Advertisements
Advertisements
प्रश्न
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
पर्याय
12
4
4!
64
MCQ
उत्तर
4!
shaalaa.com
Definite Integrals
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^2 \left( \frac{x - 1}{x^2} \right) e^x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]
\[\int\limits_0^{\pi/2} 2 \sin x \cos x \tan^{- 1} \left( \sin x \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} dx\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
\[\int\limits_0^{2a} f\left( x \right) dx = 2 \int\limits_0^a f\left( x \right) dx\]
\[\int\limits_0^3 \left( x + 4 \right) dx\]
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
