Apparently that can't be solved with a finite number of so-called "elementary functions". You can get the beginning of the series expansion here: http://www.wolframalpha.com/input/?i=integrate+x^x
Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C
integrate of x is 1/2x^2. integrate of 1 is x
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
For a discrete probability distribution, you add up x*P(x) over all possible values of x, where P(x) is the probability that the random variable X takes the value x. For a continuous distribution you need to integrate x*P(x) with respect to x.
root x=x^(1/2) and 4x =4 x^1 you add the exponents then integrate as usual. The answer you should get is 4.
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
In calculus, "to integrate" means to find the indefinite integrals of a particular function with respect to a certain variable using an operation called "integration". Synonyms for indefinite integrals are "primitives" and "antiderivatives". To integrate a function is the opposite of differentiating a function.
The integral of cot(x)dx is ln|sin(x)| + C
x-1 = 1/x ∫1/x dx = ln x + C
d/dx âˆ« f(x) dx = f(x)
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
Assuming you are integrating with respect to one of the three variables, you integrate normally. For example: ∫(x+y+z)dx = ∫ x dx + ∫ y dx + ∫ z dx (Integral of the sum is the sum of the integrals) = x^2/x + yx + zx + C Or a harder one: ∫ (sin^2(y)+sqrt(z))/x dx = (sin^2(y) + sqrt(z))*∫ 1/x dx (Factor out constants) = ln(x)*(sin^2(y) + sqrt(z)) tl;dr: just do it normally with normal integration rules
The integral of X 4Y X 8Y 2 With respect to X is 2ln(10/9).
partial of u with respect to x = partial of v with respect to y partial of u with respect to y = -1*partial of v with respect to x
I can't integrate a-x /x-3 ?
Well the number e, raised to 6 (e^6) is just a number (a constant), so you integrate a constant times dx gives you that constant times x + C --> x*e^6 + C
When you with respect to the x-axis then this is like saying with reference to the x-axis. You are using the x-axis as a guide.