0
Add your answer:
Earn +20 pts
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
How do you take the antiderivative of 1 over x?
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.
What is 4 times 4 5ths?
5/6 x 4/5 is 2/3.
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 5 6ths times 4 5ths?
5/6 x 4/5 is 2/3.
What is the antiderivative of x to the -1?
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
What is the antiderivative of e to the -x?
-e-x + C.
What is the antiderivative of exp -x?
It is -exp (-x) + C.
How do you take the antiderivative of 1 over x?
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.
How do you get the answer of three wholes times 2 5ths?
3 x 2/5 = 6/5
What is 4 times 4 5ths?
5/6 x 4/5 is 2/3.
What is the antiderivative 7x5-Cos x?
If the first term is 7x^5, ∫7x^5 -cox dx is the expression. You can split this up into two integrals if that helps you visualize the terms. ∫7x^5dx - ∫cox dx. We know that the antiderivative of cosx is sinx, so that is our second term. In the first term, we must undo the power rule, adding one to the power and multiplying by the reciprocal of the power. This gives us (7/6)x^6. So, our final antiderivative expression is (7/6)x^6-sinx+C, with C being an arbitrary constant.
What is the antiderivative of logx?
X(logX-1) + C
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 square root of x?
(2/3)*x^(3/2)
What is 5 6ths times 4 5ths?
5/6 x 4/5 is 2/3.
