Question

Let $F:[3,5]\rightarrow R$ be a twice differential function in (3,5) such that

$F(x)=e^{-x}\int_{3}^{x}(3t^{2}+2t+4F'(t))dt$. Then $F'(x)=$

• A. $\frac{(3x^2+2x)(e^x-4) - e^x(x^3+x^2-36)}{(e^x-4)^2}$
• B. $\frac{e^x F(x) - 3x^{2} - 2x}{4 - e^x}$
• C. $\frac{3x^2+2x}{e^x-4} - \frac{e^x(x^3+x^2-36)}{(e^x-4)^2}$
• D. $\frac{x^3+x^2-36}{e^x-4}$

Answer 1

Answer: (A)

$\frac{(3x^2+2x)(e^x-4) - e^x(x^3+x^2-36)}{(e^x-4)^2}$

Answer 2

We are given the equation: $F(x)=e^{-x}\int_{3}^{x}(3t^{2}+2t+4F'(t))dt$

Multiply both sides by $e^x$: $e^x F(x) = \int_{3}^{x}(3t^{2}+2t+4F'(t))dt$

Differentiate both sides with respect to $x$. Using the product rule on the left side and the Fundamental Theorem of Calculus on the right side: $\frac{d}{dx}(e^x F(x)) = e^x F(x) + e^x F'(x)$ $\frac{d}{dx}\left(\int_{3}^{x}(3t^{2}+2t+4F'(t))dt\right) = 3x^{2}+2x+4F'(x)$

Equating the derivatives: $e^x F(x) + e^x F'(x) = 3x^{2}+2x+4F'(x)$

To find $F(x)$, we can split the integral: $e^x F(x) = \int_{3}^{x}(3t^{2}+2t)dt + \int_{3}^{x}4F'(t)dt$ $e^x F(x) = [t^3+t^2]_{3}^{x} + 4[F(t)]_{3}^{x}$ $e^x F(x) = (x^3+x^2) - (3^3+3^2) + 4(F(x) - F(3))$

From the original equation, $F(3) = e^{-3} \int_{3}^{3} (\dots) dt = 0$. So, $e^x F(x) = x^3+x^2 - (27+9) + 4F(x)$ $e^x F(x) = x^3+x^2 - 36 + 4F(x)$ $F(x)(e^x - 4) = x^3+x^2 - 36$ $F(x) = \frac{x^3+x^2-36}{e^x-4}$

Now, substitute this $F(x)$ back into the differentiated equation: $e^x \left(\frac{x^3+x^2-36}{e^x-4}\right) + e^x F'(x) = 3x^{2}+2x+4F'(x)$

Rearrange to solve for $F'(x)$: $e^x F'(x) - 4F'(x) = 3x^{2}+2x - e^x \left(\frac{x^3+x^2-36}{e^x-4}\right)$ $F'(x)(e^x - 4) = 3x^{2}+2x - \frac{e^x(x^3+x^2-36)}{e^x-4}$ $F'(x) = \frac{3x^{2}+2x}{e^x-4} - \frac{e^x(x^3+x^2-36)}{(e^x-4)^2}$

To combine these terms, find a common denominator: $F'(x) = \frac{(3x^{2}+2x)(e^x-4)}{(e^x-4)^2} - \frac{e^x(x^3+x^2-36)}{(e^x-4)^2}$ $F'(x) = \frac{(3x^{2}+2x)(e^x-4) - e^x(x^3+x^2-36)}{(e^x-4)^2}$