The maximum value of the function f(x) = intl | vedclass

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(B) We evaluate the integral using integration by parts: $\int u dv = uv - \int v du$. Let $u = t$ and $dv = \sin(x + \pi t) dt$. Then $du = dt$ and $

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$\frac{1}{{{\pi ^2}}}\sqrt {{\pi ^2} + 4} $

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(B) We evaluate the integral using integration by parts: $\int u dv = uv - \int v du$. Let $u = t$ and $dv = \sin(x + \pi t) dt$. Then $du = dt$ and $v = -\frac{1}{\pi} \cos(x + \pi t)$.$f(x) = \left[ -\frac{t}{\pi} \cos(x + \pi t) \right]_0^1 + \frac{1}{\pi} \int_0^1 \cos(x + \pi t) dt$$f(x) = \left( -\frac{1}{\pi} \cos(x + \pi) - 0 \right) + \frac{1}{\pi} \left[ \frac{1}{\pi} \sin(x + \pi t) \right]_0^1$Since $\cos(x + \pi) = -\cos x$,we have $f(x) = \frac{1}{\pi} \cos x + \frac{1}{\pi^2} (\sin(x + \pi) - \sin x)$.Since $\sin(x + \pi) = -\sin x$,we have $f(x) = \frac{1}{\pi} \cos x - \frac{2}{\pi^2} \sin x$.The maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$.Here $a = \frac{1}{\pi}$ and $b = -\frac{2}{\pi^2}$.$f(x)_{\max} = \sqrt{\left(\frac{1}{\pi}\right)^2 + \left(-\frac{2}{\pi^2}\right)^2} = \sqrt{\frac{1}{\pi^2} + \frac{4}{\pi^4}} = \sqrt{\frac{\pi^2 + 4}{\pi^4}} = \frac{\sqrt{\pi^2 + 4}}{\pi^2}$.

The maximum value of the function $f(x) = \int\limits_0^1 {t\,\sin \left( {x + \pi t} \right)} dt$ for $x \in R$ is

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