Definite Integration Question 447
Question: The area bounded by the curve $ y=f(x) $ , x-axis and ordinates x = 1 and $ x=b $ is $ \frac{5}{24}\pi $ , then $ f(x) $ is
[RPET 2000]
Options:
A) $ 3(x-1)\cos (3x+4)+\sin (3x+4) $
B) $ (b-1)\sin (3x+4)+3\cos (3x+4) $
C) $ (b-1)\cos (3x+4)+3\sin (3x+4) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Area bounded by the curve $ y=f(x), $ x-axis and the ordinates $ x=1 $ and $ x=b $ is $ \int_1^{b}{f(x)dx} $
From the question $ \int_1^{b}{f(x)dx=(b-1)\sin (3b+4)} $
Differentiate with respect to b, we get $ f(b).1=3(b-1)\cos (3b+4)+\sin (3b+4) $
$ f(x)=3(x-1)\cos (3x+4)+\sin (3x+4) $ .
BETA
