0
int(- X2 dx)
= (- 1/3)X3 + C
================or
= - X3/3 + C
===================I like to pull the fraction out front. Add 1 to the exponent and then divide by that number. The power rule.
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What is the antiderivative of 2x?
The antiderivative of 2x is x2.
What is the anti-derivative of x2 1x?
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
How integration is reverse process of differentiation?
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.
What is the antiderivative of 2x minus 4?
x2-2x+C, where C is some arbitrary constant.
What is the anti derivative of the square root of 1-x2?
-1
What is the anti derivative of x divided by x squared minus 1?
The antiderivative of x/(x2-1) is ln(x2-1)/2. Proof: (ln(x2-1)/2)' = (1/(x2-1))*(x2-1)'/2=1/(x2-1)*(2x/2)=x/(x2-1).
How do you solve g x equals -3x plus 1?
If: x = -3x+1 Then: x+3x = 1 => 4x =1 So: x = 1/4 or 0.25 ----------- I notice that the question requests a solution for g x = -3x + 1. It seems possible that parentheses around the 'x' after the 'g' have gone missing, along with a prime indicating the derivative of the function g. This being the case, we would be seeking the antiderivative of -3x + 1. The antiderivative of a sum is the sum of the antiderivatives. So we can look at -3x and +1 separately. The derivative of x2 is 2x. Therefore, the antiderivative of x is x2/2, and the antiderivative of -3x is -3x2/2. The antiderivative of 1 is x. Overall, the solution is the antiderivative -3x2/2 + x + C, where C is an arbitrary constant.
What is the antiderivative of e to the power of one divided by -x?
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
How do you take the integral from negative infinity to 4 of 1 over1 plus x2?
Recall that the antiderivative of 1/(1+x2) is arctan(x). arctan(negative infinity) = -pi/2. arctan(4) = approximately 1.325818. The answer then is arctan(4) - (pi/2) = approximately -0.244979
What is the antiderivative of zero?
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
How do you Integrate the following S4x3-2x2 x-1dx in Calculus?
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C
What is the antiderivative of 70x?
35x2
