One can use integration by parts to solve this. The answer is (x-1)e^x.
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
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.
The antiderivative of the function (200 e^{-0.01x}) can be found using the formula for the antiderivative of an exponential function. The result is given by: [ \int 200 e^{-0.01x} , dx = -\frac{200}{0.01} e^{-0.01x} + C = -20000 e^{-0.01x} + C ] where (C) is the constant of integration.
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 antiderivative of 2x is x2.
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The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
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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.
Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.
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.