Consider I1=∫0π/4ex2dx, I2=∫0π/4exdx, I3=∫0π/4ex2cos x dx, I4=∫0π/4ex2sin x dx. Then which of the following is true

x>x2,∀x∈(0,π4)⇒ex>ex2∀x∈(0,π4)cos x>sin x ∀x∈(0,π4)⇒ex2cos x>ex2 sin x⇒ex>ex2>ex2cos x>ex2sin x∀∈ (0,π4)⇒I2>I1>I3>I4

Consider I1=∫0π/4ex2dx, I2=∫0π/4exdx, I3=∫0π/4ex2cos x dx, I4=∫0π/4ex2sin x dx. Then which of the following is true

x>x2,∀x∈(0,π4)⇒ex>ex2∀x∈(0,π4)cos x>sin x ∀x∈(0,π4)⇒ex2cos x>ex2 sin x⇒ex>ex2>ex2cos x>ex2sin x∀∈ (0,π4)⇒I2>I1>I3>I4

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Consider $I_{1} = \int_{0}^{π / 4} e^{x^{2}} d x, I_{2} = \int_{0}^{π / 4} e^{x} d x, I_{3} = \int_{0}^{π / 4} e^{x^{2}} \cos x d x, I_{4} = \int_{0}^{π / 4} e^{x^{2}} \sin x d x$ . Then which of the following is true

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$I_{1} > I_{2} > I_{3} > I_{4}$

$I_{3} > I_{1} > I_{2} > I_{4}$

$I_{2} > I_{1} > I_{3} > I_{4}$

$I_{4} > I_{1} > I_{3} > I_{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} x > x^{2}, \forall x \in (0, \frac{π}{4}) \Rightarrow e^{x} > e^{x^{2}} \forall x \in (0, \frac{π}{4}) \\ \cos x > \sin x \forall x \in (0, \frac{π}{4}) \\ \Rightarrow e^{x^{2}} \cos x > e^{x^{2}} \sin x \\ \Rightarrow e^{x} > e^{x^{2}} > e^{x^{2}} \cos x > e^{x^{2}} \sin x \forall \in (0, \frac{π}{4}) \\ \Rightarrow I_{2} > I_{1} > I_{3} > I_{4} \end{array}$