Hi there!

To find the indefinite integral, we must integrate by parts.
Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":
u = 4x
du = 4
dv = cos(2 - 3x)
v = 1/3sin(2 - 3x)
Write into the format:
∫udv = uv - ∫vdu
Thus, utilize the solved for expressions above:
4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx
Simplify:
-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx
Integrate the integral:
∫4/3(sin(2 - 3x)dx
u = 2 - 3x
du = -3dx ⇒ -1/3du = dx
-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C
Combine:

In the given statement above, in this case, the answer would be TRUE. It is true that the inequality x + 2y ≥ 3 is satisfied by point (1, 1). In order to prove this, we just have to plug in the values. 1 + 2(1) <span> ≥ 3
So the result is 1 + 2 </span> ≥ 3. 3 <span> ≥ 3, which makes it true, because it states that it is "more than or equal to", therefore, our answer is true. Hope this answer helps.</span>
Answer: 0.970 ??
Step-by-step explanation: