If for all real triplets a-b-c- fx-a-bx-cx2- then -01fx dx is equal to

Step 1: Given informationf(x)=a+bx+cx^2f(0)=a+b(0)+c(0)^20ex0exf(1)=a+b(1)+c(1)^20ex0exf(1)=a+b+c0ex0exf(1/2)=a+b(1/2)+c(1/2)^20ex0exf(1/2)=a+b/2+c/40ex0ex∫_0^1 ...

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Byju's Answer

Standard XII

Mathematics

Continuity of a Function

If for all re...

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

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

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

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

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

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Solution

The correct option is D
$(\frac{1}{6}) (f (0) + f (1) + 4 f (\frac{1}{2}))$

Step 1: Given information
$f (x) = a + bx + {cx}^{2}$
$f (0) = a + b (0) + c {(0)}^{2} f (1) = a + b (1) + c {(1)}^{2} f (1) = a + b + c f (\frac{1}{2}) = a + b (\frac{1}{2}) + c {(\frac{1}{2})}^{2} f (\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4}$
$\int_{0}^{1} f (x) dx$
Step 2: Taking integration for the given equation with respect to x
$\begin{array}{rcl} \int x dx & = & \frac{x^{2}}{2} + c \\ \int_{0}^{1} f (x) dx & = & \int_{0}^{1} (a + bx + {cx}^{2}) dx \\ = & a + \frac{b}{2} + \frac{c}{3} \\ = & \frac{6 a + 3 b + 2 c}{6} \\ = & \frac{1}{6} (6 a + 3 b + 2 c) \end{array}$
we can split, $(6 a + 3 b + 2 c)$ as $(a + (a + b + c) + (4 a + 2 b + c))$ ,
$\int_{0}^{1} f (x) dx = \frac{1}{6} (a + (a + b + c) + (4 a + 2 b + c))$
$\int_{0}^{1} f (x) dx = \frac{1}{6} ((a + (a + b + c) + 4 (a + \frac{b}{2} + \frac{c}{4}))$
$\int_{0}^{1} f (x) dx = \frac{1}{6} (f (0) + f (1) + 4 f (\frac{1}{2})) ∵ a = f (0), a + b + c = f (1), a + (\frac{b}{2}) + (\frac{c}{4})$
Hence, the correct option is option (D).

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If for all real triplets a,b,c,f(x)=a+bx+cx2, then ∫01f(x)dx is equal to

If for all real triplets $(a, b, c), f (x) = a + bx + {cx}^{2}$ , then $\int_{0}^{1} f (x) dx$ is equal to