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

Correct option is 416f(0)+f1+4f12

Questions

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

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23f(1)+2f12

12f(1)+3f12

13f(0)+f12

16f(0)+f1+4f12

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detailed solution

Correct option is D

∫01a+bx+cx2dx=ax+bx22+cx3301=a+b2+c3 We have f(1)=a+b+c,f(0)=a,f12=a+b2+c4 Now 16f(1)+f(0)+4f12=16a+b+c+a+4a+b2+c4 =16(6a+3b+2c)=a+b2+c3Therefore, it is matching with the option 4

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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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