Definite Integration Question 171
Question: The area bounded by the curves $ y=f(x), $ the x-axis, and the ordinates $ x=1 $ and $ x=b $ is $ (b-1)\sin (3b+4) $ . Then $ f(x) $ is
Options:
A) $ (x-1)\cos (3x+4) $
B) $ sin(3x+4) $
C) $ \sin (3x+4)+3(x-1)\cos (3x+4) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ \int_1^{b}{f(x)dx=(b-1)\sin (3b+4)} $ Differentiating both sides w.r.t. b, we get
$ \Rightarrow f(b)=3(b-1)\cos (3b+4)+\sin (3b+4) $
$ \Rightarrow f(x)=\sin (3x+4)+3(x-1)\cos (3x+4) $ .
BETA
