Definite Integration Question - None

Question

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

Accepted Answer

Solution: It is given that $f(x) = a + bx + cx^2$, then

$$\int_0^1 f(x)\, dx = \int_0^1 (a + bx + cx^2)\, dx$$

$$= \left[ax + \frac{bx^2}{2} + \frac{cx^3}{3}\right]_0^1 = a + \frac{b}{2} + \frac{c}{3}$$

Now evaluating the option (d):

$$\frac{1}{6}\left\{f(0) + f(1) + 4f\left(\frac{1}{2}\right)\right\}$$

$$= \frac{1}{6}\left\{a + (a + b + c) + 4\left(a + \frac{b}{2} + \frac{c}{4}\right)\right\}$$

$$= \frac{1}{6}\left\{a + a + b + c + 4a + 2b + c\right\} = \frac{1}{6}\left\{6a + 3b + 2c\right\} = a + \frac{b}{2} + \frac{c}{3}$$

∴ Answer is (d).

Question: