evaluate the integral.\n int_{1}^{2}sqrt{4 - x^{2}}mathrm{d}x=square \text{(type an exact answer.)}

Explanation:

Step1: Use trigonometric substitution

Let $x = 2\sin\theta$, then $dx=2\cos\theta d\theta$. When $x = 1$, $\sin\theta=\frac{1}{2}$, so $\theta=\frac{\pi}{6}$; when $x = 2$, $\sin\theta = 1$, so $\theta=\frac{\pi}{2}$. And $\sqrt{4 - x^{2}}=\sqrt{4-4\sin^{2}\theta}=2\cos\theta$.

Step2: Rewrite the integral

The integral $\int_{1}^{2}\sqrt{4 - x^{2}}dx$ becomes $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}2\cos\theta\cdot2\cos\theta d\theta=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}4\cos^{2}\theta d\theta$. Since $\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}$, the integral is $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}4\cdot\frac{1+\cos(2\theta)}{2}d\theta=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(2 + 2\cos(2\theta))d\theta$.

Step3: Integrate term - by - term

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(2 + 2\cos(2\theta))d\theta=\left[2\theta+\sin(2\theta)\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$.

Step4: Evaluate the definite integral

$2\cdot\frac{\pi}{2}+\sin(\pi)-\left(2\cdot\frac{\pi}{6}+\sin\left(\frac{\pi}{3}\right)\right)=\pi+0-\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$.

Answer:

$\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$