Question

∫ |cos x| dx=? . limit is from - 2π to π

Answer 1

Answer on Question #51706 – Math – Integral Calculus

$$\int_{-2\pi}^{\pi} |\cos x| \, dx = ?$$

Solution

To find this definite integral, we must know subsets of $[-2\pi; \pi]$, where function $\cos x$ is negative and positive. We can see it on a picture below:

Now, using definition of the absolute value $|\cos x| = \begin{cases} \cos x, & \cos x \geq 0 \\ -\cos x, & \cos x < 0 \end{cases}$, we can rewrite our integral as follows:

$$\begin{aligned} \int_{-2\pi}^{\pi} |\cos x| \, dx &= \int_{-2\pi}^{-3\pi/2} \cos x \, dx - \int_{-3\pi/2}^{-\pi/2} \cos x \, dx + \int_{-\pi/2}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{\pi} \cos x \, dx = \\ &= \sin x \left| \frac{-\frac{3\pi}{2}}{2} - \sin x \right| \frac{-\frac{\pi}{2}}{-\frac{3\pi}{2}} + \sin x \left| \frac{\frac{\pi}{2}}{2} - \sin x \right| \frac{\pi}{2} = \sin \left( -\frac{3\pi}{2} \right) - \sin (-2\pi) - \sin \left( -\frac{\pi}{2} \right) + \\ &\sin \left( -\frac{3\pi}{2} \right) + \sin \left( \frac{\pi}{2} \right) - \sin \left( -\frac{\pi}{2} \right) - \sin (\pi) + \sin \left( \frac{\pi}{2} \right) = 1 - 0 - (-1) + 1 + 1 - (-1) - 0 + \\ &+ 1 = 6. \end{aligned}$$

The second fundamental theorem of calculus (or Newton-Leibniz axiom), the fact that the antiderivative of cosine is sine plus an integration constant, some values of sine function were applied there.

Answer:

$$\int_{-2\pi}^{\pi} |\cos x| \, dx = 6.$$

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