Question

F(x) = du on the interval (5p/4, 4p/3] is -

• A. 3/2 – $\sqrt { 3 }$/2
• B. $\frac { 5 - 4 \sqrt { 3 } } { 2 }$
• C. $\frac { 7 - 4 \sqrt { 3 } } { 2 }$
• D. $\frac { 9 - 4 \sqrt { 3 } } { 2 }$

Answer 1

Answer: (D)

$\frac { 9 - 4 \sqrt { 3 } } { 2 }$

Answer 2

We have F¢(x) = 3 sin x + 4 cos x. Since sin x and cos x assume negative values in the third quadrant, we have F¢(x) < 0 for all x Ī (5p/4, 4p/3) so F(x) assumes the least value at the point x = 4p/3. Thus the least value is

F(4p/3) = $\int _ { 0 } ^ { 4 \pi / 3 } ( 3 \sin u + 4 \cos u ) d u$

= (–3 cos u + $4 \sin u ) \left. \right| _ { 0 } ^ { 4 \pi / 3 }$

= –3 cos $\frac { 4 \pi } { 3 }$– (–3)

= $\frac { 9 - 4 \sqrt { 3 } } { 2 }$