हिंदी

∫ ( X − 1 ) E − X D X is Equal to - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\left( x - 1 \right) e^{- x} dx\] is equal to

विकल्प

  • − xex + C

  • xex + C

  • − xex + C

  • xex + C

MCQ

उत्तर

− xe−x + C

 

\[\int \left( x - 1 \right)_I {e^{- x}}_{II} dx\]
\[ = \left( x - 1 \right)\int e^{- x} dx - \int\left\{ \frac{d}{dx}\left( x - 1 \right)\int e^{- x} dx \right\}dx\]
\[ = \left( x - 1 \right) \cdot e^{- x} \left( - 1 \right) - \int1 \cdot e^{- x} \times - 1 dx\]
\[ = - \left( x - 1 \right) e^{- x} + \int e^{- x} dx\]
\[ = \left( 1 - x \right) e^{- x} + \frac{e^{- x}}{- 1} + C\]
\[ \Rightarrow \left( 1 - x - 1 \right) e^{- x} + C\]
\[ = - x e^{- x} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - MCQ [पृष्ठ २००]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
MCQ | Q 9 | पृष्ठ २००

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]

\[\int\sqrt{x}\left( 3 - 5x \right) dx\]

 


\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

\[\int\frac{1}{1 + \cos 2x} dx\]

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]

\[\int\frac{x^2 + x + 5}{3x + 2} dx\]

` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


Integrate the following integrals:

\[\int\text { sin  x  cos  2x     sin 3x   dx}\]

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

`  =  ∫ root (3){ cos^2 x}  sin x   dx `


\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]

\[\int\frac{1}{a^2 x^2 + b^2} dx\]

\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]

\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]

\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]

`  ∫ \sqrt{"cosec x"- 1}  dx `

\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]

\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]

\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]

\[\int\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} \text{ dx }\]

\[\int x^2 \text{ cos x dx }\]

\[\int\frac{\log \left( \log x \right)}{x} dx\]

\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]

\[\int x \sin x \cos 2x\ dx\]

\[\int\sqrt{3 - x^2} \text{ dx}\]

\[\int\frac{x}{\left( x - 1 \right)^2 \left( x + 2 \right)} dx\]

\[\int\frac{x^3 - 1}{x^3 + x} dx\]

\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then


\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]

\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]

\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]

\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]


\[\int \tan^3 x\ \sec^4 x\ dx\]

\[\int \log_{10} x\ dx\]

\[\int\frac{\log \left( \log x \right)}{x} \text{ dx}\]

\[\int\frac{x}{x^3 - 1} \text{ dx}\]

\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×