evaluate (integral) xe^2x dx
A. 1/6x^2 e^3x+C
B. 1/2xe^2x-1/2e^2x+C
C. 1/2xe^2x-1/4e^2x+C
D. 1/2x^2-1/8e^4x+C
Evaluate (integral) e^3x cosh 2x dx
A. 1/2e^5x + 1/2e^x +C
B. 1/10e^5x + 1/2e^x + C
C. 1/4e^3x + 1/2x + C
D. 1/10e^5x + 1/5x + C
e^3x cosh 2x dx
cosh 2x = (1/2)(e^2x + e^-2x)
so
(1/2)(e^3x)(e^2x + e^-2x)dx
= (1/2)(e^5x + e^x)dx
(1/10) e^5x + (1/2) e^x + c
To evaluate the integrals, we will use integration techniques. Let's start with the first integral:
∫ xe^(2x) dx
We can solve this integral using integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
In this case, we can set:
u = x (differentiable)
du = dx
dv = e^(2x) dx
v = 1/2 e^(2x)
Using the formula for integration by parts, we have:
∫ xe^(2x) dx = 1/2 x e^(2x) - ∫ 1/2 e^(2x) dx
The integral on the right-hand side is easy to evaluate:
∫ 1/2 e^(2x) dx = 1/2 ∫ e^(2x) dx = 1/2 * (1/2) e^(2x) = 1/4 e^(2x)
Substituting this back into the equation, we have:
∫ xe^(2x) dx = 1/2 x e^(2x) - 1/4 e^(2x) + C
So the correct option is A. 1/6x^2 e^3x+C.
Now, let's move on to the second integral:
∫ e^(3x) cosh(2x) dx
To solve this integral, we can use integration by substitution. The first step is to identify the function to substitute. In this case, let's choose u = e^(3x). Now, we need to find du/dx and solve for dx:
du/dx = 3e^(3x)
dx = du / (3e^(3x))
Substituting into the integral, we have:
∫ e^(3x) cosh(2x) dx = ∫ u * cosh(2x) * (du / (3e^(3x)))
Next, we can simplify the expression by canceling out the e^(3x):
∫ u * cosh(2x) * (du / (3e^(3x))) = 1/3 ∫ u * cosh(2x) du
Now, we can integrate with respect to u:
1/3 ∫ u * cosh(2x) du = 1/3 ∫ u * (e^(2x) + e^(-2x)) du
Breaking the integral into two terms, we have:
1/3 ∫ u * (e^(2x) + e^(-2x)) du = 1/3 ∫ u * e^(2x) du + 1/3 ∫ u * e^(-2x) du
Now, we can integrate each term separately:
1/3 ∫ u * e^(2x) du = 1/6 ∫ (2x) e^(2x) du (applying integration by parts)
= 1/6 * [(2x) * e^(2x) - ∫ e^(2x) dx]
= 1/6 * [(2x) * e^(2x) - 1/2 * e^(2x)] + C1
= 1/6 * [(4x - 1) * e^(2x)] + C1
Similarly,
1/3 ∫ u * e^(-2x) du = -1/6 * [(4x + 1) * e^(-2x)] + C2
Combining both terms, we get:
∫ e^(3x) cosh(2x) dx = 1/6 * [(4x - 1) * e^(2x) - (4x + 1) * e^(-2x)] + C
Simplifying further, we have:
∫ e^(3x) cosh(2x) dx = 1/6 * [(4x - 1) * e^(2x) - (4x + 1) * e^(-2x)] + C
= 1/6 * [(4x - 1) * e^(2x) - (4x + 1) / e^(2x)] + C
= 1/6 * [(4x * e^(2x) - e^(2x) - 4x / e^(2x) - 1)] + C
= 1/6 * [(4x * e^(2x) - 4x / e^(2x)) - (e^(2x) + 1)] + C
= 1/6 * [4x * (e^(2x) - 1 / e^(2x)) - (e^(2x) + 1)] + C
So the correct option is D. 1/2x^2 - 1/8e^4x + C
(1/2) (2 x e^2x dx)
let z = 2 x
then dz = 2 dx or dx = dz/2
so
(1/2) z e^z (dz/2)
(1/4) z e^z dz
but
int z e^z dz = e^z (z-1) + c
so
(1/4)e^z(z-1) + c
(1/4) e^2x (2x-1) + c
(1/2) x e^2x - (1/4) e^2x + c
so c