What is #int e^x dx# and #int xe^(x^2)dx#?

Recall that the derivative of #e^x# is #e^x#. The function #e^x# is an exponential function with the irrational Euler's Number as base. Here is why the derivative of #e^x# is #e^x#:

#y = e^x#

Take the natural logarithm of both sides.

#lny = ln(e^x)#

Apply logarithm rules.

#lny = xlne#

#lny = x#

Recall that #d/dx(lnx) = 1/x#. Use implicit differentiation.

#1/y(dy/dx) = 1#

#dy/dx = y#

Since the original function was #y = e^x#

#dy/dx =e^x#

Knowing that #d/dx(e^x) = e^x#, we can conclude that the antiderivative of #e^x# is #e^x#. This is just like saying the derivative of #x# is #1#, so the antiderivative of #1# is #x#.

So one of the more important rules in calculus is the fact that #d/dx(e^x) = e^x# and that #int (e^x) dx = e^x + C#. Make sure to remember this one.

You can use this fact, and a simple u-substitution, to find the integral. Whenever I see an integral involving #e^f(x)#, I always look to see if it is of the form #af'(x)e^(f(x))#. This would mean that it can easily be integrated.

In our example, #f(x) = x^2#. The derivative of #f(x)# is #2x#, so indeed it is of the correct form.

We let #u = x^2#. Then #du = 2x dx# and #dx = (du)/(2x)#.

#int xe^(x^2)dx = intxe^u * (du)/(2x) = int1/2e^u du =1/2e^u + C = 1/2e^(x^2) + C#

If you take the derivative of #1/2e^(x^2) + C#, you should get #xe^(x^2)#, which is what we started with.

Hopefully this helps!

We're asked to find the antiderivatives (the first integrals) of two functions, which involve #e#.

Here's a fun fact: the integral of #e^x# is #e^x#! (plus the arbitrary constant, #C#)

Therefore,

#int e^x dx = color(red)(e^x + C#

For the second expression, we can replace #x^2# with the variable #u#, and then #du = 2x dx#:

#1/2inte^u du#

The integral of #e^u# is #e^u#:

#1/2inte^u du = (e^u)/2#

Substituting back the #x^2# in for #u#, we have

#intxe^(x^2) dx = color(blue)((e^(x^2))/2 + C#