Calculate the integral: integral from 0 to R sqrt(R²-x²)

Calculate the integral: $\int_0^R \sqrt{R^2-x^2} \, dx$.

We will use trigonometric substitution to evaluate this integral.

Step 1: Recognize the geometric connection

The expression $\sqrt{R^2-x^2}$ comes from the equation of a circle: $x^2 + y^2 = R^2$, which gives $y = \sqrt{R^2-x^2}$ for the upper semicircle of radius $R$ centered at the origin.

Step 2: Use trigonometric substitution

Let $x = R\sin\theta$. Then $dx = R\cos\theta \, d\theta$. Also:

\[\sqrt{R^2-x^2} = \sqrt{R^2(1-\sin^2\theta)} = R|\cos\theta|\]

For $\theta \in [0, \frac{\pi}{2}]$, we have $\cos\theta \geq 0$, so $|\cos\theta| = \cos\theta$.

Step 3: Change the limits of integration

When $x = 0$: $\theta = 0$. When $x = R$: $\theta = \frac{\pi}{2}$.

Step 4: Rewrite the integral

\[\int_0^R \sqrt{R^2-x^2} \, dx = \int_0^{\frac{\pi}{2}} R\cos\theta \cdot R\cos\theta \, d\theta = R^2 \int_0^{\frac{\pi}{2}} \cos^2\theta \, d\theta\]

Step 5: Use the half-angle identity

\[\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\]

Therefore:

\[R^2 \int_0^{\frac{\pi}{2}} \cos^2\theta \, d\theta = \frac{R^2}{2} \left[\int_0^{\frac{\pi}{2}} 1 \, d\theta + \int_0^{\frac{\pi}{2}} \cos(2\theta) \, d\theta\right]\]

Step 6: Integrate

\[\int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}, \qquad \int_0^{\frac{\pi}{2}} \cos(2\theta) \, d\theta = \left[\frac{\sin(2\theta)}{2}\right]_0^{\frac{\pi}{2}} = 0\]

Step 7: Combine the results

\[\int_0^R \sqrt{R^2-x^2} \, dx = \frac{R^2}{2}\left[\frac{\pi}{2} + 0\right] = \frac{\pi R^2}{4}\]

Geometric interpretation:

The integral $\int_0^R \sqrt{R^2-x^2} \, dx$ represents the area under the curve $y = \sqrt{R^2-x^2}$ from $x = 0$ to $x = R$, which is exactly the area of a quarter circle with radius $R$:

\[\text{Area of quarter circle} = \frac{1}{4} \times \pi R^2 = \frac{\pi R^2}{4}\]

This confirms our result!

Alternative verification using the general formula:

For the general integral $\int \sqrt{a^2-x^2} \, dx$, the antiderivative is:

\[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + C\]

Using this with $a = R$:

\[\int_0^R \sqrt{R^2-x^2} \, dx = \left[\frac{x}{2}\sqrt{R^2-x^2} + \frac{R^2}{2}\arcsin\left(\frac{x}{R}\right)\right]_0^R = \left(0+\frac{R^2}{2}\cdot\frac\pi2\right) - 0 = \frac{\pi R^2}{4}\]

Final answer:

\[\boxed{\int_0^R \sqrt{R^2-x^2} \, dx = \frac{\pi R^2}{4}}\]

This represents the area of a quarter circle with radius $R$.