int_-1^1|x|dx=? 

Explanation:

Another Explanation (5):

সমাধান:

আমরা জানি, \(|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}\) অতএব, \(\int_{-1}^{1} |x| dx\) কে দুটি অংশে ভাগ করা যায়: \(\int_{-1}^{1} |x| dx = \int_{-1}^{0} |x| dx + \int_{0}^{1} |x| dx\) এখন, \(\int_{-1}^{0} |x| dx = \int_{-1}^{0} -x dx = -\int_{-1}^{0} x dx = -\left[\frac{x^2}{2}\right]_{-1}^{0} = -\left[\frac{0^2}{2} - \frac{(-1)^2}{2}\right] = -\left[0 - \frac{1}{2}\right] = \frac{1}{2}\) এবং, \(\int_{0}^{1} |x| dx = \int_{0}^{1} x dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}\) সুতরাং, \(\int_{-1}^{1} |x| dx = \frac{1}{2} + \frac{1}{2} = 1\) 🎉 সুতরাং, \(\int_{-1}^{1} |x| dx = 1\) 🥳