Apply the Newton-Leibniz formula to calculate

Apply the Newton-Leibniz formula to calculate the following integral: $\int_0^1 \frac{dx}{1+x^2}$.

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: Identify the antiderivative

We recognize that:

\[\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}\]

Therefore, the antiderivative is:

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

Step 2: Apply the Newton-Leibniz formula

\[\int_0^1 \frac{dx}{1+x^2} = \left[\arctan x\right]_0^1\]

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

\[= \frac{\pi}{4} - 0\]

\[= \frac{\pi}{4}\]

Geometric interpretation:

The integral $\int_0^1 \frac{dx}{1+x^2} = \frac{\pi}{4}$ has several interesting geometric interpretations:

• Area under the curve: The integral represents the area under the curve $y = \frac{1}{1+x^2}$ from $x = 0$ to $x = 1$. This curve is part of the Witch of Agnesi (when scaled appropriately) and decreases from $y = 1$ at $x = 0$ to $y = \frac{1}{2}$ at $x = 1$.
• Angular interpretation: Since $\arctan x$ gives the angle whose tangent is $x$, the integral represents the change in angle as we move from $x = 0$ to $x = 1$ along the curve. Specifically, at $x=0$: $\arctan(0)=0$ radians; at $x=1$: $\arctan(1)=\frac{\pi}{4}$ radians.
• Relation to $\pi$: This integral provides a way to compute $\pi$: \[ \pi = 4 \int_0^1 \frac{dx}{1+x^2} \] This is the basis for various series expansions for $\pi$.
• Symmetry property: The function $\frac{1}{1+x^2}$ is even, and over the interval $[-1, 1]$: \[ \int_{-1}^1 \frac{dx}{1+x^2} = 2\int_0^1 \frac{dx}{1+x^2} = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

Alternative verification using series:

The integral can also be related to the Leibniz formula for $\pi$:

\[\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots\]

This comes from integrating the geometric series expansion of $\frac{1}{1+x^2}$.

Final answer:

\[\boxed{\int_0^1 \frac{dx}{1+x^2} = \frac{\pi}{4}}\]

This result shows that the area under $y = \frac{1}{1+x^2}$ from 0 to 1 equals exactly one-fourth of $\pi$.