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.
-e-x + C.
-1
You can't, unless it's an initial value problem. If f(x) is an antiderivative to g(x), then so is f(x) + c, for any c at all.
For example, the derivate of x2 is 2x; then, an antiderivative of 2x is x2. That is to say, you need to find a function whose derivative is the given function. The antiderivative is also known as the indifinite integral. If you can find an antiderivative for a function, it is fairly easy to find the area under the curve of the original function - i.e., the definite integral.
The general formula for powers doesn't work in this case, because there will be a zero in the denominator. The antiderivative of 1/x is ln(x), that is, the natural logarithm of x.
The antiderivative of 2x is x2.
That means that either the function is equal to zero everywhere (y = 0), or it is the exponential function (y = ex).
35x2
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
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.
(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)
-e-x + C.
It is -exp (-x) + C.