On the interval left[ frac{5pi}{3}, frac{7pi} | vedclass

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(B) Given $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$.By the Fundamental Theorem of Calculus,$f'(x) = 6\cos x - 2\sin x$.In the interval $\left

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$.By the Fundamental Theorem of Calculus,$f'(x) = 6\cos x - 2\sin x$.In the interval $\left[ \frac{5\pi}{3}, \frac{7\pi}{4} \right]$,$\cos x > 0$ and $\sin x  0$.Since $f'(x) > 0$,$f(x)$ is an increasing function on the given interval.Therefore,the greatest value occurs at the upper limit $x = \frac{7\pi}{4}$.$f\left( \frac{7\pi}{4} \right) = \int_{5\pi/3}^{7\pi/4} (6\cos t - 2\sin t) \, dt = [6\sin t + 2\cos t]_{5\pi/3}^{7\pi/4}$.$= \left( 6\sin\frac{7\pi}{4} + 2\cos\frac{7\pi}{4} \right) - \left( 6\sin\frac{5\pi}{3} + 2\cos\frac{5\pi}{3} \right)$.$= \left( 6\left(-\frac{1}{\sqrt{2}}\right) + 2\left(\frac{1}{\sqrt{2}}\right) \right) - \left( 6\left(-\frac{\sqrt{3}}{2}\right) + 2\left(\frac{1}{2}\right) \right)$.$= \left( -\frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}} \right) - (-3\sqrt{3} + 1)$.$= -\frac{4}{\sqrt{2}} + 3\sqrt{3} - 1 = 3\sqrt{3} - 2\sqrt{2} - 1$.

On the interval $\left[ \frac{5\pi}{3}, \frac{7\pi}{4} \right]$,the greatest value of the function $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$ is:

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