Explanation: 
Another Explanation (5):
প্রশ্ন: \(\int_{-1}^{1} |x| \, dx\)
সমাধান:
আমরা জানি, \(|x| = \begin{cases} -x, & \text{যদি } x < 0 \\ x, & \text{যদি } x \geq 0 \end{cases}\)
সুতরাং, \(\int_{-1}^{1} |x| \, dx\) কে দুইটি 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 \(\int_{0}^{1} |x| \, dx = \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}\) 🥳
তাহলে, \(\int_{-1}^{1} |x| \, dx = \frac{1}{2} + \frac{1}{2} = 1\) ✨
সুতরাং, \(\int_{-1}^{1} |x| \, dx = 1\)
উত্তর: 1
