Integral Calculus
Definite Integrals
MMTS_Full_Test_11
Grade 12
Question:
Let $I_1=\int_0^x e^{tx}\cdot e^{-t^2}\,dt$ and $I_2=\int_0^x e^{-t^2/4}\,dt$ where $x>0$. Then the value of $\dfrac{I_1}{I_2}$ is
$e^{-x^2/2}$
$e^{x^2/4}$
$e^{-x^2/4}$
$e^{x^2/2}$
Step-by-Step Solution
Key Concept: Complete the square in the exponent of $I_1$
$I_1=e^{x^2/4}\int_0^x e^{-(t-x/2)^2}dt=e^{x^2/4}\cdot I_2$. So $I_1/I_2=e^{x^2/4}$.
Correct Answer: 2