Using the Newton-Leibniz formula, calculate integral

Using the Newton-Leibniz formula, calculate $\int_{-1}^1\left(1-x^2\right) \, dx$.

We will use the Newton-Leibniz formula (Fundamental Theorem of Calculus):

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

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

Step 1: Find the antiderivative

We need to find $F(x)$ such that $F'(x) = 1 - x^2$.

\[F(x) = \int (1 - x^2) \, dx = \int 1 \, dx - \int x^2 \, dx\]

\[= x - \frac{x^3}{3} + C\]

Step 2: Apply the Newton-Leibniz formula

Using the antiderivative $F(x) = x - \frac{x^3}{3}$:

\[\int_{-1}^1 (1 - x^2) \, dx = \left[x - \frac{x^3}{3}\right]_{-1}^1\]

\[= \left(1 - \frac{1^3}{3}\right) - \left((-1) - \frac{(-1)^3}{3}\right)\]

\[= \left(1 - \frac{1}{3}\right) - \left(-1 - \frac{-1}{3}\right)\]

\[= \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right)\]

\[= 1 - \frac{1}{3} + 1 - \frac{1}{3}\]

\[= 2 - \frac{2}{3} = \frac{6 - 2}{3} = \frac{4}{3}\]

Geometric interpretation:

The integral represents the area under the parabola $y = 1 - x^2$ from $x = -1$ to $x = 1$. This parabola opens downward with vertex at $(0, 1)$ and intersects the $x$-axis at $x = \pm 1$.

Final answer:

\[\boxed{\int_{-1}^1 (1 - x^2) \, dx = \frac{4}{3}}\]