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Mathematical Expression Editor
We learn a new technique, called substitution, to help us solve problems involving
integration.
Computing antiderivatives is not as easy as computing derivatives. One
issue is that the chain rule can be difficult to “undo.” We have a general
method called “integration by substitution” that will somewhat help with this
difficulty.
If the functions are differentiable, we can apply the chain rule and obtain
\[ \ddx f(g(x)) = f'(g(x))g'(x) \]
If the
derivatives are continuous, we can use this equality to evaluate a definite integral.
Namely,
Since the right hand sides of these equalities are equal, the left hand sides must be
equal, too. So, it follows that
\[ \int _a^b f'(g(x))g'(x) \d x = \int _{g(a)}^{g(b)} f'(u)\d u . \]
This simple observation leads to the following theorem.
Integral Substitution Formula Let \(g'\) be the derivative of a function \(g\) and let \(f'\) be the
derivative of a function \(f\). If \(g'\) is continuous on the interval \([a,b]\) and \(f'\) is continuous on the
interval \([g(a),g(b)]\), then
\[ \int _a^b f'(g(x)) g'(x) \d x =\int _{g(a)}^{g(b)} f'(u) \d u. \]
The integral on the right appears to be much simpler than the original integral. So,
when evaluating a difficult integral, we try to apply this theorem and replace the
original integral with a simpler one. We can do this as long as the conditions of the
theorem are satisfied.
We will apply the Integral Substitution Formula (ISF) to the following example.
Compute:
\[ \int _0^2 x e^{x^2}\d x \]
Why do we think that the ISF can be useful in this example?
Does this integral have the structure
\[ \int _a^b f'(g(x)) g'(x) \d x , \]
so that we can apply the ISF?
We set \(g(x) =x^2\), so \(g'(x)dx =2xdx\), and note that
\[ \int _0^2 x e^{x^2}\d x =\frac {1}{2}\int _0^{2} e^{\underbrace {x^2}}\underbrace {2x\d x}=\frac {1}{2}\int _0^{2} e^{g(x)}g'(x)\d x \]
So we wrote the integral in a form suitable for application of the ISF. Therefore, we
can substitute \(u\) for \(g(x)\) and \(\d u\) for \(g'(x)\d x\) and write
\[ \int _0^2 x e^{x^2}\d x=\frac {1}{2}\int _0^{2} e^{g(x)}g'(x)\d x=\frac {1}{2}\int _{g(0)}^{g(2)} e^{u}\d u \]
\[ \int _0^2 x e^{x^2}\d x=\frac {1}{2}\int _{0}^{4} e^{u}\d u \]
The figure below illustrates both integrals.
The two shaded
areas in the figure, \(A= \int _0^2 x e^{x^2}\d x\) and \(B=\frac {1}{2}\int _{0}^{4} e^{u}\d u\), appear to be equal, and this confirms what was established
by computation before.
Therefore, whenever we are faced with a problem of evaluating a difficult integral, we
try to replace it with a simpler one, as long as both integrals represent the same net
area.
We can solve a problem like this in a slightly different way. Let’s do the same
example again, this time we will think in terms of differentials.
Compute:
\[ \int _0^2 x e^{x^2}\d x \]
Here we will set
\[ u=g(x) = x^2. \]
Then
\[ \d u =g'(x)\d x=2x\d x. \]
Here, we are thinking in terms of differentials. We
can solve for \(\d x\) to get
\[ \d x = \frac {\d u}{2x}. \]
Using this equality, we can write
\begin{align*} \int _0^2 x e^{x^2}\d x &= \int _{g(0)}^{g(2)} x e^{u} \frac {\d u}{2x}\\ &= \int _{0}^{4} \frac {e^u}{2}\d u . \end{align*}
Notice that we have substituted in the new variable \(u\), in both the integrand and in
the limits of integration. When we make the substitution we have to update
the limits of integration to rewrite them in terms of the new substitution
variable.
At this point, we can continue as we did before and write
Finally, sometimes we simply want to deal with the antiderivative on its
own. We will repeat the example one more time in oder to demonstrate this
approach.
Compute:
\[ \int _0^2 x e^{x^2}\d x \]
We can apply the Second Fundamental Theorem of Calculus
and, in order to do that, we first need to compute an indefinite integral.
\[ \int x e^{x^2} \d x. \]
As before, we set \(u=g(x)=x^2\), and compute \(\d u = 2x \d x\), thinking in terms of differentials. Now we see that
\[ \int x e^{x^2} \d x = \int x e^{u} \frac {\d u}{2x} = \int \frac {e^{u}}{2}\d u . \]
Hence
\[ \int x e^{x^2} \d x = \frac {e^{u}}{2}+C . \]
This does not look right, since we need to find an antiderivative
of the function \( x e^{x^2},\) and our result is a function of \(u\). We can easily fix this, by
simply substituting \(u\) with \(g(x)=x^2\). Therefore,
\[ \int xe^{x^2}\d x =\frac {e^{u}}{2}+C= \frac {e^{x^2}}{2}+C . \]
Finally, we can apply the SFTOC
\begin{align*} \int _0^2 x e^{x^2}\d x &=\eval {\frac {e^{x^2}}{2}}_0^2\\ &= \frac {\answer [given]{e^4} -e^0}{2}. \end{align*}
In this case, because we used substitution to find the indefinite integral, we did not
need to modify the limits of integration when calculating the definite integral.
However, we did have the extra step of rewriting the antiderivative back in terms of
the original variable. If you follow this method, notice that you are performing two
separate calculations. The first uses substitution to calculate the antiderivative
(without any limits of integration), then using that result to calculate the definite
integral using SFTOC.
1 More examples
With some experience, it is (usually) not too hard to see what to substitute as \(u\). We
will work through the following examples in the same way that we did for
Example 4.
Compute:
\[ \int x^4(x^5+1)^{9} \d x \]
Here we set \(u = \answer [given]{x^5+1}\), so \(\d u = \answer [given]{5x^4} \d x\). Then it follows that \(\d x=\frac {\d u}{5x^4}\). Therefore
Notice that this example is an indefinite integral and not a definite integral, meaning
that there are no limits of integration. So we do not need to worry about
changing the endpoints of the integral. However, we do need to back-substitute
into our answer, so that our final answer is a function of \(x\). Recalling that \(u= x^5+1\),
we have our final answer
Reminder: you can always verify your result by
differentiating.
If substitution works to calculate an integral (and that is not always the case!), a
common trick to find what to substitute for is to locate the “ugly” part of
the function being integrated. We then substitute for the “inside” of this
ugly part. While this technique is certainly not rigorous, it can prove to
be very helpful. This is especially true for students new to the technique
of substitution. The next two problems are really good examples of this
philosophy.
Compute:
\[ \int _{-1}^0 12x^3 \cos ({x^4}) \d x \]
The “ugly” part of the function being integrated is \(\cos ({x^4})\). The “inside” of this
term is then \(x^4\). So a good possibility is to try
\[ u =g(x)= x^4. \]
Then
\[ \d u = (4x^3) \d x \qquad \Rightarrow \qquad \d x = \answer [given]{\frac {1}{4x^3}} \d u \]
To summarize, if we suspect that a given function is the derivative of another via
the chain rule, we introduce a new variable \(u=g(x)\), where \(g\) is a likely candidate
for the inner function. We rewrite the integral entirely in terms of \(u\), with
no \(x\) remaining in the expression. If we can integrate this new function of \(u\),
then the antiderivative of the original function is obtained by replacing \(u\) by
\(g(x)\).
![[Picture]](digInIdeaOfSubstitution.online-2676421353299a36a79de3bad843b7ca.svg)
![[Picture]](digInIdeaOfSubstitution.online-99d737657e32e37cae1ce1895aab305f.svg)