If $H(x) = \sin(x^2)$, whose range is $[-1, 1]$. $a = -1, b = 1 \Rightarrow a + 2b = 1$
Step-by-Step Solution
Key Concept: Use the formula $\int e^t(f(t) + f'(t))dt = e^t f(t) + C$ to identify the function and its range.
We know that $\int e^t(f(t) + f'(t))dt = e^t f(t) + C$. Thus, $\int e^x(x^2 + 2x - 2 - 2 + 2)dx = \int e^x(x^2 - 2x + 2) + e^t(2x - 2)dx = e^t(x^2 - 2x + 2) + C$. Since $f(x) = x^2 - 2x + 2 = (x-1)^2 + 1$, the range is $[1, \infty)$. Therefore $a = 1, b = -1$, so $\frac{a}{4} = 0.25$.
Correct Answer: 0.25