∫_-5^5≤ft(3x^5+√5 x^3)e^x^2 dx is equal to :

The correct answer is Option (1) → 0

Let \(f(x) = \left(3x^5 + \sqrt{5}\,x^3\right)e^{x^2}\). Replacing x by −x:

    \(f(-x) = \left(-3x^5 - \sqrt{5}\,x^3\right)e^{x^2}\)
    \(f(-x) = -f(x)\)

So f is an odd function. For an odd function, \(\int_{-a}^{a} f(x)\,dx = 0\), because the areas on the two sides of the y-axis cancel. Hence

    \( \displaystyle \int_{-5}^{5}\left(3x^5 + \sqrt{5}\,x^3\right)e^{x^2}\,dx = 0 \)