The least value of the function F(x) = int_{5 | vedclass

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(B) Given $F(x) = \int_{5\pi /4}^x (3\sin u + 4\cos u) du$. By the Fundamental Theorem of Calculus,$F'(x) = 3\sin x + 4\cos x$. In the interval $\left

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$-2\sqrt{3} + \frac{3}{2} + \frac{1}{\sqrt{2}}$

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(B) Given $F(x) = \int_{5\pi /4}^x (3\sin u + 4\cos u) du$. By the Fundamental Theorem of Calculus,$F'(x) = 3\sin x + 4\cos x$. In the interval $\left[ \frac{5\pi}{4}, \frac{4\pi}{3} ight]$,both $\sin x$ and $\cos x$ are negative. Specifically,$3\sin x + 4\cos x = 5 \left( \frac{3}{5}\sin x + \frac{4}{5}\cos x ight) = 5\sin(x + \alpha)$,where $	an \alpha = 4/3$. Since $F'(x) Therefore,the least value occurs at the right endpoint,$x = \frac{4\pi}{3}$. $F\left( \frac{4\pi}{3} ight) = \int_{5\pi /4}^{4\pi /3} (3\sin u + 4\cos u) du = [-3\cos u + 4\sin u]_{5\pi /4}^{4\pi /3}$. $= \left( -3\cos\frac{4\pi}{3} + 4\sin\frac{4\pi}{3} ight) - \left( -3\cos\frac{5\pi}{4} + 4\sin\frac{5\pi}{4} ight)$. $= \left( -3(-\frac{1}{2}) + 4(-\frac{\sqrt{3}}{2}) ight) - \left( -3(-\frac{1}{\sqrt{2}}) + 4(-\frac{1}{\sqrt{2}}) ight)$. $= (\frac{3}{2} - 2\sqrt{3}) - (\frac{3}{\sqrt{2}} - \frac{4}{\sqrt{2}}) = \frac{3}{2} - 2\sqrt{3} - (-\frac{1}{\sqrt{2}}) = \frac{3}{2} - 2\sqrt{3} + \frac{1}{\sqrt{2}}$.

The least value of the function $F(x) = \int_{5\pi /4}^x {(3\sin u + 4\cos u)\,du} $ on the interval $\left[ \frac{5\pi}{4}, \frac{4\pi}{3} \right]$ is

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