Let f:[0,∞)→R be a continuous and strictly increasing function such that f4(x)=∫0xtf3(t)dt ∀x>0. The area enclosed by y=f(x), X−axis and Ordinate at x=22 is

4f3f1=xf3, 4∫f1(x)=x=∫xdx, 4f(x)=x22+c, c=0, f(x)=x28

Let f:[0,∞)→R be a continuous and strictly increasing function such that f4(x)=∫0xtf3(t)dt ∀x>0. The area enclosed by y=f(x), X−axis and Ordinate at x=22 is

4f3f1=xf3, 4∫f1(x)=x=∫xdx, 4f(x)=x22+c, c=0, f(x)=x28

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Let $f : [0, \infty) \to R$ be a continuous and strictly increasing function such that $f^{4} (x) = \int_{0}^{x} t f^{3} (t) d t \forall x > 0 .$ The area enclosed by $y = f (x), X -$ axis and Ordinate at $x = 2 \sqrt{2}$ is

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$2 \sqrt{2}$

$\frac{2 \sqrt{2}}{5}$

$\frac{2 \sqrt{2}}{3}$

$\frac{8}{3}$

answer is C.

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$4 f^{3} f^{1} = x f^{3}, 4 \int f^{1} (x) = x = \int x d x, 4 f (x) = \frac{x^{2}}{2} + c, c = 0, f (x) = \frac{x^{2}}{8}$

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Let f:[0,∞)→R be a continuous and strictly increasing function such that f4(x)=∫0xtf3(t)dt ∀x>0. The area enclosed by y=f(x), X−axis and Ordinate at x=22 is

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Let $f : [0, \infty) \to R$ be a continuous and strictly increasing function such that $f^{4} (x) = \int_{0}^{x} t f^{3} (t) d t \forall x > 0 .$ The area enclosed by $y = f (x), X -$ axis and Ordinate at $x = 2 \sqrt{2}$ is