Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_0^1 x \left( 1 - x \right)^5 d x . Then, \]
\[I = \int_0^1 \left( x - 1 + 1 \right) \left( 1 - x \right)^5 d x\]
\[ \Rightarrow I = \int_0^1 \left[ - \left( 1 - x \right)^6 + \left( 1 - x \right)^5 \right] d x\]
\[ \Rightarrow I = \left[ \frac{\left( 1 - x \right)^7}{7} \right]_0^1 - \left[ \frac{\left( 1 - x \right)^6}{6} \right]_0^1 \]
\[ \Rightarrow I = - \frac{1}{7} + \frac{1}{6}\]
\[ \Rightarrow I = \frac{1}{42}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate the following integral:
Prove that:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
