Question

Evaluate \(\int_{-2}^{0} \sqrt{4-\mathrm{x}^{2}} \mathrm{dx}\)

Short answer

The value of the definite integral is \(2 \pi\).

Step-by-step solution

Identify the integral function

We are given the definite integral: \(\int_{-2}^{0} \sqrt{4-x^2} dx\). Our function is \(f(x) = \sqrt{4-x^2}\), and we need to find its antiderivative and then evaluate it at the boundaries.

Integration Substitution

To solve the integral, we can use the substitution method. Let's denote \(x = 2\sin u\), so \(dx = 2\cos u\, du\). When \(x = -2\), \(u = -\frac{\pi}{2}\), and when \(x = 0\), \(u = 0\). So, we rewrite our integral in terms of \(u\): \(\int_{-\frac{\pi}{2}}^0\sqrt{4 - (2\sin u)^2} \cdot 2\cos u\, du\).

Simplify the Integral Expression

Now, we simplify the new integral expression: \begin{align*} \int_{-\frac{\pi}{2}}^0\sqrt{4 - (2\sin u)^2} \cdot 2\cos u\,du &= \int_{-\frac{\pi}{2}}^0\sqrt{4(1 - \sin^2 u)} \cdot 2\cos u\, du \\ &= \int_{-\frac{\pi}{2}}^0 2\sqrt{4\cos^2 u} \cdot 2\cos u\, du \\ &= \int_{-\frac{\pi}{2}}^0 4\cos u \cdot 2\cos u\, du \\ &= 8\int_{-\frac{\pi}{2}}^0 \cos^2 u\, du \end{align*}

Use Power Reduction Formula

To integrate the \(\cos^2 u\) term, we can use the power reduction formula: \(\cos^2 u = \frac{1 + \cos 2u}{2}\). The integral becomes: $$8\int_{-\frac{\pi}{2}}^0 \frac{1 + \cos 2u}{2}\,du$$

Integration

Now, we integrate each term separately. $$8\int_{-\frac{\pi}{2}}^0 \frac{1 + \cos 2u}{2}\,du = 8\int_{-\frac{\pi}{2}}^0 \frac{1}{2}\,du + 8\int_{-\frac{\pi}{2}}^0 \frac{\cos 2u}{2}\,du$$ By performing the integration, we get: $$4\left[u - \frac{1}{4}\sin(2u)\right]\Big|_{-\frac{\pi}{2}}^0$$

Evaluate at the boundaries

Now, we evaluate the expression at the boundaries: \begin{align*} 4\left[u - \frac{1}{4}\sin(2u)\right]\Big|_{-\frac{\pi}{2}}^0 &= 4\left(\left[0 - \frac{1}{4}\sin(0)\right] - \left[-\frac{\pi}{2} - \frac{1}{4}\sin(-\pi)\right]\right) \\ &= 4\left(0 + \frac{\pi}{2}\right) \\ &= 2 \pi\ \end{align*} Thus, the definite integral evaluates to \(\int_{-2}^{0} \sqrt{4-x^2} dx = 2 \pi\).