It os 1/2*e2x + c
Original function: f(x) = e2x Integrated function: F(x) = e2x/2 Evidence: F'(x) --> f(x) = e2x ; F'(x) = 2e2x/2 = e2x = f(x) Q.E.D
y = e2x+1
A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)
You integrate each element of the matrix.
I assume you mean e2x. I tried this at the Wolframalpha site (using the search term: "integrate e^(2*x)"; the site says it is e2x/2. This can be obtained by a simple substitution.
It os 1/2*e2x + c
Derivative of e2x - y = 2e2x - dy/dx
Original function: f(x) = e2x Integrated function: F(x) = e2x/2 Evidence: F'(x) --> f(x) = e2x ; F'(x) = 2e2x/2 = e2x = f(x) Q.E.D
3
y = e2x+1
coth(x) = cosh(x)/sinh(x) = (ex + e-x)/ (ex - e-x) or (e2x + 1)/ (e2x - 1)
If you mean ex squared, the answer is e2x
e2x + ex = 0e2x = -exe2x / ex = -1ex = -1x = ln(-1)Actually, that's just a restatement of the original problem, but that'sas far as you can go with it, because there's no such thing as the log(or the ln) of a negative number. Your equation has no solution.
This person just asked me to integrate the word integrate into a sentence.
A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)
You integrate each element of the matrix.