P2 38

INTEGER ANS TYPE QS

1) Let F (x) be a non- negative continuous function defined on R such that F (x) + F ( x+1/2) =

3

and the value of ∫[0--->1500] F(x)dx is 9000/a . Then the numerical value of ' a ' is .

6 Answers

1
rahul nair ·

4?

4
UTTARA ·

Ya Post ure method plzz

1
rahul nair ·

On observation,f(x) has to be a constant fn.
f(x)=c.

2c=3,c=1.5
\int_{0}^{1500}{1.5}dx=2250

4
UTTARA ·

THANKS

62
Lokesh Verma ·

a more accurate solution will be that \int_{0}^{1}{F(x)dx}=\int_{0}^{1/2}{[F(x)+F(x+1/2)]dx}

Which gives us that \\\int_{0}^{1}{F(x)dx}=\int_{0}^{1/2}F(x)dx+\int_{1/2}^{1}F(x+1/2)]dx\\=\int_{0}^{1/2}[F(x)+F(x+1/2)]dx=\int_{0}^{1/2}3dx=3/2

This can be done between any two integers using the logic above.. \\\int_{n}^{n+1}{F(x)dx}=\int_{n}^{n+1/2}F(x)dx+\int_{n+1/2}^{n+1}F(x+1/2)]dx\\=\int_{n}^{n+1/2}[F(x)+F(x+1/2)]dx=\int_{n}^{n+1/2}3dx=3/2

Hence the answer for the whole integral is 2250

4
UTTARA ·

Thanks sir

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