Question
If $$f\left( x \right) = a + bx + c{x^2},$$ then what is $$\int_0^1 {f\left( x \right)dx} $$ equal to ?
A.
$$\frac{{\left[ {f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)} \right]}}{6}$$
B.
$$\frac{{\left[ {f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)} \right]}}{3}$$
C.
$$\left[ {f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)} \right]$$
D.
$$\frac{{\left[ {f\left( 0 \right) + 2f\left( {\frac{1}{2}} \right) + f\left( 1 \right)} \right]}}{6}$$
Answer :
$$\frac{{\left[ {f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)} \right]}}{6}$$
Solution :
$$\eqalign{
& {\text{Given, }}f\left( x \right) = a + bx + c{x^2} \cr
& \therefore \,\int_0^1 {f\left( x \right)dx} \cr
& = \int_0^1 {\left( {a + bx + c{x^2}} \right)dx} \cr
& = \left[ {ax + \frac{{b{x^2}}}{2} + \frac{{c{x^3}}}{3}} \right]_0^1 \cr
& = a + \frac{b}{2} + \frac{c}{3}......\left( {\text{i}} \right) \cr
& {\text{Here, }}f\left( 0 \right) = a,\,f\left( {\frac{1}{2}} \right) = a + \frac{b}{2} + \frac{c}{4}{\text{ and }}f\left( 1 \right) = a + b + c \cr
& {\text{Now, }}\frac{{f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)}}{6} \cr
& = \frac{{a + 4\left( {a + \frac{b}{2} + \frac{c}{4}} \right) + a + b + c}}{6} \cr
& = \frac{{a + 4\left( {\frac{{4a + 2b + c}}{4}} \right) + a + b + c}}{6} \cr
& = \frac{{a + 4a + 2b + c + a + b + c}}{6} \cr
& = \frac{{6a + 3b + 2c}}{6} \cr
& = a + \frac{b}{2} + \frac{c}{3} \cr
& \therefore {\text{ From equations }}\left( {\text{i}} \right)\,{\text{and }}\left( {{\text{ii}}} \right),{\text{ we get}} \cr
& \int_0^1 {f\left( x \right)dx} = \frac{{f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)}}{6} \cr} $$