If $I = \int_3^4 \frac{1}{\sqrt[3]{\ln x}} dx$, then:
Step-by-Step Solution
Key Concept: Comparing integrals by analyzing the monotonicity of exponents and bounding factors on $(0,1)$ identifies the largest integral.
For $x > e$, we use $1 -x$, so $e^{-x^2} > e^{-x}$. Multiplying by $\cos^2 x \leq 1$ gives $\int_0^1 e^{-x^2}\cos^2 x dx < \int_0^1 e^{-x}dx$. Also, $\int_0^1 e^{-x^2}\cos^2 x dx < \int_0^1 e^{-x^2}dx = 1$ since $\cos^2 x \leq 1$. Both $\int_0^1 e^{-x^2}dx = 1$ and $\int_0^1 e^{-1/2 x^2}dx = 1$ are equivalent, making $I_4$ the greatest.
Correct Answer: 1,2,3