$\int\limits^{5}_{-5}(3x^5+\sqrt{5}x^3).e^{x^2}dx$ is equal to :
Answer & explanation
Correct answer: option 1
Given integral:
$\int_{-5}^{5}\left(3x^5+\sqrt{5}\,x^3\right)e^{x^2}\,dx$
$e^{x^2}$ is an even function, while $3x^5$ and $x^3$ are odd functions.
Product of an odd function and an even function is an odd function.
The integral of an odd function over symmetric limits $[-a,a]$ is zero.
final answer: The value of the integral is 0