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Rough meaning of grades:
29-34 is A
23-28 is B
17-22 is C
11-16 is D
Regarding the quiz--if you do every homework problem that was assigned,
you'll have a severe case of deja vu on the quiz! On the exam, we do
not restrict ourselves like this, but you get to have a sheet of paper.
The first homework problem is to compute
Your first idea might be to do some sort of
substitution, e.g., , but
is nowhere to be seen and this simply doesn't work.
Likewise, integration by parts gets us nowhere.
However, a technique called ``inverse trig substitutions''
and a trig identity easily dispenses with the
above integral and several similar ones!
Here's the crucial table:
||Relevant Trig Identity
Inverse substitution works as follows. If we write , then
This is not the same as substitution. You can just apply
inverse substitution to any integral directly--usually you get
something even worse, but for the integrals in this section using
a substitution can vastly improve the situation.
If is a function, then you can even use inverse substitution
for a definite integral. The limits of integration are obtained as follows.
To help you understand this, note that as varies from
to , the function varies
, so is being integrated
over exactly the same values. Note also that (5.3.2) once
again illustrates Leibniz's brilliance in designing the notation
Let's give it a shot with (5.3.1).
From the table we use the inverse substition
Wow! That was like magic. This is really an amazing technique.
Let's use it again to find the area of an ellipse.
Consider an ellipse with radii
, so it
. An equation
for the part of an ellipse in the first quadrant is
Thus the area of the entire ellipse is
is because the integral computes
th of the area
of the whole ellipse.
So we need to compute
Obvious substitution with
...? nope. Integration by parts...? nope.
Let's try inverse substitution.
The table above suggests using
Thus the area is
If the ellipse is a circle, i.e.,
, this is
which is a well-known formula for the area of a circle.
Trigonometric substitution is useful for functions that
but not all at once!
. See the above table for how
to do each.
One other important technique is to use completing the square.
We complete the square
We do a usual substitution to get rid of the
Now we have an integral that we can do; it's almost
identical to the previous example, but with
(and this is an indefinite integral).
Of course, we must transform
back into a function in
that's a little tricky.
Here we use that
Also, to compute
draw a right triangle with side lengths
To compute this, we complete the square, etc.
[[Draw triangle with sides
Back to the integral, we have