The integral of 2 is "who gives a $%&#." You need to know what 2 is relating to and what the 2 means. If the question is "What is the integral of 2 dx" the answer would be "2x + c," with c being a constant. If you instead wish to know what the integral of 2 dy, the answer is very different. (2y +c)
First, draw the region/solid being bounded by parameters say: y^2 + z^2 = 9, x = -2, and x = 2 Now analyze what possible iterated integrals can be used to find this region. the two "main" iterated integrals are: the triple integral from [-2,2] [-3,3] [-sqrt(9-y^2),sqrt(9-y^2)] dz dy dx and [-2,2] [-3,3] [-sqrt(9-z^2),sqrt(9-z^2)] dy dz dx Now, instead of sketching every region to find the different possible integrals, using the rules of triple integration, they will essentially be any legal alteration of the order of the "main" integrals. essentially, the first main integral can be rewritten as dx dz dy, and dz dx dy the second can be written as dx dy dz and dy dx dz.
Int = 3x^(2) dy y = 3x^(3) / 3 + c y = x^(3) + C
If x = g(y) ∫ f(x) dx = ∫ f(g(y))g'(y) dy This is called change of variables.
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
it isn't good enough, you forgot to put in the '^'s between the 'x' and 'y' and the 'dy'
It comes from Calculus. You are integrating a function which has some form of y²dy {dy is the height of a tiny horizontal slice, which is the shape of a square which is area y²}. The integral of y²dy is y³/3
dy/dx = 3 integral = (3x^2)/2
The integral of 2 is "who gives a $%&#." You need to know what 2 is relating to and what the 2 means. If the question is "What is the integral of 2 dx" the answer would be "2x + c," with c being a constant. If you instead wish to know what the integral of 2 dy, the answer is very different. (2y +c)
First, draw the region/solid being bounded by parameters say: y^2 + z^2 = 9, x = -2, and x = 2 Now analyze what possible iterated integrals can be used to find this region. the two "main" iterated integrals are: the triple integral from [-2,2] [-3,3] [-sqrt(9-y^2),sqrt(9-y^2)] dz dy dx and [-2,2] [-3,3] [-sqrt(9-z^2),sqrt(9-z^2)] dy dz dx Now, instead of sketching every region to find the different possible integrals, using the rules of triple integration, they will essentially be any legal alteration of the order of the "main" integrals. essentially, the first main integral can be rewritten as dx dz dy, and dz dx dy the second can be written as dx dy dz and dy dx dz.
For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
Int = 3x^(2) dy y = 3x^(3) / 3 + c y = x^(3) + C
We have:int int (x * sin(y)) dx dyIntegrate x first:int(x)dx = 1/2 * x2 + CNow integrate sin(y):int(sin(y))dy = -cos(y) + CMultiply:-1/2 * x2 * cos(y) + C
∫x3ex4 dx = 1/4ex4 + c To solve, let y = x4, then: dy = 4x3 dx ⇒ 1/4dy = x3 dx ⇒ ∫x3ex4 dx =∫ex4 x3 dx = ∫ey 1/4 dy = 1/4ey + c but y = x4, thus: = 1/4ex4 + c
Sixto Dy was born on August 6, 1934, in Tarlac Province, Philippines.
Paolo Dy's birth name is Paolo Crisostomo Arbiol Dy.
he was assinated