JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Definite Integrals

If for all real triplets

(a, b, c)

f (x) = a + b x + c x^{2}

; then

\int_{0}^{1} f (x) d x

is equal to:

Answer: (B)

Solution

Integrating directly,

\int_{0}^{1} (a + b x + c x^{2}) d x = a + \frac{b}{2} + \frac{c}{3}

. Now

f (0) = a

f (1) = a + b + c

and

f (\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4}

, so

\frac{1}{6} [f (0) + f (1) + 4 f (\frac{1}{2})] = \frac{1}{6} (6 a + 3 b + 2 c) = a + \frac{b}{2} + \frac{c}{3}

, which is exactly the integral. This is Simpson's rule, exact for a quadratic.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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