Well, darling, the antiderivative of 70x is 35x^2 + C, where C is the constant of integration. It's as simple as that, honey. Just integrate like there's no tomorrow and don't forget that little "+ C" at the end.
252-70x+49 = (5x-7)(5x-7) when factored
I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
The antiderivative of a constant, such as (8\pi), is found by multiplying the constant by the variable of integration and adding a constant of integration (C). Thus, the antiderivative of (8\pi) with respect to (x) is (8\pi x + C), where (C) represents any constant.
The antiderivative of 2x is x2.
252-70x+49 = (5x-7)(5x-7) when factored
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
Yes because 25x2-70x+49 = (5x-7)(5x-7) when factored
Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.
I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
The fundamental theorum of calculus states that a definite integral from a to b is equivalent to the antiderivative's expression of b minus the antiderivative expression of a.
Yes.
It is -exp (-x) + C.