Let I 1 = 0 x e t x . e - t 2 d t and I 2 = 0 x e - t 2 / 4 d t where x 0 , then the value of I 1 I 2 is

Question

Let $I_{1} = \int_{0}^{x} e^{t x} . e^{- t^{2}} d t$ and $I_{2} = \int_{0}^{x} e^{- t^{2} / 4} d t$ where $x > 0,$ then the value of $\frac{I_{1}}{I_{2}}$ is

Smart Achievers · Accepted Answer

(2)

$I_{1} = \int_{0}^{x} e^{\frac{x^{2}}{4} - {(t - \frac{x}{2})}^{2}} d t = e^{\frac{x^{2}}{4}} \int_{- \frac{x}{2}}^{\frac{x}{2}} e^{- z^{2}} d z t - \frac{x}{2} = z$

$= 2 e^{\frac{x^{2}}{4}} \int_{0}^{\frac{x}{2}} e^{- z^{2}} d z$

$I_{2} = \int_{0}^{x} e^{- \frac{t^{2}}{4} d t}$

$Let \frac{t}{2} = z = \int_{0}^{\frac{x}{2}} e^{- z^{2}} \times 2 d z$

Solution

Q.
Let $I_{1} = \int_{0}^{x} e^{t x} . e^{- t^{2}} d t$ and $I_{2} = \int_{0}^{x} e^{- t^{2} / 4} d t$ where $x > 0,$ then the value of $\frac{I_{1}}{I_{2}}$ is

1 $e^{- x^{2} / 2}$

2 $e^{x^{2} / 4}$

3 $e^{- x^{2} / 4}$

4 $e^{x^{2} / 2}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let I1=∫0xetx.e-t2dt and I2=∫0xe-t2/4dt where x>0, then the value of I1I2 is

1 e-x2/2

2 ex2/4

3 e-x2/4

4 ex2/2