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Integration cos3 x?
Take f(x) = cos(3x) ∫ f(x) dx = ∫ cos(3x) dx Take u=3x → du = 3dx = ∫ 1/3*cos(u) du = 1/3*∫ cos(u) du = 1/3*sin(u) + C, C ∈ ℝ = 1/3*sin(3x) + C
How do i solve Y squared equals negative 3 x?
That depends on what you want to "solve" it for. Do you want to plot it on a graph? Are you looking for the value of y with respect to x? The value of x with respect to y? The derivative of each? The integral of each? You need to clarify your question.In answer to all of those interpretations:It describes a horizontally aligned parabola that proceeds indefinitely to the "left" (ie. negative) whose vertex lies at the point 0, 0y = (-3x)1/2dy/dx = (-3x/2)-1/2∫(-3x)1/2 dx = (-x/2)3/2 + Cx = -y2/3dx/dy = -2y/3∫(-y2/3) dy = -y3/9 + C
Math Algebra question If one of the roots of the quadratic equation dx to the 2 plus 5x minus 3 equals 0 is x equals -3 A Determine the value of d and B Determine the other root Please help?
Answers: A) d = 2 B) x = 1/2 Why? We have dx2 + 5x - 3 = 0, and one solution is x = -3. The equation is quadratic, so it can be factored into two linear terms (things that look like ax+b). Since, x = -3 x + 3 = -3 +3 x + 3 = 0 So x + 3 is one of the terms. So dx2 + 5x - 3 = 0 is equivalent to (?x + ?)(x + 3) = 0 The first ? must be d since ?x*x = dx2. The second ? must be -1 since ?*3 = -3. So we have (dx - 1)(x + 3) = 0 When we distribute (students often know as F.O.I.L.), we find that dx2+ 3dx - x - 3 = 0 dx2+ (3d-1)x - 3 = 0 (factor out x, or think of combining like terms) This is the same as: dx2 + 5x - 3 = 0 So we know 3d - 1 = 5, an equation we can solve for d. 3d - 1 = 5 3d - 1 + 1 = 5 + 1 3d = 6 3d/3 = 6/3 d = 2 Yay! Part A of your problem is complete: d = 2. Next, Part B, we must find the other root, or solution. So now we have 2x2 + 5x - 3 = 0 is the same as (2x - 1)(x + 3) = 0 so we know 2x - 1 = 0 2x - 1 + 1 = 0 + 1 2x = 1 2x/2 = 1/2 x = 1/2 So for Part B, we have x = 1/2 is the other root.
