Integral Calculus Question 255
Question: $ \int_{{}}^{{}}{( 1+x+\frac{x^{2}}{2\ !}+\frac{x^{3}}{3\ !}+………. )\ dx=} $
Options:
A) $ -e^{x}+c $
B) $ e^{x}+c $
C) $ {e^{-x}}+c $
D) $ -{e^{-x}}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{( 1+x+\frac{x^{2}}{2,!}+\frac{x^{3}}{3,!}+……. )dx=\int_{{}}^{{}}{e^{x}dx=e^{x}+c.}} $
BETA
