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To integrate e^(-2x)dx, you need to take a u substitution.
u=-2x
du=-2dx
Since the original integral does not have a -2 in it, you need to divide to get the dx alone.
-(1/2)du=dx
Since the integral of e^x is still e^x, you get:
y = -(1/2)e^(-2x)
Well, that was one method. I usually solve easier functions like this by thinking how the function looked like before it was differentiated. I let f(x) stand for the given function and F(x) stand for the primitive function; the function we had before differentiation (the integrated function).
f(x)= e-2x <-- our given function
F(x)= e-2x/-2 <-- our integrated function
Evidence: F'(x)= -2e-2x/-2 = e-2x = f(x) Q.E.D
It's as simple as that.
What else can I help you with?
How do you integrate x power x?
e^x/1-e^x
How do you differentiate cos squared x?
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
How do you solve 3e to the power 2x -1 equals 8?
you need to know natural logarithms3e to the 2x-1 power = 8(2x-1) ln e = ln (8/3)ln e = 1(2x-1) = ln(8/3) = 0.982x = 1.98x = 0.99
What is the second derivative of xsqaured e to the power of x?
f(x) = (x^2)(e^x)f'(x) = e^x((x^2)+2x) - i thinkf"(x) = ?--------f(x) = (x^2)(e^x)apply the power rulef'(x) = (x^2)(e^x) + (2x)(e^x)apply the power rule to the first part and apply the power rule to the second part, then add those togetherf''(x) = [(x^2)(e^x) + (2x)(e^x)] + [(2x)(e^x) + (2)(e^x)]simplifyf''(x) = (e^x)(x^2 + 4x +2)I got it right. It checked out on my calculator.
Integrate of ln x squared?
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
How do you integrate e 2x?
e 2x = (1/2) e 2x + C ============
How do you integrate of e power 2x plus e power minus 2x?
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 ====================
How do you integrate e-2x?
= inegrate (e-2x) / derivate -2x = (e-2x)/-2-> integral esomething = esomething , that's why (e-2x) don't change-> (-2x)' = -2
How do you integrate e powerintegral2x-1?
2x
How do you integrate of e -2x for x0?
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
How do you integrate e power minus 2x?
∫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
How do you intergrate ex2?
To integrate ( e^{2x} ), you use the basic rule for integrating exponential functions. The integral is given by: [ \int e^{2x} , dx = \frac{1}{2} e^{2x} + C ] where ( C ) is the constant of integration. This result comes from the fact that the derivative of ( e^{2x} ) is ( 2e^{2x} ), so you divide by the coefficient of ( x ) in the exponent during integration.
How do you integrate functions?
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
How to integrate of e to the power of -2x with x to the power of 2?
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.
How you integrate xxxx 1 dx?
.2x^5+x+C
D dx e 2x?
d/dx(e^2x) = 2xe^2x
What is the integral of 2-2x with limits 0 to t?
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
