1) y=4-2x ⇒ dx= -0,5dy
-1/2· ∫∛y dy = -1/2 ·∫y^1/3 dy = -0,5·[y^(1/3+1)]/(1/3+1) =
=-2/3·∛y^4= -2/3 · ∛[2(2-x)]^4= -4/3·(2-x)·∛(4-2x)
2) A=5/2 ·∫(7x²-1)^(-1/2) dx² = 5/14·∫(7x²-1)^(-1/2) d(7x²-1) =
=5/14·[(7x²-1)^(-1/2+1)]/(-1/2+1) = 5/7· √(7x²-1)
3 A= ∫d(2x)/[(2x)² -(√3)²] =
извините ... мне пора ...