JEE Main 2020 (Online) 9th January Morning ShiftQuestion 1
If for all real triplets $(a,b,c)$, $f(x)=a+bx+cx^2$; then $\int_0^1f(x)\,dx$ is equal to :
Definite Integration
Solution
Concept. Compute $\int_0^1f$ directly and express it with $f(0)$, $f\left(\frac12\right)$, $f(1)$. Formulas used. (1) $\int_0^1(a+bx+cx^2)\,dx=a+\dfrac b2+\dfrac c3$Step 1 - The integral. By (1), $\int_0^1f=\dfrac{6a+3b+2c}{6}$. Step 2 - The values. $f(0)=a$, $f(1)=a+b+c$, $f\left(\frac12\right)=a+\frac b2+\frac c4$. So $$f(0)+f(1)+4f\left(\tfrac12\right)=a+(a+b+c)+(4a+2b+c)=6a+3b+2c.$$ Step 3 - Compare. $\int_0^1f(x)\,dx=\dfrac16\left\{f(0)+f(1)+4f\left(\tfrac12\right)\right\}$. Answer: (D)