Advertisements
Advertisements
प्रश्न
विकल्प
− xex + C
xex + C
− xe−x + C
xe−x + C
उत्तर
− xe−x + C
\[\int \left( x - 1 \right)_I {e^{- x}}_{II} dx\]
\[ = \left( x - 1 \right)\int e^{- x} dx - \int\left\{ \frac{d}{dx}\left( x - 1 \right)\int e^{- x} dx \right\}dx\]
\[ = \left( x - 1 \right) \cdot e^{- x} \left( - 1 \right) - \int1 \cdot e^{- x} \times - 1 dx\]
\[ = - \left( x - 1 \right) e^{- x} + \int e^{- x} dx\]
\[ = \left( 1 - x \right) e^{- x} + \frac{e^{- x}}{- 1} + C\]
\[ \Rightarrow \left( 1 - x - 1 \right) e^{- x} + C\]
\[ = - x e^{- x} + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
` ∫ sin x \sqrt (1-cos 2x) dx `
Integrate the following integrals:
` = ∫ root (3){ cos^2 x} sin x 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
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]