Definite Integration Question 197
Question: The line y = mx bisects the area enclosed by lines $ x=0,y=0 $ and $ x=3/2 $ and the curve $ y=1+4x-x^{2} $ . Then the value of m is
Options:
A) $ \frac{13}{6} $
B) $ \frac{13}{2} $
C) $ \frac{13}{5} $
D) $ \frac{13}{7} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y=1+4x-x^{2}=5-{{(x-2)}^{2}} $ We have $ \int\limits_0^{3/2}{(1+4x-x^{2})dx=2\int\limits_0^{3/2}{mxdx}} $
$ =\frac{3}{2}+2( \frac{9}{4} )-\frac{1}{3}( \frac{27}{8} )=m.\frac{9}{4} $
On solving we get $ m=\frac{13}{6} $
BETA
