Euler's constant, e, has some basic rules when used in conjunction with logs. e raised to x?æln(y),?æby rule is equal to (e raised to ln(y) raised to x). e raised to ln (y) is equal to just y. Thus it becomes equal to y when x = 1 or 0.
e-2 = 1/e2 ≈ 0.1353
ln is the inverse of e. So the e and the ln cancel each other out and you are left with 2. eln2 = 2
The integral of esec(x) dx is not a function that may be expressed in terms of well-studied mathematical functions, elementary or nonelementary. In general, it must be evaluated by numerical methods.
e2ln2 + 2 = 22 + 2 = 4 + 2 = 6
e raised to the 0 power is 1
e^x/1-e^x
int[e(2X) +e(- 2X)] integrate term by term 1/22 e(2X) - 1/22 e(- 2X) + C (1/4)e(2X) - (1/4)e(- 2X) + C ====================
e 2x = (1/2) e 2x + C ============
Well the number e, raised to 6 (e^6) is just a number (a constant), so you integrate a constant times dx gives you that constant times x + C --> x*e^6 + C
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
The first derivative of e to the x power is e to the power of x.
e-2x^2 cannot be integrated, only approximated unless there is an additional x attached to the front of the e, otherwise this function is not integrable Actually, it can be integrated, but it requires multivariable calculus and a conversion from cartesian to polar form to do so.
It is possible.
1.e+54
Y = ex(x + 2) Y = ex/(X + 2) =========
= inegrate (e-2x) / derivate -2x = (e-2x)/-2-> integral esomething = esomething , that's why (e-2x) don't change-> (-2x)' = -2