Advertisements
Advertisements
प्रश्न
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
विकल्प
0
1
2
`1/2`
MCQ
Advertisements
उत्तर
`1/2`
shaalaa.com
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]
\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]
\[\int\limits_0^1 \log\left( \frac{1}{x} - 1 \right) dx\]
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]
\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
