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
It os 1/2*e2x + c
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)
Integration by Parts is a special method of integration that is often useful when two functions.
Assuming integration is with respect to a variable, x, the answer is 34x + c where c is the constant of integration.
It os 1/2*e2x + c
Derivative of e2x - y = 2e2x - dy/dx
3
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.
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.
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)
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
INTEGRATION
Given the limitations of the browser that is used by this site, it is difficult to be certain about your question. If I misunderstand your question, (a) please accept my apologies and (b) resubmit your question spelling out the symbols as "plus", "minus", "times", "equals", "squared", "cubed" etc. I am assuming that you wish to solve e2x = 3x2 The only way that I can think of solving this is numerically and, in that case, the Newton Raphson method is quick and easy (for differentiable functions). To start with, your question is equivalent to finding a solution to e2x - 3x2 = 0, so define f(x) = e2x - 3x2. Therefore f'(x) = 2e2x - 6x. Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f'(xn) for n = 0, 1, 2, … Continue until you get convergence of xn. If you plot a graph of f(x) against x, you will know that the root of f(x) lies between -1 and 0. This helps to select a suitable value for x0. If you start with -1, your error is around 8 per billion at x4. If you start with 0, it is less than 1 in 1.5 billion at x4. The answer is -0.390646381 (approx).