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Integration by Substitution
“Integration by Substitution” (also called “u-Substitution” or “The Reverse Chain Rule”) is a method to find an integral, but only when it can be set up in a special way.
The first and most vital step is to be able to write our integral in this form:
Note that we have g(x) and its derivative g'(x)
Like in this example:
Here f=cos, and we have g=x2 and its derivative 2x
This integral is good to go!
When our integral is set up like that, we can do this substitution:
Then we can integrate f(u), and finish by putting g(x) back as u.
Like this:
Example: ∫cos(x2) 2x dx
We know (from above) that it is in the right form to do the substitution:
Now integrate:
∫cos(u) du = sin(u) + C
And finally put u=x2 back again:
sin(x2) + C
So ∫cos(x2) 2x dx = sin(x2) + C
That worked out really nicely! (Well, I knew it would.)
But this method only works on some integrals of course, and it may need rearranging:
Example: ∫cos(x2) 6x dx
Oh no! It is 6x, not 2x like before. Our perfect setup is gone.
Never fear! Just rearrange the integral like this:
∫cos(x2) 6x dx = 3∫cos(x2) 2x dx
(We can pull constant multipliers outside the integration, see Rules of Integration.)
Then go ahead as before:
3∫cos(u) du = 3 sin(u) + C
Now put u=x2 back again:
3 sin(x2) + C
Done!
Now let’s try a slightly harder example:
Example: ∫x/(x2+1) dx
Let me see … the derivative of x2+1 is 2x … so how about we rearrange it like this:
∫x/(x2+1) dx = ½∫2x/(x2+1) dx
Then we have:
Then integrate:
½∫1/u du = ½ ln(u) + C
Now put u=x2+1 back again:
½ ln(x2+1) + C
And how about this one:
Example: ∫(x+1)3 dx
Let me see … the derivative of x+1 is … well it is simply 1.
So we can have this:
∫(x+1)3 dx = ∫(x+1)3 · 1 dx
Then we have:
Then integrate:
∫u3 du = (u4)/4 + C
Now put u=x+1 back again:
(x+1)4 /4 + C
We can take that idea further like this:
Example: ∫(5x+2)7 dx
If it was in THIS form we could do it:
∫(5x+2)7 5 dx
So let’s make it so by doing this:
15 ∫(5x+2)7 5 dx
The 15 and 5 cancel out so all is fine.
And now we can have u=5x+2
And then integrate:
15 ∫u7 du = 15 u88 + C
Now put u=5x+2 back again, and simplify:
(5x+2)840 + C
Now get some practice, OK?
In Summary
When we can put an integral in this form:
Definite Integrals
You might like to read Introduction to Integration first!
Integration
Notation
The symbol for “Integral” is a stylish “S” (for “Sum”, the idea of summing slices):
After the Integral Symbol we put the function we want to find the integral of (called the Integrand).And then finish with dx to mean the slices go in the x direction (and approach zero in width).
INTEGRATION BY INSPECTION
Definite Integral
A Definite Integral has start and end values: in other words there is an interval .
a and b (called limits, bounds or boundaries) are put at the bottom and top of the “S”, like this:
(from a to b)
(no specific values)
We find the Definite Integral by calculating the Indefinite Integral at a, and at b, then subtracting:
Example: What is
We are being asked for the Definite Integral, from 1 to 2, of 2x dx
First we need to find the Indefinite Integral.
Using the Rules of Integration we find that ∫2x dx = x2 + C
Now calculate that at 1, and 2:
Subtract:
And “C” gets cancelled out … so with Definite Integrals we can ignore C.
Result:
Check: with such a simple shape, let’s also try calculating the area by geometry:
A = 2+42 × 1 = 3
Yes, it does have an area of 3.
(Yay!)
Notation: We can show the indefinite integral (without the +C) inside square brackets, with the limits a and b after, like this:
Example (continued)
A good way to show your answer:
= [ x2 ]
Let’s try another example:
Example:
The Definite Integral, from 0.5 to 1.0, of cos(x) dx:
(Note: x must be in radians)
The Indefinite Integral is: ∫cos(x) dx = sin(x) + C
We can ignore C for definite integrals (as we saw above) and we get:
= [ sin(x) ]
And another example to make an important point:
Example:
The Definite Integral, from 0 to 1, of sin(x) dx:
The Indefinite Integral is: ∫sin(x) dx = −cos(x) + C
Since we are going from 0, can we just calculate the integral at x=1?
−cos(1) = −0.540…
What? It is negative? But it looks positive in the graph.
Well … we made a mistake!
Because we need to subtract the integral at x=0. We shouldn’t assume that it is zero.
So let us do it properly, subtracting one from the other:
= [ −cos(x) ]
That’s better!
But we can have negative regions, when the curve is below the axis:
Example:
The Definite Integral, from 1 to 3, of cos(x) dx:
Notice that some of it is positive, and some negative.
The definite integral will work out the net value.
Let us do the calculations:
= [ sin(x) ]
So there is more negative than positive with a net result of −0.700….
So we have this important thing to remember:
Try integrating cos(x) with different start and end values to see for yourself how positives and negatives work.
Positive Area
But sometimes we want all area treated as positive (without the part below the axis being subtracted).
In that case we must calculate the areas separately, like in this example:
Example: What is the total area between y = cos(x) and the x-axis, from x = 1 to x = 3?
This is like the example we just did, but now we expect that all area is positive (imagine we had to paint it).
So now we have to do the parts separately:
The curve crosses the x-axis at x = π/2 so we have:
From 1 to π/2:
= sin(π/2) − sin(1)
From π/2 to 3:
= sin(3) − sin(π/2)
That last one comes out negative, but we want it to be positive, so:
Total area = 0.159… + 0.859… = 1.018…
This is very different from the answer in the previous example.
Continuous
Oh yes, the function we are integrating must be Continuous between a and b: no holes, jumps or vertical asymptotes (where the function heads up/down towards infinity).
Example:
A vertical asymptote between a and b affects the definite integral.
Properties
Area above − area below
The integral adds the area above the axis but subtracts the area below, for a “net value”:
Adding Functions
The integral of f+g equals the integral of f plus the integral of g:
Reversing the interval
Reversing the direction of the interval gives the negative of the original direction.
Interval of zero length
When the interval starts and ends at the same place, the result is zero:
Adding intervals
We can also add two adjacent intervals together:
Summary
The Definite Integral between a and b is the Indefinite Integral at b minus the Indefinite Integral at a.
Assignment
ASSIGNMENT : INTEGRATION ASSIGNMENT MARKS : 20 DURATION : 1 week, 3 days