Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_0^\infty e^{- x} d\ x\ . Then, \]
\[I = \left[ - e^{- x} \right]_0^\infty \]
\[ \Rightarrow I = - e^{- \infty} + e^0 \]
\[ \Rightarrow I = 0 + 1\]
\[ \Rightarrow I = 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
Evaluate each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following:
Γ(4)