Definite-Integration Question 518
Question: The least value of the function $ F(x)= $ $ \int_{5\pi /4}^{x}{(3\sin u+4\cos u),du} $ on the interval $ [ \frac{5\pi }{4},\frac{4\pi }{3} ] $ is
Options:
A) $ \sqrt{3}+\frac{3}{2} $
B) $ -2\sqrt{3}+\frac{3}{2}+\frac{1}{\sqrt{2}} $
C) $ \frac{3}{2}+\frac{1}{\sqrt{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ F’(x)=3\sin x+4\cos x $ Since in $ [ \frac{5\pi }{4},\frac{4\pi }{3} ],F’(x)<0, $ so assume the least value at the point $ x=\frac{4\pi }{3}. $ Thus, $ f( \frac{4\pi }{3} )=\int_{5\pi /4}^{4\pi /3}{(3\sin u+4\cos u)du} $ $ =\frac{3}{2}-2\sqrt{3}+\frac{1}{\sqrt{2}} $ .
BETA
