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प्रश्न
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उत्तर
\[Let\ I = \int_4^{12} x \left( x - 4 \right)^\frac{1}{3} d x . \]
\[Let\ x - 4 = t . Then, dx = dt\]
\[When\ x = 4, t = 0\ and\ x\ = 12, t = 8\]
\[ \therefore I = \int_0^8 \left( t + 4 \right) t^\frac{1}{3} dt\]
\[ \Rightarrow I = \int_0^8 \left( t^\frac{4}{3} + 4 t^\frac{1}{3} \right) dt\]
\[ \Rightarrow I = \left[ \frac{3}{7} t^\frac{7}{3} + \frac{3}{1} t^\frac{4}{3} \right]_0^8 \]
\[ \Rightarrow I = \frac{384}{7} + 48\]
\[ \Rightarrow I = \frac{720}{7}\]
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