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Question: The least value of the function f(x) = \(\int_{0}^{x}{(3\sin t + 4\cos t)dt}\) on the interval \(\l...

The least value of the function f(x) = ∫0x(3sin⁡t+4cos⁡t)dt\int_{0}^{x}{(3\sin t + 4\cos t)dt}

on the interval (5π4,4π3)\left( \frac{5\pi}{4},\frac{4\pi}{3} \right) is

A

9−432\frac{9 - 4\sqrt{3}}{2}

B

9+432\frac{9 + 4\sqrt{3}}{2}

C

5−432\frac{5 - 4\sqrt{3}}{2}

D

5+432\frac{5 + 4\sqrt{3}}{2}

Answer

9−432\frac{9 - 4\sqrt{3}}{2}

Explanation

Solution

By NL in [5π4,4π3]\left\lbrack \frac{5\pi}{4},\frac{4\pi}{3} \right\rbrack Ž 3rd quadrant

f ¢(x) = 3 sin x + 4 cos x < 0

Ž function is Æ

\ least value f(4π3)\left( \frac{4\pi}{3} \right)= ∫04π/3\int_{0}^{4\pi/3}{} (3sin t + 4cos t) dt

= 92−432\frac{9}{2} - \frac{4\sqrt{3}}{2}