Apply the Newton-Leibniz formula to calculate

Apply the Newton-Leibniz formula to calculate the following integral: $\int_1^2 \frac{1}{x^x} \, dx$.

We need to evaluate $\int_1^2 \frac{1}{x^x} \, dx = \int_1^2 x^{-x} \, dx$.

Step 1: Analyze the integrand

The function $f(x) = x^{-x} = \frac{1}{x^x}$ does not have an elementary antiderivative. This means we cannot express the antiderivative in terms of standard functions (polynomials, exponentials, logarithms, trigonometric functions, etc.).

Step 2: Note on the Newton-Leibniz formula

While the Newton-Leibniz formula states that:

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

where $F'(x) = f(x)$, this formula can only be applied directly when we can find an explicit form for $F(x)$.

For $f(x) = x^{-x}$, no such elementary function $F(x)$ exists.

Step 3: Alternative approaches

Since we cannot use the Newton-Leibniz formula directly with elementary functions, we must resort to:

1. Numerical integration: Using numerical methods such as Simpson's rule, the trapezoidal rule, or Gaussian quadrature.
2. Series expansion: We can write $x^{-x} = e^{-x \ln x}$ and expand as a Taylor series: \[ e^{-x \ln x} = \sum_{n=0}^{\infty} \frac{(-x \ln x)^n}{n!} \] Then integrate term by term (though convergence must be verified).
3. Special function representation: The integral can be expressed using the Sophomore's dream integral or related special functions, but these are not elementary.

Step 4: Numerical approximation

Using numerical integration methods, we can approximate:

\[\int_1^2 x^{-x} \, dx \approx 0.6388\]

More precisely:

\[\int_1^2 x^{-x} \, dx \approx 0.638817...\]

Properties of the integrand:

• At $x = 1$: $1^{-1} = 1$
• At $x = 2$: $2^{-2} = \frac{1}{4} = 0.25$
• The function $x^{-x}$ is continuous and decreasing on $[1, 2]$
• We can write $x^{-x} = e^{-x \ln x}$, which shows the function decays exponentially

Geometric interpretation:

The integral $\int_1^2 x^{-x} \, dx$ represents the area under the curve $y = x^{-x}$ from $x = 1$ to $x = 2$. This curve starts at height $y=1$ when $x=1$, decreases to $y=0.25$ when $x=2$, is concave (bending downward), and has the interesting property that since $x^x$ has a minimum at $x = \frac{1}{e} \approx 0.368$, the function $x^{-x}$ has a maximum there.

Final answer:

\[\boxed{\int_1^2 x^{-x} \, dx = \int_1^2 \frac{1}{x^x} \, dx \text{ has no elementary closed form}}\]

The integral must be evaluated numerically:

\[\boxed{\int_1^2 \frac{1}{x^x} \, dx \approx 0.6388}\]

Note: This integral is an example of a definite integral that exists and has a finite value, but cannot be expressed in terms of elementary functions. The Newton-Leibniz formula is still valid in principle, but the antiderivative $F(x)$ cannot be written using standard mathematical functions.