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
Yes, the chain rule: If u= 2x, then
But the integral goes the other way. To integrate , let u= 2x so that du= 2dx, so the integral becomes .
Differentiating requires, as you say, a multiplication by 2. Integration, being the inverse of differentiation, requires a division by 2.