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  • If for all real triplets   <img alt="f(x)=a+bx+cx^{2};\: \: then\: \: \int_{0}^{1}f(x)dx" src

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

 

Option: 1

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


Option: 2

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


Option: 3

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

 


Option: 4

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


Answers (1)

best_answer

 

 

Integration as Reverse Process of Differentiation -

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx. 

By definition,

\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}

Rules of integration:

f(x) and g(x) are functions with antiderivatives ∫ f(x) and  ∫ g(x) dx. Then,

\\ {\text { (a) } \int \operatorname{kf}(x) d x=k \int f(x) d x \text { for any constant } k .} \\ {\text { (b) } \int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x} \\ {\text { (c) } \int(f(x)-g(x)) d x=\int f(x) d x-\int g(x) d x}

\\\mathrm{Since,\;\;\frac{d}{dx}\left ( F(x) \right )=f(x)\;\;\Leftrightarrow \;\;\int f(x)dx=F(x)+c}

Based on this definition and various standard formulas (which we studied in Limit, Continuity and Differentiability) we obtained the following important integration formulae, 

\\\mathrm{1.\;\;\frac{d}{d x}\left({kx}\right)=k \Rightarrow \int kd x=kx+C, \text{ where k is a constant}}\\\\\mathrm{2.\;\;\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^{n}, n \neq-1 \Rightarrow \int x^{n} d x=\frac{x^{n+1}}{n+1}+C, n \neq-1}\\\\\mathrm{3.\;\;\frac{d}{d x}(\log |x|)=\frac{1}{x} \Rightarrow \int \frac{1}{x} d x=\log |x|+C, \text { when } x \neq 0}\\\\\mathrm{4.\;\;\frac{d}{d x}\left(e^{x}\right)=e^{x} \Rightarrow \int e^{x} d x=e^{x}+C}\\\\\mathrm{5.\;\;\frac{d}{d x}\left(\frac{a^{x}}{\log _{e} a}\right)=a^{x}, a>0, a \neq 1\;\;\Rightarrow \int a^{x} d x=\frac{a^{x}}{\log _{e} a}+C}

 

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Definite Integration -

Working Rule to evaluate definite Integral \int_{a}^{b} f(x) d x

\\\mathrm{1.\;\;\;\;First\;find\;the\;indefinite\;integration\;\int f(x)\;dx \;and\;suppose}\\\mathrm{\;\;\;\;\;\;\;the\;result\;be\;F(x)}\\\\\mathrm{2.\;\;\;\;Next\;find\;the\;F(b)\;and\;F(a)}\\\\\mathrm{3.\;\;\;\;And, finally\;value\;of\;definite \;integral\;is\;obtained\;by\;subtracting}\\\mathrm{\;\;\;\;\;\;\;F(a)\;from\;F(b).}\\\mathrm{\;\;\;\;\;\;\text { Thus, } \quad \int_{a}^{b} f(x) d x=[F(x)]_{a}^{b}=F(b)-F(a).}

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

f(1) = a + b + c

f(0) = a

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

Option (4) satisfies the given condition

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Anam Khan

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