int_-1^1|x|dx=? 

Explanation:

Another Explanation (5):

সমাধান:

আমরা জানি, পরম মান ফাংশন \( |x| \) কে নিম্নরূপে সংজ্ঞায়িত করা হয়: \[ |x| = \begin{cases} x, & \text{যদি } x \geq 0 \\ -x, & \text{যদি } x < 0 \end{cases} \] অতএব, \( \int_{-1}^{1} |x| dx \) কে দুইটি অংশে ভাগ করে লেখা যায়: \[ \int_{-1}^{1} |x| dx = \int_{-1}^{0} |x| dx + \int_{0}^{1} |x| dx \] এখন, \( -1 \leq x < 0 \) এর জন্য \( |x| = -x \) এবং \( 0 \leq x \leq 1 \) এর জন্য \( |x| = x \). সুতরাং, \[ \int_{-1}^{0} |x| dx = \int_{-1}^{0} (-x) dx = -\int_{-1}^{0} x dx = -\left[ \frac{x^2}{2} \right]_{-1}^{0} = -\left( \frac{0^2}{2} - \frac{(-1)^2}{2} \right) = -\left( 0 - \frac{1}{2} \right) = \frac{1}{2} \] 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, \[ \int_{-1}^{1} |x| dx = \frac{1}{2} + \frac{1}{2} = 1 \] অতএব, \( \int_{-1}^{1} |x| dx = 1 \) 🥳🎉.