Let F:[3,5] rightarrow R be a twice different | vedclass

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(B) Given $F(x) = e^{-x} \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$. Note that $F(3) = 0$.Multiply by $e^{x}$: $e^{x}F(x) = \int_{3}^{x} (3t^{2}+2t

Vedclass · Accepted Answer

$16$

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(B) Given $F(x) = e^{-x} \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$. Note that $F(3) = 0$.Multiply by $e^{x}$: $e^{x}F(x) = \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$.Differentiating both sides with respect to $x$ using the Leibniz rule:$e^{x}F(x) + e^{x}F^{\prime}(x) = 3x^{2}+2x+4F^{\prime}(x)$.Rearranging terms: $(e^{x}-4)F^{\prime}(x) + e^{x}F(x) = 3x^{2}+2x$.This is a linear differential equation of the form $\frac{d}{dx} [F(x)(e^{x}-4)] = 3x^{2}+2x$.Integrating both sides with respect to $x$: $F(x)(e^{x}-4) = \int (3x^{2}+2x) \,dx = x^{3}+x^{2}+C$.Since $F(3) = 0$,we have $0 = 3^{3}+3^{2}+C \Rightarrow C = -36$.Thus,$F(x) = \frac{x^{3}+x^{2}-36}{e^{x}-4}$.Now,$F^{\prime}(x) = \frac{(3x^{2}+2x)(e^{x}-4) - (x^{3}+x^{2}-36)e^{x}}{(e^{x}-4)^{2}}$.At $x=4$: $F^{\prime}(4) = \frac{(3(16)+2(4))(e^{4}-4) - (64+16-36)e^{4}}{(e^{4}-4)^{2}} = \frac{56(e^{4}-4) - 44e^{4}}{(e^{4}-4)^{2}} = \frac{12e^{4}-224}{(e^{4}-4)^{2}}$.Comparing with $\frac{\alpha e^{\beta}-224}{(e^{\beta}-4)^{2}}$,we get $\alpha=12$ and $\beta=4$.Therefore,$\alpha+\beta = 12+4 = 16$.

Let $F:[3,5] \rightarrow R$ be a twice differentiable function on $(3,5)$ such that $F(x)=e^{-x} \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$. If $F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{(e^{\beta}-4)^{2}}$,then $\alpha+\beta$ is equal to $....$

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