$\int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} $ का मान है

  • A
    $0$
  • B
    $2\int_0^1 {\frac{{\sin x}}{{3 - |x|}}\,dx} $
  • C
    $2\int_0^1 {\frac{{ - {x^2}}}{{3 - |x|}}} \,dx$
  • D
    $2\int_0^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} $

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Similar Questions

मान लीजिए $g_i: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}, i=1, 2$,और $f: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}$ ऐसे फलन हैं कि $g_1(x)=1, g_2(x)=|4x-\pi|$ और $f(x)=\sin^2 x$,सभी $x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right]$ के लिए।
$S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) dx, i=1, 2$ को परिभाषित करें।
$(1)$ $\frac{16S_1}{\pi}$ का मान है।
$(2)$ $\frac{48S_2}{\pi^2}$ का मान है।

यदि $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx = \pi(\alpha \pi^2 + \beta)$,जहाँ $\alpha, \beta \in \mathbb{Z}$,तो $(\alpha + \beta)^2$ का मान ज्ञात कीजिए:

$\int_{0}^{\pi} \cos^3 x \, dx = $

$\int_{\pi}^{16\pi} |\sin x| dx = $

समाकलन $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \frac{\pi-x}{\pi+x}\right) \cos x \, dx$ का मान ज्ञात कीजिए।

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