हिंदी

∫ X ( 1 − X ) 23 D X

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प्रश्न

\[\ \int\ x \left( 1 - x \right)^{23} dx\]

 

योग
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उत्तर

\[\int\ x \left( 1 - x \right)^{23} dx\]
\[\text{Let 1 - x }= t \]
\[ \Rightarrow x = 1 - t\]
\[ \Rightarrow 1 = - \frac{dt}{dx}\]
\[ \Rightarrow dx = - dt\]
\[Now, \int\ x \left( 1 - x \right)^{23} dx\]
\[ = - \int\left( 1 - t \right) \cdot t^{23} dt\]
\[ = - \int\left( t^{23} - t^{24} \right)dt\]
\[ = \int\left( t^{24} - t^{23} \right) dt\]
\[ = \frac{t^{25}}{25} - \frac{t^{24}}{24} + C\]
\[ = \frac{24 t^{25} - 25 t^{24}}{600} + C\]
\[ = \frac{t^{24}}{600}\left[ 24t - 25 \right] + C\]
\[ = \frac{\left( 1 - x \right)^{24}}{600} \left[ 24\left( 1 - x \right) - 25 \right] + C\]
\[ = \frac{- 1}{600} \left( 1 - x \right)^{24} \left( 1 + 24x \right) + C\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Indefinite Integrals - Exercise 19.10 [पृष्ठ ६५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 18 Indefinite Integrals
Exercise 19.10 | Q 8 | पृष्ठ ६५
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