If for all real triplets (a, b, c), ƒ(x) = a + bx + cx2; then \int\limits_0^1 f(x)dx is equal to :

1 \over 6\ f(0) + f(1) + 4f( 1 \over 2 ) \

1 \over 3\ f(0) + f( 1 \over 2 ) \

1 \over 2\ f(1) + 3f( 1 \over 2 ) \

Definite Integration PYQ — JEE Main 2020 · 9 Jan · Shift 1 · Q39 with Solution

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Correct answer: A

$f(x)=a+bx+cx^2$

for all real $a,b,c$ . We need to evaluate

\int_0^1 f(x)\,dx

and compare it with the given options.

\int_0^1 (a+bx+cx^2)\,dx = a\int_0^1 dx + b\int_0^1 x\,dx + c\int_0^1 x^2\,dx

Using

\quad \int_0^1 x\,dx=\frac12, \quad \int_0^1 x^2\,dx=\frac13,$$ we get

\int_0^1 f(x),dx = a+\frac b2+\frac c3.

3. Now check option A. We need: $$f(0)=a$$ $$f(1)=a+b+c$$ $$f\left(\frac12\right)=a+\frac b2+\frac c4$$ So,

\frac16\left[f(0)+f(1)+4f\left(\frac12\right)\right]

becomes

\frac16\left[a+(a+b+c)+4\left(a+\frac b2+\frac c4\right)\right].

Now simplify inside the bracket:

a+a+b+c+4a+2b+c = 6a+3b+2c.

Therefore,

\frac16(6a+3b+2c)=a+\frac b2+\frac c3.

This matches the integral exactly. So option A is correct. 4. Since this is a single-correct MCQ, we can conclude:

\int_0^1 f(x),dx = \frac16\left{f(0)+f(1)+4f\left(\frac12\right)\right}.

This is also Simpson's rule, which is exact for polynomials up to degree $3$, hence certainly exact for a quadratic.