Advertisements
Advertisements
प्रश्न
Advertisements
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 each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
If f(2a − x) = −f(x), prove that
If f is an integrable function, show that
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Choose the correct alternative:
`Γ(3/2)`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
