The Basics
When first introduced to integration in AP Calculus, it can be rather heady. It’s effectively the same process as taking derivatives, but in reverse! Still, after a brief learning curve working out the power rule, students seldom struggle too much with it. After the power rule, all of the normal derivative rules usually come quickly and easily.
But then a new problem shows up.
It’s almost there. It’s so close to being solvable! But that 3x mucks things up just a little bit. It’s tremendously tempting to say that the integral should be cos(3x)… but, as we’ll soon see, that’s not quite correct.
So, what do we do from here? Well, we can’t find the integral of sin(3x)… but we could find the integral of sin(x). How about we force things into a shape we can more easily deal with?
That’s closer! But we can’t take the integral unless our variable (u) and our dx match variables. Let’s see if we can do some more substitution to make our dx turn into a du!
We’ll start with our u = 3x equation, then take the derivative of both sides.
From here, we can divide both sides by 3, and…
Voila! Now, we can substitute that back into our integral, replacing dx with du/3.
From here, all we have to do is substitute u back out for 3x, and we get our final answer!
When to use U-Substitution
U-Substitution has one fundamental role: it undoes the Chain Rule. If you’re seeing integrands that are similar to those you might have seen for your answer when using the Chain Rule, that’s a strong hint that it’s time to bust this technique out!
In the example above, du was especially simple to find, but in most applications of U-Substitution, you’ll be using it to remove a particularly troublesome element from the integral.
As a similar example:
Allows us to get to:
To summarize, U-Substitution is commonly useful when the derivative of one part of the integral to is the same as another part of the integral.
Advanced Applications
However, U-Substitution won’t always have applications that are so easy to spot! Especially when combined with fractions, it can be very tricky to notice the opportunities you have available. Here’s an example where we have to apply an advanced technique.
How troublesome! We can’t make x squared our u value, because then we won’t be able to make dx into du. So, what do we do?
Well, with a little trick, we can make x + 3 a helpful u value. Take a look at this rewriting of the equation:
From here, we can use u = x + 3 and du = dx to get:
It’s still not something we can evaluate! But we’re getting closer. Let’s distribute out the top:
And if we rewrite the numerator as three separate fractions, we can finally get to a point where we can evaluate:
Then we can substitute u back out for x + 3 to get to the end of our problem.
Phew! That one was a doozy. Even if a question’s got you stumped, some inventive U-Substitution might still be able to save the day!
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