0.5
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
-1/15sin5ycos3y
-1
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
0.5
The integral of 1 + x2 is x + 1/3 x3 + C.
The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.
The integral of -x2 is -1/3 x3 .
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
The indefinite integral of (1/x^2)*dx is -1/x+C.
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
non integral is type of numbers behaviour: i can say that set of numbers without any "holes inside" are integral and set of numbers with "holes inside are non integral. example : integral group "1..100" non integral group "1,4,8,67"
x=1
-1/15sin5ycos3y
The integral of 2-x = 2x - (1/2)x2 + C.