The value of f(1),given the equation int_0^{x | vedclass

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(D) Given the equation: $\int_0^{x^2} x f(t) dt = x^5 - x^3$. Since $x$ is independent of $t$,we can write it as $x \int_0^{x^2} f(t) dt = x^5 - x^3$.

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$1$

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(D) Given the equation: $\int_0^{x^2} x f(t) dt = x^5 - x^3$. Since $x$ is independent of $t$,we can write it as $x \int_0^{x^2} f(t) dt = x^5 - x^3$. Dividing by $x$ (assuming $x 
eq 0$),we get $\int_0^{x^2} f(t) dt = x^4 - x^2$. Differentiating both sides with respect to $x$ using the Leibniz integral rule: $f(x^2) \cdot \frac{d}{dx}(x^2) = \frac{d}{dx}(x^4 - x^2)$. $f(x^2) \cdot (2x) = 4x^3 - 2x$. $f(x^2) = \frac{4x^3 - 2x}{2x} = 2x^2 - 1$. To find $f(1)$,we set $x^2 = 1$,which implies $x = 1$. Substituting $x = 1$ into the expression for $f(x^2)$: $f(1) = 2(1)^2 - 1 = 2 - 1 = 1$. Thus,the value of $f(1)$ is $1$.

The value of $f(1)$,given the equation $\int_0^{x^2} x f(t) dt = x^5 - x^3$ is

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