1) evaluate
integral of xe^3x dx
2) evaluate integral form 1 to 0 of 4e^2x dx
I just need help finding the anitderivative. Would it be 2e^2x?
u = 3x
dx = (1/3) du
(1/3) u e^u * (1/3) du
= (1/9) u e^u du
= (1/9) [ ue^u - e^u]
= (1/9) [ 3x e^(3x) - e^(3x) ]
d/dx ( 2e^(2x) ) = 4 e^(2x) so yes
for number 2 how did you get 2e^2x?
To evaluate the integral of xe^3x dx, you can use the technique of integration by parts. The formula for integration by parts is:
∫ u dv = uv − ∫ v du
In this case, let's choose u = x and dv = e^3x dx. This implies du = dx and v = ∫ e^3x dx. To find v, you can use the rule of integration for e^cx, which is ∫ e^cx dx = (1/c) e^cx.
So, v = ∫ e^3x dx = (1/3) e^3x.
Next, using the formula for integration by parts, you have:
∫ xe^3x dx = uv − ∫ v du
= x * (1/3)e^3x - ∫ (1/3)e^3x dx
= (1/3)xe^3x - (1/9)e^3x + C
So the antiderivative of xe^3x is (1/3)xe^3x - (1/9)e^3x + C, where C is the constant of integration.
Now, let's move on to the second question. To evaluate the integral from 1 to 0 of 4e^2x dx, you first need to find the antiderivative of 4e^2x.
The antiderivative of 4e^2x is obtained by integrating term by term. Since the integral of e^cx is (1/c) e^cx, you have:
∫ 4e^2x dx = 4 ∫ e^2x dx = 4 * (1/2) e^2x = 2e^2x + C
So, yes, the antiderivative of 4e^2x is indeed 2e^2x + C.
I hope this helps! Let me know if you have any further questions.