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(B) Given $10 \int_1^x f(t) dt = 5x f(x) - x^5 - 9$.Differentiating both sides with respect to $x$ using the Leibniz rule:$10 f(x) = 5 f(x) + 5x f'(x)

Vedclass · Accepted Answer

$32$

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(B) Given $10 \int_1^x f(t) dt = 5x f(x) - x^5 - 9$.Differentiating both sides with respect to $x$ using the Leibniz rule:$10 f(x) = 5 f(x) + 5x f'(x) - 5x^4$.Rearranging the terms:$5 f(x) + 5x^4 = 5x f'(x)$$f'(x) - \frac{1}{x} f(x) = x^3$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = x^3$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.Multiplying by $IF$:$\frac{1}{x} f'(x) - \frac{1}{x^2} f(x) = x^2$.Integrating both sides:$\frac{f(x)}{x} = \int x^2 dx = \frac{x^3}{3} + C$.To find $C$,put $x=1$ in the original equation: $10 \int_1^1 f(t) dt = 5(1)f(1) - 1^5 - 9 \Rightarrow 0 = 5f(1) - 10 \Rightarrow f(1) = 2$.Substituting $x=1$ in $\frac{f(x)}{x} = \frac{x^3}{3} + C$:$\frac{2}{1} = \frac{1}{3} + C \Rightarrow C = 2 - \frac{1}{3} = \frac{5}{3}$.Thus,$f(x) = \frac{x^4}{3} + \frac{5x}{3}$.For $x=3$: $f(3) = \frac{3^4}{3} + \frac{5(3)}{3} = \frac{81}{3} + 5 = 27 + 5 = 32$.

Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^{x} f(t) dt = 5x f(x) - x^5 - 9$ for all $x \geq 1$,then the value of $f(3)$ is:

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