(I4) Substitution Rule (Definite)#

By the end of the lesson you will be able to:

  • calculate a definite integral using substitution rule.

Lecture Videos#

Two Methods#

Method 1

Change the limits of integration from \(x\) to \(u\).

When you go through the integration, you do not need to convert the antiderivative back to \(x\). You can leave it in terms of \(u\), and then go right to evaluating.

Method 2

Do not change the limits. Keep them in terms of \(x\).

When you go through the integration here, you do need to go a little bit further and convert the antiderivative back to the original variable \(x\).

Example 1#

Evaluate the following integral:

\[ \int_0^1 2x(x^2+1)^5 \;dx \]

Method 1#

Let’s calculate this by converting the limits of integration.

Click through the steps to see what we do.

If you want, give it a try first!

We choose \(u\) to be what’s inside the parentheses:

\[u=x^2+1\]

And then calculate \(du\)

\[du=\big[x^2+1\big]' dx \longrightarrow du = 2x\cdot dx\]

Next, we need to compare:

Our calculated \(du\)
\[du = 2x\cdot dx\]
Leftovers in the Integral
\[2x\cdot dx\]

This is a perfect match, so no other work is needed here.

Next, let’s convert the limits of integration

  • Upper limit: \(x=1 \longrightarrow u=(1)^2+1 \longrightarrow u=2\)

  • Lower limit: \(x=0 \longrightarrow u=(0)^2+1 \longrightarrow u=1\)

Now we’re ready to substitute our \(u\) and \(du\) and new limits of integration into the original integral:

\[\int_0^1 (x^2+1)^5 \cdot 2x\;dx = \int_{1}^{2} u^{5} \; du\]

We do a little power rule:

\[\int_{1}^{2} u^{5} \; du= \frac{1}{6} u^{6}\biggr\rvert_{u=1}^{u=2}\]

And then finally we evaluate:

\[\frac{1}{6} u^{6}\biggr\rvert_{u=1}^{u=2}= \left(\dfrac{1}{6}\cdot 2^6\right)-\left(\dfrac{1}{6}\cdot 1^6\right)\]

(Remember that the top number goes in first, bottom number goes in second, and we subtract.)

Method 2#

Now let’s try this example again but this time leave our limits of integration in terms of \(x\).

Click through the steps to see what we do.

If you want, give it a try first!

We choose \(u\) to be what’s inside the parentheses:

\[u=x^2+1\]

And then calculate \(du\)

\[du=\big[x^2+1\big]' dx \longrightarrow du = 2x\cdot dx\]

Next, we need to compare:

Our calculated \(du\)
\[du = 2x\cdot dx\]
Leftovers in the Integral
\[2x\cdot dx\]

This is a perfect match, so no other work is needed here.

Now we’re ready to substitute our \(u\) and \(du\) into the original integral:

\[\int_0^1 (x^2+1)^5 \cdot 2x\;dx = \int_{x=0}^{x=1} u^{5} \; du\]

We do a little power rule:

\[\int_{x=0}^{x=1} u^{5} \; du= \frac{1}{6} u^{6}\biggr\rvert_{x=0}^{x=1}\]

Convert the antiderivative back to \(x\)

\[\frac{1}{6} u^{6}\biggr\rvert_{x=0}^{x=1} = \frac{1}{6} (x^2+1)^{6}\biggr\rvert_{x=0}^{x=1}\]

And then finally we evaluate:

\[\frac{1}{6} (x^2+1)^{6}\biggr\rvert_{x=0}^{x=1}= \left(\frac{1}{6} (1^2+1)^{6}\right)-\left(\frac{1}{6} (0^2+1)^{6}\right)\]

(Remember that the top number goes in first, bottom number goes in second, and we subtract.)

Example 2#

Evaluate the integral.

\[ \int_1^2 \dfrac{dx}{(3-5x)^2} \]

Method 1#

Let’s try calculating this by converting the limits of integration.

Click through the steps to see what we do.

If you want, give it a try first!

We start by rewriting this fraction with a negative exponent.

\[\int_1^2 \dfrac{dx}{(3-5x)^2} = \int_1^2 (3-5x)^{-2} \; dx\]

(This just helps us later on, if we have to use the power rule.)

We choose \(u\) to be what’s inside the parentheses:

\[u=3-5x\]

And then calculate \(du\)

\[du=\big[3-5x\big]' dx \longrightarrow du = -5\cdot dx\]

Next, we need to compare:

Our calculated \(du\)
\[du = -5\cdot dx\]
Leftovers in the Integral
\[dx\]

Solving for the term(s) that shows up in both of these gives us:

\[du = -5\cdot dx \longrightarrow \tfrac{-1}{5}\cdot du = dx\]

Next, let’s convert the limits of integration

  • Upper limit: \(x=2 \longrightarrow u=3-5(2) \longrightarrow u=-7\)

  • Lower limit: \(x=1 \longrightarrow u=3-5(1) \longrightarrow u=-2\)

Now we’re ready to substitute our \(u\) and \(\tfrac{-1}{5}\cdot du\) and new limits of integration into the original integral:

\[\int_1^2 (3-5x)^{-2} \; dx = \int_{-2}^{-7} u^{-2} \cdot \frac{-1}{5}\; du\]

This gives us one of our basic integrals, which we can just do:

\[\frac{-1}{5} \int_{-2}^{-7} u^{-2} \; du = \frac{1}{5} u^{-1}\biggr\rvert_{u=-2}^{u=-7}\]

And then finally we evaluate:

\[\frac{1}{5u} \biggr\rvert_{u=-2}^{u=-7} = \left(\dfrac{1}{-35}\right)-\left(\dfrac{1}{-10}\right)\]

(Remember that the top number goes in first, bottom number goes in second, and we subtract.)

Method 2#

Now let’s give the other method a go, where we leave our limits of integration in terms of \(x\).

Click through the steps to see what we do.

If you want, give it a try first!

We start by rewriting this fraction with a negative exponent.

\[\int_1^2 \dfrac{dx}{(3-5x)^2} = \int_1^2 (3-5x)^{-2} \; dx\]

(This just helps us later on, if we have to use the power rule.)

We choose \(u\) to be what’s inside the parentheses:

\[u=3-5x\]

And then calculate \(du\)

\[du=\big[3-5x\big]' dx \longrightarrow du = -5\cdot dx\]

Next, we need to compare:

Our calculated \(du\)
\[du = -5\cdot dx\]
Leftovers in the Integral
\[dx\]

Solving for the term(s) that shows up in both of these gives us:

\[du = -5\cdot dx \longrightarrow \tfrac{-1}{5}\cdot du = dx\]

Now we’re ready to substitute our \(u\) and \(\tfrac{-1}{5}\cdot du\) into the original integral:

\[\int_1^2 (3-5x)^{-2} \; dx = \int_{x=1}^{x=2} u^{-2} \cdot \frac{-1}{5}\; du\]

Since we are not converting the limits, we should write them in the form \(x=\cdots\)

This gives us one of our basic integrals, which we can just do:

\[\frac{-1}{5} \int_{x=1}^{x=2} u^{-2} \; du = \frac{1}{5} u^{-1}\biggr\rvert_{x=1}^{x=2}\]

And then we convert our antiderivative back to \(x\):

\[\frac{1}{5u} \biggr\rvert_{x=1}^{x=2}= \frac{1}{5(3-5x)} \biggr\rvert_{x=1}^{x=2} \]

Now we can evaluate:

\[\frac{1}{5(3-5x)} \biggr\rvert_{x=1}^{x=2}= \left(\dfrac{1}{5(-7)}\right)-\left(\dfrac{1}{5(-2)}\right)\]

(Remember that the top number goes in first, bottom number goes in second, and we subtract.)

Example 3#

Evaluate the integral.

\[ \int_0^4 \dfrac{x}{\sqrt{2x+1}} \; dx \]

Method 1#

Click through the steps to see what we do.

If you want, give it a try first!

We start by rewriting this root function as a power function:

\[\int_0^4 \dfrac{x}{\sqrt{2x+1}} \; dx = \int_0^4 x(2x+1)^{-1/2} \; dx\]

(This just helps us later on, if we have to use the power rule.)

Since the more complicated piece of our function is \((2x+1)^{-1/2}\), we choose \(u\) to be the inner function:

\[u=2x+1\]

And then calculate \(du\)

\[du=\big[2x+1\big]' dx \longrightarrow du = 2\cdot dx\]

Next, we need to compare:

Our calculated \(du\)
\[du = 2\cdot dx\]
Leftovers in the Integral
\[x\cdot dx\]

Since \(dx\) is the only term that shows up in both, we solve for \(dx\):

\[du = 2\cdot dx \longrightarrow \tfrac{1}{2}\cdot du = dx\]

Next, let’s convert the limits of integration

  • Upper limit: \(x=4 \longrightarrow u=2(4)+1 \longrightarrow u=9\)

  • Lower limit: \(x=0 \longrightarrow u=2(0)+1 \longrightarrow u=1\)

Now we’re ready to substitute our \(u\) and \(\frac{1}{2}du\) and new limits of integration into the original integral:

\[\int_0^4 (2x+1)^{-1/2} \cdot x\; dx = \int_1^9 u^{-1/2} \cdot x \cdot \tfrac{1}{2}\; du \]

But… we can’t go any further yet since we still have some \(x\)-terms that we haven’t gotten rid of yet. So we turn to our “solve for \(x\) strategy.”

\[ u=2x+1\longrightarrow x=\dfrac{u-1}{2} \]

And then we plug this in for \(x\):

\[\int_1^9 u^{-1/2} \cdot x \cdot \tfrac{1}{2}\; du = \int_1^9 u^{-1/2} \cdot \dfrac{u-1}{2} \cdot \tfrac{1}{2}\; du \]

````{tab-item} Step 3
This gives us a basic integral, which can be solved by multiplying out the integrand and then using power rule:

```{math}
\begin{split}
\dfrac{1}{4}\int_1^9 u^{-1/2} \cdot (u-1)\;  du  &= \dfrac{1}{4} \cdot \int_1^9 u^{1/2}-u^{-1/2} \; du \\
& = \dfrac{1}{4}\cdot \left( \dfrac{2}{3}u^{3/2}-2u^{1/2}\right)\biggr\rvert_{u=1}^{u=9}
\end{split}
```

And then evaluate!

\[\tfrac{1}{6}u^{3/2}-\tfrac{1}{2}u^{1/2}\biggr\rvert_{u=1}^{u=9} = \bigg(\tfrac{1}{6}\cdot 9^{3/2}-\tfrac{1}{2}\cdot 9^{1/2}\bigg)- \bigg(\tfrac{1}{6}\cdot 1^{3/2}-\tfrac{1}{2}\cdot 1^{1/2}\bigg)\]

Simplify as desired!

Method 2#

Click through the steps to see what we do.

If you want, give it a try first!

We start by rewriting this root function as a power function:

\[\int_0^4 \dfrac{x}{\sqrt{2x+1}} \; dx = \int_0^4 x(2x+1)^{-1/2} \; dx\]

(This just helps us later on, if we have to use the power rule.)

Since the more complicated piece of our function is \((2x+1)^{-1/2}\), we choose \(u\) to be the inner function:

\[u=2x+1\]

And then calculate \(du\)

\[du=\big[2x+1\big]' dx \longrightarrow du = 2\cdot dx\]

Next, we need to compare:

Our calculated \(du\)
\[du = 2\cdot dx\]
Leftovers in the Integral
\[x\cdot dx\]

Since \(dx\) is the only term that shows up in both, we solve for \(dx\):

\[du = 2\cdot dx \longrightarrow \tfrac{1}{2}\cdot du = dx\]

Now we’re ready to substitute our \(u\) and \(\frac{1}{2}du\) into the original integral:

\[\int_0^4 (2x+1)^{-1/2} \cdot x\; dx = \int_{x=0}^{x=4} u^{-1/2} \cdot x \cdot \tfrac{1}{2}\; du \]

But… we can’t go any further yet since we still have some \(x\)-terms that we haven’t gotten rid of yet. So we turn to our “solve for \(x\) strategy.”

\[ u=2x+1\longrightarrow x=\dfrac{u-1}{2} \]

And then we plug this in for \(x\):

\[\int_{x=0}^{x=4} u^{-1/2} \cdot x \cdot \tfrac{1}{2}\; du = \int_{x=0}^{x=4} u^{-1/2} \cdot \dfrac{u-1}{2} \cdot \tfrac{1}{2}\; du \]

This gives us a basic integral, which can be solved by multiplying out the integrand and then using power rule:

\[\begin{split}\begin{split} \dfrac{1}{4}\int_{x=0}^{x=4} u^{-1/2} \cdot (u-1)\; du &= \dfrac{1}{4} \cdot \int_{x=0}^{x=4} u^{1/2}-u^{-1/2} \; du \\ & = \dfrac{1}{4}\cdot \left( \dfrac{2}{3}u^{3/2}-2u^{1/2}\right)\biggr\rvert_{x=0}^{x=4} \end{split}\end{split}\]

Convert the antiderivative back to \(x\)

\[\dfrac{1}{6}u^{3/2}-\dfrac{1}{2}u^{1/2}\biggr\rvert_{x=0}^{x=4} = \dfrac{1}{6}\cdot (2x+1)^{3/2}-\dfrac{1}{2}\cdot (2x+1)^{1/2}\biggr\rvert_{x=0}^{x=4}\]

And then evaluate!

\[\begin{split}\begin{split} \tfrac{1}{6}\cdot (2x+1)^{3/2} &-\tfrac{1}{2}\cdot (2x+1)^{1/2}\biggr\rvert_{x=0}^{x=4} \\ & = \bigg(\tfrac{1}{6}\cdot 9^{3/2}-\tfrac{1}{2}\cdot 9^{1/2}\bigg)- \bigg(\tfrac{1}{6}\cdot 1^{3/2}-\tfrac{1}{2}\cdot 1^{1/2}\bigg) \end{split}\end{split}\]

Simplify as desired!