Calculus
posted by mathstudent .
Integrate e^(x^2/2) dx
What branch of calculus is this? Is this differential equations?

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

Calculus 
mathstudent
That's what I needed. Thanks so much for the help!
Respond to this Question
Similar Questions

Integral calculus
Please do help solve the followings 1) Integrate e^4 dx 2) Integrate dx/sqrt(90^24x^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 
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 … 
calculus
Can you give me a good website on the topic, slope fields and differential equations? 
calculus
is y = x^3 a solution to the differential equation xy'3y=0? 
Differential Calculus
Given that p=(2q^25)^2. when q=3, it is increased by 0.7%. find the appropriate percentage in p. solution dp = 2(2q^25) (2dq) dp = (4x(185)(0.7%) dp = 4(13)(0.7%) am lost from here is dp=52x0.7/100= 0.364% Responses Differential … 
Calculus AB
Verify if the given equations are solutions to the differential equation. dy/dt=k(ya) i) y=1ae^(kt) ii) y=a+Ce^(kt) 
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 … 
calculus
calculus. suppose dR/dt= (d/R)^(2) and P(1) 4. separate the differential equation, integrate both sides 
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. 
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...