int_-1^1|x|dx=? 

সমাধান

আমরা জানি, \( |x| = \begin{cases} -x, & x < 0 \\ x, & x \geq 0 \end{cases} \)

সুতরাং, \( \int_{-1}^{1} |x| dx \) কে দুটি অংশে ভাগ করা যায়:

\( \int_{-1}^{1} |x| dx = \int_{-1}^{0} |x| dx + \int_{0}^{1} |x| dx \)

এখন, \( \int_{-1}^{0} |x| dx = \int_{-1}^{0} -x dx \)

\( = \left[ -\frac{x^2}{2} \right]_{-1}^{0} = -\frac{0^2}{2} - \left( -\frac{(-1)^2}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2} \)

এবং, \( \int_{0}^{1} |x| dx = \int_{0}^{1} x dx \)

\( = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} - 0 = \frac{1}{2} \)

অতএব, \( \int_{-1}^{1} |x| dx = \frac{1}{2} + \frac{1}{2} = 1 \) 🎉

সুতরাং, উত্তর: 1