^2}{2e}}\]

Or in decimal form:

\[\boxed{\int_0^1 \sinh x \, dx \approx 0.5431}\]