Comparison Testj

[pliant sphinx]

Solutions neededj

what are the solution for the above twoj

j

[vernal minnow]

compare with the integrals of simple exponentials sir

[pliant sphinx]

j

howj

j

[serene tapir]

for a), use $$\dfrac{e^{-x}}{\ln x}\le e^{-x}$$ for $x\ge e$ because we have $\ln x\ge 1$ for $x\ge e$.

for b), use $$e^{x+x^2+x^3}\ge e^x$$ for $x\ge 0$.

[gentle spear]

Kevin S