Definite Integration Question 148
Question: If $ f(x)=a+bx+cx^{2} $ , where $ c>0 $ and $ b^{2}-4ac<0 $ , then the area enclosed by the coordinate axes, the line $ x=2 $ and the curve $ y=f(x) $ is given by
Options:
A) $ \frac{1}{3}{4f(1)+f(2)} $
B) $ \frac{1}{2}{f(0)+4f(1)+f(2)} $
C) $ \frac{1}{2}{f(0)+4f(1)} $
D) $ \frac{1}{3}{f(0)+4f(1)+f(2)} $
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Answer:
Correct Answer: D
Solution:
[d] Area of $ OABL=\int_0^{2}{ydx} $
$ =\int_0^{2}{(a+bx+cx^{2})dx=2a+2b+\frac{8}{3}c} $
$ =\frac{1}{3}[6a+6b+8c]….(i) $ But, $ f(x)=a+bx+cx^{2};f(0)=a,f(1)=a+b+c $
$ f(2)=a+2b+4c\Rightarrow \frac{1}{3}{f(0)+4f(1)+f(2)} $
$ =\frac{1}{3}{a+4(a+b+c)+(a+2b+4c)} $
$ =\frac{1}{3}{6a+6b+8c} $
BETA
