The integral of x^2:
f(x)= A . x^n
f´(x)= A . n . x^(n-1) = x^2
Hence, n must be 3 for (n-1)=2 which automatically means that A must be 1/2 in order for A . n= 1
Thus, f(x)=0,5 . x^3 (antiderivative)
0,5e . x^3
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
3
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
∫ (x2+3) = ∫x2 + ∫3(1/3)X3 + 3X + C
X/1 is just X. so (1/2)X2 + C or X2/2 + C
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
I suppose you mean of e-x? It is -e-x + C.
2x
The anti-derivative of sqrt(x) : sqrt(x)=x^(1/2) The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2) The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x) The anti-derivative of -sin(x) is cos(x) Adding, the anti-derivative is (2/3) x^(3/2) + 4 e^x + cos(x) + C
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).
2x is the first derivative of x2.
2x is the first derivative of x2.