Sneaky Integral

This integral can be done by parts, but I didn't call it "sneaky" for nothing!

Take a look at the polynomial: $x^3+3x^2+4x+4.$ Rearrange it so it is $x^3+4x+3x^2+4$ and let $u=x^3+4x.$ Note that $u'=3x^2+4.$ Adding $u$ and $u'$ gives the original polynomial, so we can say this. $\int_0^2e^x(x^3+3x^2+4x+4)\text{ }dx=\int_0^2e^x(u+u')\text{ }dx$ Let's see if there is an easy way to find the antiderivative of $e^x(u+u').$ Let $v=e^x.$ Note that $v'=e^x.$ We are now trying to find the antiderivative of $uv+u'v.$ Knowing that $v=v'\text{,}$ we can say we are trying to find the antiderivative of $uv'+u'v.$ This, by the Product Rule, is equal to $(uv)'.$ Substituting $u$ and $v$ back in, we can say this. $\int e^x(x^3+3x^2+4x+4)\text{ }dx=e^x(x^3+4x)+C$ All we need to now is plug in $0$ and $2.$ $\left.e^x(x^3+4x)\right]^2_0 =e^2(8+8)-e^0(0+0) =16e^2$ $N=16e^2\approx118.22489\ldots\text{,}$ so $\lfloor N\rfloor=\boxed{118}$

Let $g(x)= e^x f(x)$

Hence $g'(x)=e^x(f(x)+f'(x))$

Now in our case $f(x)=x^3+4x$

Hence $\int_0^2 e^x(x^3+3x^2+4x+4) dx=(e^x(x^3+4x))\vert _0^2=16e^2$