Question

बदलू की बेटियां की कलाइयों पर लाख की चूड़िया सजने का क्या रहस्य था​

Answer 1

Answer:

Answer:

$\huge \frac{13}{420}$

Step-by-step explanation:

$\int_{0}^{1} \int_{ {y}^{2} }^{1 - y } \int_{0}^{1 - x} xdzdxdy \\ \\ = \int_{0}^{1} \int_{ {y}^{2} }^{1 - y } \int_{0}^{1 - x} \{ xz \}dxdy \\ \\ = \int_{0}^{1} \int_{ {y}^{2}} ^{1 - y} x({1 - x} )dxdy \\ \\ = \int_{0}^{1} \int_{ {y}^{2} }^{1 - y }( \frac{ {x}^{2} }{2} - \frac{ {x}^{3} }{3} )dy \\ \\ = \int_{0}^{1} (\frac{ {(1 - y)}^{2} }{2} - \frac{ {(1 - y)}^{3} }{3} - \frac{ {(y {}^{2} )}^{2} }{2} + \frac{ {( {y}^{2}) }^{3} }{3} )dy dy \\ \\using \: integral \: property \: in \: first \: two \: terms \\ \int _{0} ^{1} f(y) \: dy = \int _{0} ^{1} f(1 - y) \: dy \\ \\ = \ \int _{0} ^{1} ( \frac{ {y}^{2} }{2} - \frac{ {y}^{3} }{3} - \frac{ {y}^{4} }{2} + \frac{ {y}^{6} }{3} )dy \\ \\ = \frac{ {y}^{3} }{6} - \frac{ {y}^{4} }{12} - \frac{ {y}^{5} }{10} + \frac{ {y}^{7} }{21} \: \: \{ \: lim \: \: 0 \: to \: 1 \\ \\ = \frac{1}{6} - \frac{1}{12} - \frac{1}{10} + \frac{1}{21} \\ \\ = \frac{70 - 35 - 42 + 20}{420} \\ \\ = \frac{13}{420}$

COPIED;)