If $f(x)=a+b x+c x^2$ where $a, b, c \in R$, then $\int_0^1 f(x) d x$ is equal to

$\frac{1}{3}\left\{f(1)+f(0)+2 f\left(\frac{1}{2}\right)\right\}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Understanding +4 | -1

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

where

a, b, c \in R

, then

\int_0^1 f(x) d x

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by evaluating the integral

\int_0^1 f(x) dx

and comparing it with the given options, we find that the correct expression is

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

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