Discuss the geometric interpretation of integrating odd

Discuss the geometric interpretation of integrating odd functions over symmetric intervals, then apply the Newton-Leibniz formula to calculate $\int_0^1 x^n \, dx$.

Geometric interpretation of Problem 7

The results from Problem 7 can be interpreted geometrically using the concept of symmetry.

For odd functions: When a function $f(x)$ is odd (i.e., $f(-x) = -f(x)$), its graph is symmetric with respect to the origin. This means that if we reflect the graph across both the $x$-axis and $y$-axis, we get the same graph.

When we integrate an odd function over a symmetric interval $[-a, a]$:

• The area under the curve from $-a$ to $0$ (below the $x$-axis when $f(x) < 0$) has the same magnitude as the area from $0$ to $a$ (above the $x$-axis when $f(x) > 0$).
• These two areas have opposite signs (one is positive, one is negative).
• Therefore, they cancel each other out, giving a net integral of zero.

Specific interpretations:

1. $\int_{-1}^1 \arcsin x \, dx = 0$: The graph of $y = \arcsin x$ is symmetric about the origin. The area under the curve from $-1$ to $0$ (which is negative) exactly cancels the area from $0$ to $1$ (which is positive).
2. $\int_{-a}^a \sinh x \, dx = 0$: The hyperbolic sine function $y = \sinh x$ has origin symmetry. The signed area on the left side of the $y$-axis cancels the signed area on the right side.
3. $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan x \, dx = 0$: The tangent function $y = \tan x$ is odd and passes through the origin. The negative area from $-\frac{\pi}{4}$ to $0$ cancels the positive area from $0$ to $\frac{\pi}{4}$.

Calculate $\int_0^1 x^n \, dx$ using Newton-Leibniz formula

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: Find the antiderivative

For $n \neq -1$, we have:

\[F(x) = \int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]

Step 2: Apply the Newton-Leibniz formula

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

\[= \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}\]

\[= \frac{1}{n+1} - 0\]

\[= \frac{1}{n+1}\]

Special cases:

• For $n = 0$: $\int_0^1 1 \, dx = \frac{1}{0+1} = 1$ (area of unit square)
• For $n = 1$: $\int_0^1 x \, dx = \frac{1}{1+1} = \frac{1}{2}$ (area of right triangle)
• For $n = 2$: $\int_0^1 x^2 \, dx = \frac{1}{2+1} = \frac{1}{3}$
• For $n = 3$: $\int_0^1 x^3 \, dx = \frac{1}{3+1} = \frac{1}{4}$

Geometric interpretation:

The integral $\int_0^1 x^n \, dx$ represents the area under the curve $y = x^n$ from $x = 0$ to $x = 1$. As $n$ increases, the curve $y = x^n$ becomes more concave (bends more toward the $x$-axis), so the area decreases. The formula $\frac{1}{n+1}$ shows that as $n \to \infty$, the area approaches $0$.

Final answer:

\[\boxed{\int_0^1 x^n \, dx = \frac{1}{n+1} \quad \text{for } n \neq -1}\]