If for all real triplets (a,b,c), f(x)=a+bx+cx^2; then _0^1f(x),dx is equal to:

Question

If for all real triplets $(a,b,c)$, $f(x)=a+bx+cx^2$; then $\displaystyle\int_0^1f(x)\,dx$ is equal to:

Accepted Answer

$$\int_0^1(a+bx+cx^2)dx=a+\frac b2+\frac c3$$

With $f(1)=a+b+c$, $f(0)=a$, $f\left(\dfrac12\right)=a+\dfrac b2+\dfrac c4$:
$$\frac16\left[f(0)+f(1)+4f\left(\frac12\right)\right]=\frac16\left[a+b+c+a+4a+2b+c\right]=\frac16(6a+3b+2c)=a+\frac b2+\frac c3$$

$$\therefore\int_0^1f(x)dx=\frac16\left[f(0)+f(1)+4f\left(\frac12\right)\right]$$

Solutions · 1