Integral Calculus Question 253
Question: If $ f(x)=a+bx+cx^{2}, $ then what is $ \int_0^{1}{f(x)dx} $ equal to?
Options:
A) $ [f(0)+4f(1/2)+f(1)]/6 $
B) $ [f(0)+4f(1/2)+f(1)]/3 $
C) $ [f(0)+4f(1/2)+f(1)] $
D) $ [f(0)+2f(1/2)+f(1)]/6 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given, $ f(x)=a+bx+cx^{2} $
$ \therefore \int_0^{1}{f(x)dx=\int_0^{1}{(a+bx+cx^{2})dx}} $ $ =[ ax+\frac{bx^{2}}{2}+\frac{cx^{3}}{3} ]_0^{1} $ $ =a+\frac{b}{2}+\frac{c}{3}…(i) $ Here, $ f(0)=a,f( \frac{1}{2} )=a+\frac{b}{2}+\frac{c}{4} $ and $ f(1)=a+b+c $ Now, $ \frac{f(0)+4f( \frac{1}{2} )+f(1)}{6} $ $ =\frac{a+4( a+\frac{b}{2}+\frac{c}{4} )+a+b+c}{6} $ $ =\frac{a+4( \frac{4a+2b+c}{4} )+a+b+c}{6} $ $ =\frac{a+4a+2b+c+a+b+c}{6}=\frac{6a+3b+2c}{6} $ $ =a+\frac{b}{2}+\frac{c}{3} $
$ \therefore $ From equations (i) and (ii), we get $ \int_0^{1}{f(x)dx=\frac{f(0)+4f( \frac{1}{2} )+f(1)}{6}} $
BETA
