Advertisements
Advertisements
प्रश्न
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
विकल्प
1
`2 int_0^1 x^3 "e"^(x^4) "d"x`
0
`"e"^(x^4)`
MCQ
Advertisements
उत्तर
0
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]
\[\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
\[\int\limits_0^\pi x \sin^3 x\ dx\]
\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]
\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is
\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
