Show that \(\frac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^2+1\right)e^{2x-1}\right]=2e^{2x-1}\lef

Question

Show that $\frac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^2+1\right)e^{2x-1}\right]=2e^{2x-1}\left(x^2+x+1\right)$.[3] Hence, show that $\int -2e^{2x-1}\left(x^2+x\right) \mathrm{d}x = -x^2e^{2x-1}+c$.[4]

Accepted Answer

(a) $\frac{\mathrm d}{\mathrm dx}[(x^2+1)e^{2x-1}]=2e^{2x-1}(x^2+x+1)$. (b) $\int-2e^{2x-1}(x^2+x)\,\mathrm dx=-x^2e^{2x-1}+C$.

TKSS 4AM 2022 Prelims P2 Q8

TKSS 4AM 2022 Prelims P2 Q8

Video Solution:

Solution:

Solution locked