If for all real triplets a , b , c , f x = a + b x + c x 2 ; then ∫ 0 1 f x d x is equal to:

∫ 0 1 a + b x + c x 2 d x = a x + b x 2 2 + c x 3 3 0 1 = a + b 2 + c 3 f 1 = a + b + c f 0 = a f 1 2 = a + b 2 + C 4 Now 1 6 f 1 + f 0 + 4 f 1 2 = 1 6 a + b + c + a + 4 a + b 2 + c 4 = 1 6 6 a + 3 b + 2 c = a + b 2 + c 3

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationGeneralHard2 minPYQ_2020

MathematicsHardsingle choice

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:

Options:

Answer:

Solution:

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

$f (1) = a + b + c$

$f (0) = a$

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

Now

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

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:General

⏱ 2mℹ️ Source: PYQ_2020

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