Question

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

Answer 1

See below.

Explanation:

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!

Answer 2

`int e^x dx = e^x + C`

`int xe^(x^2) dx = (e^(x^2))/2 + C`

Explanation:

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`