If int e^x(x^3+x^2-x+4) dx = e^x f(x) + c, th | vedclass

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(D) We know that $\int e^x(g(x) + g'(x)) dx = e^x g(x) + c$. Given $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$, we have $f(x) + f'(x) = x^3+x^2-x+4$. Le

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$3$

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(D) We know that $\int e^x(g(x) + g'(x)) dx = e^x g(x) + c$. Given $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$, we have $f(x) + f'(x) = x^3+x^2-x+4$. Let $f(x) = ax^3 + bx^2 + cx + d$. Then $f'(x) = 3ax^2 + 2bx + c$. Substituting these into the equation: $ax^3 + (a+b)x^2 + (b+c)x + (c+d) = x^3 + x^2 - x + 4$. Comparing coefficients: $a = 1$. $a + b = 1 \Rightarrow 1 + b = 1 \Rightarrow b = 0$. $b + c = -1 \Rightarrow 0 + c = -1 \Rightarrow c = -1$. $c + d = 4 \Rightarrow -1 + d = 4 \Rightarrow d = 5$. Thus, $f(x) = x^3 - x + 5$. Calculating $f(1) = 1^3 - 1 + 5 = 5$. Wait, let us re-evaluate the derivative approach: If $f(x) = x^3 - 2x^2 + 3x + 1$, then $f'(x) = 3x^2 - 4x + 3$. $f(x) + f'(x) = x^3 - 2x^2 + 3x + 1 + 3x^2 - 4x + 3 = x^3 + x^2 - x + 4$. This matches the integrand. Therefore, $f(1) = 1^3 - 2(1)^2 + 3(1) + 1 = 1 - 2 + 3 + 1 = 3$.

If $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$, then $f(1) =$

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