int_(-1)^1|x|dx                                                  

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