"Let f(x) and g(x) be two functions satisfying f(x2) + g(4 $-$ x) = 4x3 and g(4 $-$ x) + g(x) = 0, then the value of $\int\limits_{ - 4}^4 {f{{(x)}^2}dx}$ is"

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Accepted Answer

"$I = 2\int\limits_0^4 {f({x^2})dx}$ ............(1)

$\Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx}$ ..............(2)

Adding equation (1) & (2)

$2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx$ ............(3)

Now using $f{(x^2)} + g(4 - x) = 4{x^3}$ ............. (4)

$x \to 4 - x$

$f({(4 - x)^2}) + g(x) = 4{(4 - x)^3}$ ..............(5)

Adding equation (4) & (5)

$f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]$

$\Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]$

Now, $I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512}$"

Let f(x) and g(x) be two functions satisfying f(x²) + g(4 $-$ x) = 4x³ and g(4 $-$ x) + g(x) = 0, then the value of $\int\limits_{ - 4}^4 {f{{(x)}^2}dx}$ is

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Let f(x) and g(x) be two functions satisfying f(x2) + g(4 $-$ x) = 4x3 and g(4 $-$ x) + g(x) = 0, then the value of $\int\limits_{ - 4}^4 {f{{(x)}^2}dx}$ is

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Let f(x) and g(x) be two functions satisfying f(x²) + g(4 $-$ x) = 4x³ and g(4 $-$ x) + g(x) = 0, then the value of $\int\limits_{ - 4}^4 {f{{(x)}^2}dx}$ is