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The Substitution Rule
Calculus 2 · Chapter 1: Review & Advanced Integration TechniquesSection 1.1 · lesson 1 of 5412 min · free
Next lesson: Integration by Parts →
In this lesson
The home base for all elementary antidifferentiation.
- Separate the indefinite integral (a family of functions) from the definite integral (a number)
- Read the chain rule backward to get the substitution rule
- Apply the four-move procedure to radicals, exponentials and trig arguments
- Handle definite integrals two ways - and never mix the limits
Transcript
Everything said in the video, section by section (1,879 words). Read along, or search it.
MAT-265 - Calculus 2 · Section 1.1
The Substitution Rule
A NASA test sled accelerates down a straight track. Engineers want the total distance it covers during a two second burn. They already have a formula for its velocity — but it's a mess. A polynomial, multiplied by the square root of another polynomial. Integrating that directly looks hopeless. And yet, within a single line of work, the whole thing collapses into something you could have handled in your first calculus course. That's not luck. That's substitution. It is the single most important integration technique in all of calculus, and every method in this course is built on top of it. So let's make sure you own it.
Foundation
Antiderivatives and the Two Kinds of Integral
Before the technique, one distinction that quietly causes more confusion than anything else in this course. An antiderivative of f is any function big F whose derivative hands you back f. Now here's the thing. The derivative of a constant is zero. So antiderivatives are never unique. If big F works, so does big F plus C. That entire family is what the indefinite integral means. Its output is a function, and it always carries plus C. The definite integral is a completely different animal. It carries a lower limit and an upper limit, and it evaluates to one single number. Big F of b, minus big F of a. That's the Fundamental Theorem of Calculus, part two. Notice what happens to the constant in that subtraction — it cancels. Which is exactly why a definite integral never carries a plus C. One of these is a function. The other is a number. Keep them apart, and most of the confusion later in this course never happens to you.
The Technique
Substitution Is the Chain Rule, Read Backward
So where does substitution actually come from? It isn't a trick someone invented. It's the chain rule, read backward. The chain rule says the derivative of a composite, big F of g of x, is big F prime of g of x, times g prime of x. Integrate both sides of that, and you've recovered the substitution rule. Here's the statement. If u equals g of x is differentiable, and big F is an antiderivative of f, then the integral of f of g of x times g prime of x d x becomes just the integral of f of u, d u. In plain words: when the integrand contains an inner function together with a constant multiple of its own derivative, replacing that inner function by u turns a hard problem into an easy one. The procedure is four moves. Choose u — the inner function. Inside a power, a root, an exponent, or a trig argument. Differentiate, to get d u, and solve for the grouping that actually appears in the integral. Rewrite the whole integral in u. No x may survive. Then antidifferentiate, and put g of x back in at the end. And here's the test for whether you picked a good u. A good u is one whose derivative is already sitting there in the integrand — up to a constant factor you can balance.
Worked Example 1
A Polynomial Inside a Radical
Let's watch that happen. Evaluate the integral of two x, times the square root of x squared plus one, d x. Step one. What's the awkward part? The inner function under the root. So let u equal x squared plus one. Step two. Differentiate. d u equals two x d x. Now look closely at the original integral. That factor, two x d x, is sitting right there. It is exactly d u. So the entire integral becomes the integral of root u, d u. Which is the integral of u to the one half. That's the whole point of substitution. A problem you couldn't touch is now a power rule. Raise the exponent by one, divide by the new exponent, and you get two thirds u to the three halves, plus C. Back-substitute, and you're done. Two thirds, times x squared plus one, to the three halves, plus C. Now — how do you know that's right? Differentiate it. The chain rule gives you two thirds, times three halves, times x squared plus one to the one half, times two x. The two thirds and the three halves cancel, and you're left with the original integrand. Differentiating your answer is always the fastest way to check a substitution. Use it every time.
Worked Examples 2 and 3
An Exponent, and a Trig Argument
You just saw the derivative sitting in the integrand perfectly. Now watch what happens when it's only almost there. Evaluate the integral of x, times e to the minus x squared, d x. Let u be the exponent — minus x squared. Then d u is minus two x d x. But the integrand only gives us x d x. Not minus two x d x. That's fine. Solve for what you have: x d x equals minus one half d u. The derivative was present up to a constant factor, and a constant factor is something you can always balance. The answer is minus one half, e to the minus x squared, plus C. And that factor, e to the minus x squared, is the heart of the normal distribution. So this exact substitution shows up constantly in probability work. Third example, and it's the one you'll use most often. The integral of cosine of five x, d x. Let u be the argument. u equals five x, so d u equals five d x, and d x is one fifth d u. You get one fifth sine of five x, plus C. Here's the pattern worth memorizing outright. Whenever the inner function is linear — a x — the integral picks up a factor of one over a. Alternating current, sound waves, seasonal models. You'll see this one everywhere.
Definite Integrals
Two Correct Routes - Worked Example 4
Now put limits on it, and something new has to be handled. Take the same integral as before, from zero to two, and the same substitution: u equals x squared plus one. There are two correct routes here, and choosing well saves you time. Method A. Change the limits. When x is zero, u is one. When x is two, u is five. So the new integral runs from one to five, and you evaluate entirely in u. You get two thirds, times five root five minus one. About six point seven nine. Method B. Back-substitute first. Find the antiderivative in u, return to x, and then apply the original limits, zero and two. Same number. Six point seven nine. As it must be. Method A is usually faster, because it skips the back-substitution step entirely. But now here is the mistake — and it is the single most common error in this whole section. The limits zero and two belong to x. The limits one and five belong to u. If you take the u antiderivative, root u, and evaluate it at x equals zero and x equals two, you have blended two different coordinate systems. The number you get is meaningless. Not slightly off. Meaningless. Pick a method, and commit to it all the way through.
Figure 1.1
Substitution Transports the Region of Integration
I want to show you why that error is so easy to make, because the picture makes it obvious. On the left is the original problem. The function two x root x squared plus one, over the interval from zero to two, in x. On the right is what substitution turned it into. Root u, over the interval from one to five. These look like completely different pictures. Different curves, different intervals, different scales on both axes. But those two shaded regions have exactly the same area. So here's the idea to hold onto. Substitution does not merely rename your variable. It picks up the entire region of integration and transports it somewhere else. The interval zero to two doesn't stay behind. It becomes the interval one to five. The limits have to travel with the substitution. Once you can see that, you'll never mix them up again.
Reference
Four High-Frequency Substitution Patterns
Let's compress everything you've seen into four patterns you'll recognize on sight. First. Something times x, with x squared plus a constant inside. Take u to be x squared plus a. d u is two x d x, and you pick up a one half. Second. e to the k x. Take u to be k x, and you pick up one over k. Third. Cosine or sine of a x. Take u to be a x, and again you pick up one over a. Fourth, and this is the general shape all three are special cases of. Some function g of x raised to a power, multiplied by g prime of x. Take u to be g of x, and the whole thing becomes the integral of u to the n. In every single row, the same thing happens. The inner function becomes u, and one constant factor gets balanced. Now let me leave you with something to think about. u equals x squared plus one worked beautifully for the integral of two x root x squared plus one, because that factor of two x was already there. But suppose the integrand were just root x squared plus one, with no x out front. Would the same substitution still work? Ask yourself why the presence of g prime of x is the deciding factor — and what kind of technique you'd need when it's missing.
Avoiding Errors
The Four Mistakes That Cost the Most Points
Four mistakes. These are the ones that actually cost students points. One. Forgetting the d u. You replace the inner function with u, but you leave the d x sitting there untouched. Writing the integral of cosine five x equals sine five x plus C, with no one fifth, is wrong for exactly this reason. The d x was never converted. Two. Mismatched limits. We just covered this one, and it's worth hearing twice. The moment the integral is rewritten in u, the original x limits no longer apply. Change them, or go all the way back to x first. Three. A leftover x. If any x survives the substitution, you are not finished. Sometimes you can rescue it — solve the u equation for x and substitute that in. But if you can't, that's your signal that the choice of u was poor, and you need a different one. Four. Dropping the plus C. This is not a formality. An indefinite integral is a family of functions, and omitting the constant throws away every member of that family but one. It's a real error, and it's graded like one.
Key Takeaways
What to Carry Into the Rest of the Course
So, what do you carry out of this section? An indefinite integral returns a family of functions, with plus C. A definite integral returns a single number, and the constant cancels. Substitution reverses the chain rule. It works whenever the integrand contains an inner function together with a constant multiple of its derivative. The procedure never changes. Choose u. Compute d u. Rewrite entirely in u. Antidifferentiate. Back-substitute. For definite integrals, either change the limits to u values, or return to x before evaluating. Never mix the two. And guard the four classics. A forgotten d u. Mismatched limits. A leftover x. A missing plus C. Substitution is the home base for everything ahead of you. Next up is section one point two, integration by parts — which handles the products that substitution simply cannot touch.
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