The Idea of Integration by Parts

The Idea of Integration by Parts

The idea is to use the product rule in reverse.

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ =?

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x)

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x)+

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx+

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

What happens when you integrate a derivative?

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

What happens when you integrate a derivative?∫[u(x)v(x)]′ dx =?

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

What happens when you integrate a derivative?∫[u(x)v(x)]′ dx =?

By definition of indefinite integral, you get back the functionyou’re taking the derivative of:

The Idea of Integration by Parts

The idea is to use the product rule in reverse. Remember theproduct rule?

[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)

Let’s now integrate both sides:∫[u(x)v(x)]′ dx =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

What happens when you integrate a derivative?∫[u(x)v(x)]′ dx =?

By definition of indefinite integral, you get back the functionyou’re taking the derivative of:∫

[u(x)v(x)]′ dx = u(x)v(x)

The Idea of Integration by Parts

The Idea of Integration by Parts

So, our formula becomes:

The Idea of Integration by Parts

So, our formula becomes:

u(x)v(x) =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

The Idea of Integration by Parts

So, our formula becomes:

u(x)v(x) =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

Now, if we subtract∫u′(x)v(x)dx to both sides we get:

The Idea of Integration by Parts

So, our formula becomes:

u(x)v(x) =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

Now, if we subtract∫u′(x)v(x)dx to both sides we get:

u(x)v(x) −∫

u′(x)v(x)dx =

∫u(x)v ′(x)dx

The Idea of Integration by Parts

So, our formula becomes:

u(x)v(x) =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

Now, if we subtract∫u′(x)v(x)dx to both sides we get:

u(x)v(x) −∫

u′(x)v(x)dx =

∫u(x)v ′(x)dx

Or:

The Idea of Integration by Parts

So, our formula becomes:

u(x)v(x) =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

Now, if we subtract∫u′(x)v(x)dx to both sides we get:

u(x)v(x) −∫

u′(x)v(x)dx =

∫u(x)v ′(x)dx

Or: ∫u(x)v ′(x)dx = u(x)v(x) −

∫u′(x)v(x)dx

The Idea of Integration by Parts

So, our formula becomes:

u(x)v(x) =

∫u′(x)v(x)dx +

∫u(x)v ′(x)dx

Now, if we subtract∫u′(x)v(x)dx to both sides we get:

u(x)v(x) −∫

u′(x)v(x)dx =

∫u(x)v ′(x)dx

Or: ∫u(x)v ′(x)dx = u(x)v(x) −

∫u′(x)v(x)dx

This is the formula of integration by parts.

Example 1

Example 1

Let’s try to find the integral:

Example 1

Let’s try to find the integral: ∫xexdx

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex∫xexdx =

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

∫��u(x)

xexdx =

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

∫x��>

v ′(x)exdx =

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex∫xexdx =

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex∫xexdx = xex

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

∫xexdx =��

u(x)xex

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

∫xexdx = x��>

v(x)ex

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex∫xexdx = xex−

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex∫xexdx = xex −

∫1.exdx

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

∫xexdx = xex −

∫���

u′(x)

1.exdx

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex

∫xexdx = xex −

∫1.��>

v(x)exdx

Example 1

Let’s try to find the integral: ∫xexdx

This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫

u(x)v ′(x)dx = u(x)v(x) −∫

u′(x)v(x)dx

So, in this case we can make:

u(x) = x v ′(x) = ex

u′(x) = 1 v(x) = ex∫xexdx = xex −

∫1.exdx

Example 1

Example 1

So, we only need to solve:

Example 1

So, we only need to solve:∫xexdx = xex −

∫exdx

Example 1

So, we only need to solve:∫xexdx = xex −

∫exdx

The integral in the second term is easy to solve:

Example 1

So, we only need to solve:∫xexdx = xex −

∫exdx

The integral in the second term is easy to solve:∫xexdx = xex − ex

Example 1

So, we only need to solve:∫xexdx = xex −

∫exdx

The integral in the second term is easy to solve:∫xexdx = xex − ex

∫xexdx = ex (x − 1)

Example 1

So, we only need to solve:∫xexdx = xex −

∫exdx

The integral in the second term is easy to solve:∫xexdx = xex − ex

∫xexdx = ex (x − 1)

That’s the final answer.