A Calculus Problem

Since $\dfrac{\pi}{4}<1$ we have for $x\in (0,\dfrac{\pi}{4})$ , $\displaystyle e^x > e^{x^2} > e^{x^2}\cos x > e^{x^2}\sin x$ since $\cos x > \sin x$ in $x\in (0,\dfrac{\pi}{4})$ . Now integrate all sides to get $I_2>I_1>I_3>I_4$ .

Lets focus on the limit of the integrals. Clearly x can take values from 0 to $\pi$ /4.

In this interval $\sin \theta$ trails $\cos \theta$ . Thus, $I_{3}$ > $I_{4}$ We have reduced our search for the answer to two options with this.

Now, since $\pi$ /4 < 1, we infer $x^{2}$ < x in this interval. Thus, $I_{2}$ > $I_{1}$

Voila, we have the answer :)

$\pi /4 < 1 \Rightarrow \forall x \in [ 0, \pi/4 ], x > x^2 \Rightarrow \int_{0}^{\pi/4} e^x dx > \int_{0}^{\pi/4} e^{x^2} dx. \\ \forall x \in [ 0, \pi/4 ] \ 0 < cos(x) < 1 \Rightarrow 0 < \int_{0}^{\pi/4} e^{x^2}cos(x) dx < \int_{0}^{\pi/4} e^{x^2} dx \\(cos(x)-sin(x))' = -sin(x) - cos(x) <= 0 \ \forall x \in [ 0, \pi/4 ] \\ \Rightarrow \forall x \in [ 0, \pi/4 ] cos(x)-sin(x) >= cos(\pi/4) - sin(\pi/4) = 0 \\ \Rightarrow cos(x) > sin(x) \Rightarrow \int_{0}^{\pi/4} e^{x^2}cos(x) dx > \int_{0}^{\pi/4} e^{x^2}sin(x) dx$