Discuss the geometric interpretation of integral from 0

Discuss the geometric interpretation of $\int_0^1 x^n \, dx = \frac{1}{n+1}$, then apply the Newton-Leibniz formula to calculate $\int_0^\pi \sin^2 x \, dx$.

Geometric interpretation of $\int_0^1 x^n \, dx = \frac{1}{n+1}$

The integral $\int_0^1 x^n \, dx$ represents the area under the curve $y = x^n$ from $x = 0$ to $x = 1$ in the unit square $[0,1] \times [0,1]$.

Key observations:

• For $n = 0$: $y = x^0 = 1$ is a horizontal line at height 1. The area is the entire unit square: $\frac{1}{0+1} = 1$.
• For $n = 1$: $y = x$ is a straight line from $(0,0)$ to $(1,1)$. The area is that of a right triangle with base 1 and height 1: $\frac{1}{1+1} = \frac{1}{2}$.
• For $n = 2$: $y = x^2$ is a parabola. The curve is concave up and lies below the line $y = x$. The area is $\frac{1}{2+1} = \frac{1}{3}$, which is less than $\frac{1}{2}$.
• As $n$ increases: The curve $y = x^n$ becomes more curved and stays closer to the $x$-axis for $0 < x < 1$, approaching the $x$-axis everywhere except near $x = 1$. Thus the area decreases as $\frac{1}{n+1} \to 0$ when $n \to \infty$.
• The ratio: The area under $y = x^n$ is always $\frac{1}{n+1}$ of the unit square's area, showing how the power function "fills" less of the square as $n$ increases.

Calculate $\int_0^\pi \sin^2 x \, dx$ using Newton-Leibniz formula

Step 1: Use the half-angle identity

We use the trigonometric identity:

\[\sin^2 x = \frac{1 - \cos(2x)}{2}\]

Therefore:

\[\int_0^\pi \sin^2 x \, dx = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx\]

\[= \frac{1}{2} \int_0^\pi (1 - \cos(2x)) \, dx\]

\[= \frac{1}{2} \left[\int_0^\pi 1 \, dx - \int_0^\pi \cos(2x) \, dx\right]\]

Step 2: Find the antiderivatives

For the first integral:

\[\int 1 \, dx = x + C\]

For the second integral, let $u = 2x$, so $du = 2dx$:

\[\int \cos(2x) \, dx = \frac{1}{2}\sin(2x) + C\]

Step 3: Apply the Newton-Leibniz formula

\[\int_0^\pi \sin^2 x \, dx = \frac{1}{2} \left[\left[x\right]_0^\pi - \left[\frac{\sin(2x)}{2}\right]_0^\pi\right]\]

\[= \frac{1}{2} \left[(\pi - 0) - \left(\frac{\sin(2\pi)}{2} - \frac{\sin(0)}{2}\right)\right]\]

\[= \frac{1}{2} \left[\pi - \left(\frac{0}{2} - \frac{0}{2}\right)\right]\]

\[= \frac{1}{2} \left[\pi - 0\right]\]

\[= \frac{\pi}{2}\]

Geometric interpretation:

The integral $\int_0^\pi \sin^2 x \, dx = \frac{\pi}{2}$ represents the area under the curve $y = \sin^2 x$ from $x = 0$ to $x = \pi$.

Key observations:

• The function $\sin^2 x$ is always non-negative and ranges from 0 to 1.
• At $x = 0$ and $x = \pi$, we have $\sin^2 x = 0$.
• At $x = \frac{\pi}{2}$, we have $\sin^2 x = 1$ (maximum).
• The curve is symmetric about $x = \frac{\pi}{2}$.
• Interestingly, $\int_0^\pi \sin^2 x \, dx = \int_0^\pi \cos^2 x \, dx = \frac{\pi}{2}$, and their sum equals $\int_0^\pi (\sin^2 x + \cos^2 x) \, dx = \int_0^\pi 1 \, dx = \pi$.

Final answer:

\[\boxed{\int_0^\pi \sin^2 x \, dx = \frac{\pi}{2}}\]