Integral sign with lower limit of 0 and upper limit of 1
x^2e^2x dx
So I found u = e^2x v=1/2(x)^3
du=2e^2x dv= x^2 dx
answer is 1/3(x^3e^2x)-1/12(x^4)(2e^2x) +c.
NOT sure what to do with the limits.
Do I have this correct above??
thanks again
Integral sign with lower limit of 0 and upper limit of 1
x^2e^2x dx
So I found u = e^2x v=1/2(x)^3
du=2e^2x dv= x^2 dx
answer is 1/3(x^3e^2x)-1/12(x^4)(2e^2x) +c.
NOT sure what to do with the limits.
Do I have this correct above??
thanks again
You need to swap your u and v around. You'll end up in a loop if you continue to set $\displaystyle u = e^{2x}$
$\displaystyle u = x^2 \Leftrightarrow u' = 2x$
$\displaystyle v' = e^{2x} \Leftrightarrow v = \dfrac{1}{2}e^{2x}$
$\displaystyle \dfrac{1}{2}x^2e^{2x} - \int \dfrac{1}{2}e^{2x} \cdot 2x = \dfrac{1}{2}x^2e^{2x} - \int xe^{2x}$
Now do integration by parts again.
Yes, the chain rule: If u= 2x, then $\displaystyle f(x)= e^{2x}= f(u)= e^u$
$\displaystyle \frac{df}{dx}= \frac{df}{du}{\frac{du}{dx}= e^{u}(2)= 2e^{2x}$
But the integral goes the other way. To integrate $\displaystyle \int e^{2x}dx$, let u= 2x so that du= 2dx, $\displaystyle dx= \frac{1}{2}du$ so the integral becomes $\displaystyle \frac{1}{2}\int e^u du= \frac{1}{2}e^u+ C= \frac{1}{2}e^{2x}+ C$.
Differentiating $\displaystyle e^{2x}$ requires, as you say, a multiplication by 2. Integration, being the inverse of differentiation, requires a division by 2.