int_0^5sqrt(25-x^2dx এর মান হলো-

সমাধান:

আমরা জানি, \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \)

এখানে, \( a^2 = 25 \), সুতরাং \( a = 5 \)।

তাহলে, \( \int_0^5 \sqrt{25 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{25}{2} \sin^{-1}\left(\frac{x}{5}\right) \right]_0^5 \)

= \( \left( \frac{5}{2} \sqrt{25 - 5^2} + \frac{25}{2} \sin^{-1}\left(\frac{5}{5}\right) \right) - \left( \frac{0}{2} \sqrt{25 - 0^2} + \frac{25}{2} \sin^{-1}\left(\frac{0}{5}\right) \right) \)

= \( \left( \frac{5}{2} \cdot 0 + \frac{25}{2} \sin^{-1}(1) \right) - \left( 0 + \frac{25}{2} \sin^{-1}(0) \right) \)

= \( \frac{25}{2} \cdot \frac{\pi}{2} - 0 \)

= \( \frac{25\pi}{4} \)

সুতরাং, \( \int_0^5 \sqrt{25 - x^2} \, dx = \frac{25\pi}{4} \). 🎉

অতএব, নির্ণেয় মান \( \frac{25\pi}{4} \)।