If \( f(x)=ax^2+bx+c \), then \( \int_{0}^{1}\displaystyle f(x)dx \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{6}\left\{f(0)+4f\left(\displaystyle\frac{1}{2}\right)+f(1)\right\} \)
Integrating term by term:
\( \displaystyle \int_{0}^{1}(ax^2+bx+c)\,dx=\frac{a}{3}+\frac{b}{2}+c \)
The values of f that appear in the options are
\( \displaystyle f(0)=c \)
\( \displaystyle f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c \)
\( \displaystyle f(1)=a+b+c \)
For the combination \( f(0)+4f\left(\frac{1}{2}\right)+f(1) \):
\( \displaystyle c+(a+2b+4c)+(a+b+c)=2a+3b+6c \)
\( \displaystyle \frac{1}{6}(2a+3b+6c)=\frac{a}{3}+\frac{b}{2}+c \)
This equals the integral, so Option (3) is correct.
Shortcut: The result must hold for every quadratic, so test the options with f(x) = 1, for which the integral is 1:
\( \displaystyle (1):\ \frac{3}{6}=\frac{1}{2} \)
\( \displaystyle (2):\ \frac{0}{6}=0 \)
\( \displaystyle (3):\ \frac{6}{6}=1 \)
\( \displaystyle (4):\ \frac{4}{3} \)
Only Option (3) gives 1.
Tip: This is Simpson’s rule: \( \int_{a}^{b}f(x)\,dx=\frac{b-a}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right] \), exact for every polynomial of degree up to 3.