# Calculus

posted by .

Integrate e^(-x^2/2) dx

What branch of calculus is this? Is this differential equations?

• Calculus -

Nope, just and ordinary integral. However, it is a special integral, called the error function. Look that up.

• Calculus -

That's what I needed. Thanks so much for the help!

## Similar Questions

1. ### Integral calculus

Please do help solve the followings 1) Integrate e^4 dx 2) Integrate dx/sqrt(90^2-4x^2) 3) Integrate (e^x+x)^2(e^x+1) dx 4) Integrate xe^x2 dx e^4 is a constant. 3) let u= e^x + x du= (e^x + 1)dx 4) let u= x du=dx v= e^x dv= e^x dx
2. ### calculus

1) Integrate Cos^n(x) dx 2) Integrate e^(ax)Sinbx dx 3) Integrate (5xCos3x) dx I Will be happy to critique your thinking on these. 1) Derive a recursive relation. 2) Simplest by replacing sin(bx) by Exp[i b x] and taking imaginary …
3. ### calculus

Can you give me a good website on the topic, slope fields and differential equations?
4. ### calculus

is y = x^3 a solution to the differential equation xy'-3y=0?
5. ### Differential Calculus

Given that p=(2q^2-5)^2. when q=3, it is increased by 0.7%. find the appropriate percentage in p. solution dp = 2(2q^2-5) (2dq) dp = (4x(18-5)(0.7%) dp = 4(13)(0.7%) am lost from here is dp=52x0.7/100= 0.364% Responses Differential …
6. ### Calculus AB

Verify if the given equations are solutions to the differential equation. dy/dt=k(y-a) i) y=1-ae^(kt) ii) y=a+Ce^(kt)
7. ### Calculus

dy/dx = 4ye^(5x) a) Separate the differential equation, then integrate both sides. b) Write the general solution as a function y(x). For the second part, I got y(x)=e^((5e^(5x))/(5)) + C but I don't understand how to separate differential …
8. ### calculus

calculus. suppose dR/dt= (d/R)^(2) and P(1) 4. separate the differential equation, integrate both sides
9. ### Calculus - Differential Equations

Use separation of variables to find the solution to the differential equation: 4 (du/dt) = u^2, subject to the initial condition u(0)=6.
10. ### Calculus - Differential Equations

Use separation of variables to find the solution to the differential equation: 4 (du/dt) = u^2, subject to the initial condition u(0)=6. So far, I have: 4 du = u^2 dt 4/u^2 du = dt -4/u = t+C I am unsure what to do from this point...

More Similar Questions