\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Technique of Substitution

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

The technique of substitution is something you need to understand while you learn calculus. It is used a lot, so it is important that you have a good grasp on how to use it.
There are actually two contexts when talking about substitution. First, using substitution to change an equation so that we can factor it or otherwise manipulate it more easily, which is discussed on this page. Another context is to use substitution to solve systems of equations.
You can find that discussion on the solving systems of equations page.

Concept of Substitution

The idea of substitution is to replace part of an equation with a temporary variable so that you can perform some operation more easily. One example is factoring. Let's say you have an expression that looks like this \[ e^{2n} + 2e^n + 1 \] and you are asked to factor this. I don't know about you but nothing really jumps out at me and I don't even know if it is factorable. However, let's use substitution and see if that helps. I know how to factor polynomials, so if I replace \( e^n \) with \( t \), i.e. let \( t = e^n \), the result is \( t^2 + 2t + 1 \) which factors into \( (t + 1)(t + 1) = (t+1)^2 \). Substituting back in we end up with \[ e^{2n} + 2e^n + 1 = (e^n+1)^2 \]

Details of Substitution

Now, let's discuss the details of substitution, using the last example. First, we need to use a variable different from what is given in the problem. In this case, the variable was n, so we used t. We could have used x or y or almost any other letter. We could have even used a greek letter like \( \theta \). The point is to use something other than n. Note: We never use the letter o or O since they are easily confused with the number zero.

Secondly, what to choose to substitute for is a matter of experience. However, as you are learning, try different things until something works. Then sit back and notice patterns in your work.

Third, once you have chosen the substitution, you need to substitute appropriately everywhere in the expression. So, in the previous example, it wouldn't help to substitute at one place and not the other, i.e. \( e^{2n} + 2t + 1 \) doesn't really help us and I can't really do anything with it.

Fourth, never substitute something like \( n = t \). This substitution doesn't help at all.

Finally, once you have finished all adjustments or algebra or calculus on the expression, remember to undo the substitution by substituting back, i.e. if you initially substitute \( t \) for \( e^n \), you need to go back and replace all places where you have \( t \) with \( e^n \). Your final answer should contain the same variables as the original expression.

Sometimes this last step is easy, other times it involves more work. In the example above, we just factored, so we had \( t \) in the expression and we could just do a straight substitution back. But sometimes, you may need to do some algebra on the expression to know how to substitute back.

For example, if you initially substituted \( t = e^n \) but ended up with something like \( n + 1 \), you need to do the following.

\( \begin{array}{rcl} t & = & e^n \\ \ln(t) & = & \ln(e^n) \\ \ln(t) & = & n \end{array} \)

Then substitute \( \ln(t) \) for \( n \) to get \( n + 1 = \ln(t) + 1 \). This kind of substitution is something you may see in derivatives and integrals.

Another thing to remember when doing the initial substitution is that you do not need to do anything with constants in the problem, UNLESS the constants are dependent on the variable, for example they involve specific units.

Really UNDERSTAND Precalculus

Log in to rate this page and to see it's current rating.

To bookmark this page, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

how to take good notes

Shop Amazon - Rent eTextbooks - Save up to 80%

As an Amazon Associate I earn from qualifying purchases.

I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me.

Support 17Calculus on Patreon

Search

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2022 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics