Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^\infty e^{- x} d x\]
\[ = - \left[ e^{- x} \right]_0^\infty \]
\[ = - \left( 0 - 1 \right)\]
\[ = 0 + 1\]
\[ = 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate the following integral:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
