The-least-value-of-the-function-x-5pi-4-x-3-sin-t-4-cos-t-dt-on-the-interval-5pi-4-4pi-3-is-

Question Number 25773 by Ali Makhzoum last updated on 14/Dec/17

The least value of the function

ϕ (x) = \underset{5 π / 4}{\int^{x}} (3 \sin t + 4 \cos t) d t

on the interval [\frac{5 π}{4}, \frac{4 π}{3}] is

Answered by ajfour last updated on 14/Dec/17

ϕ (x) = (4 \sin t - 3 \cos t) ∣_{5 π / 4}^{x}

= 4 \sin x - 3 \cos x - (- \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}})

= 4 \sin x - 3 \cos x + \frac{1}{\sqrt{2}}

= 5 \sin (x - α) + \frac{1}{\sqrt{2}}

w h e r e α = \sin^{- 1} (\frac{4}{5})

\frac{d [ϕ (x)]}{d x} = 5 \cos (x - α)

i f \frac{5 π}{4} ⩽ x ⩽ \frac{4 π}{3}

225 ° - 53 ° ⩽ x - α ⩽ 240 ° - 53 °

\Rightarrow 172 ° ⩽ x - α ⩽ 187 °

s o \frac{d [ϕ (x)]}{d x} = 5 \cos (x - α) < 0

f u n c t i o n ϕ (x) i s d e c r e a s i n g i n

t h i s i n t e r v a l; h e n c e i n t h i s

i n t e r v a l ϕ (x) i s l e a s t f o r x = 4 π / 3

ϕ (\frac{4 π}{3}) = 4 \sin (\frac{4 π}{3}) - 3 \cos (\frac{4 π}{3}) + \frac{1}{\sqrt{2}}

= 4 \times (- \frac{\sqrt{3}}{2}) - 3 (- \frac{1}{2}) + \frac{1}{\sqrt{2}}

= \frac{3}{2} + \frac{1}{\sqrt{2}} - 2 \sqrt{3} .

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