Calculus Weeks 3&4

Calculus Weeks 3&4 More on integrals and their applications

Integration by parts This is another way to try and integrate products. It is in fact the opposite of the product rule for derivatives: Product rule for derivatives: d(uv) = u dv + v du . Integration by parts: Use this when you have a product under the integral sign, and it appears that integrating one factor and differentiating the other will make the resulting integral easier.

Example: -- There's no other rule for this, so we try parts. If we differentiate x2 , we'll get 2x which seems simpler. And integratingex doesn't change anything. Let u = x and dv = ex dx. Then du = 2x dx and v = ex. The integration by parts formula then gives us:

Continuing: We can use parts again on this integral (with u = 2x and dv = ex dx) to get:

You try one... …contrast this with

XOXXOX There is a method called the "tic-tac-toe" method for integration by parts. Different strategies for choosing uand dv are called for to integrate xln x or x tan-1 x orsomething like ex sin x x

A problem for you... p x cos(x) dx Find the value of 0 A) p B) 2p C) 2 D) 0 E) -2 F) 1 G) 1/2 H) p/2

Last one for today! 1 x ln xdx Evaluate 0 A) -1/4 B) -1/2 C) 0 D) 1/4 E) 1/2

Trigonomeric substitutions These are just substitutions like the ordinary substitution method, but some of them are a little surprising. There are two kinds: To integrate products of powers of sine and cosine (like sin (x) cos (x) ) . You need the identities: sin (x) + cos (x) = 1 (everybody knows that one!), and the double angle formulas: cos (x) = and sin (x) = 1 1 The first, 3 6 2 2 1+cos(2x) 2 2 1-cos(2x) 2 2

The trick is as follows... (a) If both the power of sine and the power of cosine are even, then use the double angle formulas to divide both in half. Keep doing this until at least one of the powers is odd. (b) Once at least one of the powers is odd, say it's the power of cosine, then let u = sin( -- ) -- then use the Pythagorean identity to convert all but one power of cosine to sines, and the last power of cosine will be the du in a substitution.

Let’s do one... Both the powers are even, so use the double angle formula trick: In the last two terms, use the identities again to get what’s on the next slide.

Trig identity city! …and (finally!) this is something we can integrate!

…and at last: WOW!!

Here’s one with an odd power for you to do: A) 2/15 B) 4/15 C) 2/5 D) 8/15 E) 2/3 F) 4/5 G) 14/15

Try this one: A) 2 B) p C) p - 1/2 D) 2 p E) 3p/8 p cos x dx = 4 0

2 2 The second, -- and more important, kind of trig substitution happens when there is a sum or difference of squares in the integrand. Usually one of the squares is a constant and the other involves the variable. If some other substitution doesn't suggest itself first, then try the Pythagorean trig identity that has the same pattern of signs as the one in the problem. Either , , or

Example: If you think about it, the substitution u = 5 - x2 won’t work, because of the extra factor of x in the numerator. But 5 - x2is vaguely reminiscent of 1 - sin2, so let Then and

We make all the substitutions and get: This is a trig integral of the other sort . And we can use the double-angle formula to get:

Now we have to get back to x’s So far, where therefore and so Also, And from the triangle,

The conclusion is: Quite a bit of work for one integral!

Examples for you… Here are two for you to work on. Notice the subtle difference in the integrand that changes entirely the method used:

Partial Fractions Last but not least is integration by partial fractions. This method is based on an algebraic trick. It works to integrate a rational function (quotient of polynomials) when the degree of the denominator is greater than the degree of the numerator (otherwise "when in doubt, divide it out") and you can factor the denominator completely. In full generality, partial fractions works when the denominator has quadratic and repeated factors, but we will consider only the case of distinct linear factors (i.e., the denominator factors into linear factors and they're all different). It also uses the (easy) fact that

Partial Fractions (continued) The idea of partial fractions is to take a rational function of the form (where there can be more or fewer factors in the denominator, but the degree of p(x) must be less than the number of factors) and rewrite it as a sum: for some constants A, B, C (etc..).

For instance, You can rewrite , first by factoring the denominator: and then in partial fractions as:

? Two questions... 1. Why do partial fractions? 2. How to do partial fractions?

Two reasons for why: First reason: It helps us to do integrals. From the previous example, we see that we don't know how to integrate the fraction right away, but ...

Two reasons for why: Second reason: Partial fractions helps us to understand the behavior of a rational function near its "most interesting” points. For the same example, we graph the function in blue, and the partial fractions and in red: Let’s take a closer look

red and blue curves Note that one or the other of the red curves mimics the behavior of the blue one at each singularity.

OK, now for the “how” First, we give the official version, then a short-cut. Official version: It is a general fact that the original fraction will break up into a sum with one term for each factor in the denominator. So (in the example), write where A and B are to be determined. To determine A and B, pick two values of x other than x=-3 or x=-1, substitute them into the equation, and then solve the resulting two equations for the two unknowns A and B. For instance, we can put x=0, and get and for x=1, get . Solve to get A = 5/2 and B = -1/2 as we did before.

“Short cut” 1. Write the fraction with denominator in factored form, and leave blanks in the numerators of the partial fractions: 2. To get the numerator that goes over x + 3, put your hand over the x + 3 factor in the fraction on the left and set x = -3 in the rest. You should end up with 5/2. 3. To get the numerator that goes over x + 1, cover the x + 1 and set x = -1 (and you get -1/2). It's that simple.

Another example: First, the numerator and denominator have the same degree. So we have to divide it out before we can do partial fractions. and we can use partial fractions on the second term

Partial fractions gives: Use either the official or short-cut method to get A = -1 and B = 2. Therefore:

An example for you to try: 3 dx x (x-1) Find 2 A) 3/2 B) 4/3 C) ln 2 D) ln 3 E) ln (3/2) F) ln (4/3) G) ln (2/3) H) 3 ln (2)/2

Another example: 4 4x - 6dx x - 3x +2 = 2 3 A) ln (4/3) B) 2 + arctan (3) C) ln (9) D) ln (12/5) E) p/3 - arctan (1/4)

Arc length The length of a curve in the plane is generally difficult to compute. To do it, you must add up the little “pieces of arc”, ds. A good approximation to ds is given by the Pythagorean theorem: We can use this to find the length of any graph – provided we can do the integral that results!

Example Find the arclength of the parabola y = x2for x between -1 and 1. Since dy / dx = 2x, the element of arclength is so the total length is:

Conclusion So far, we have that the length is To do this integral, we will need a trig substitution. But, appealing to Maple, we get that

Can we do the integral? The arc length integral from before was: This is a trig substitution integral of the second kind: With the identity tan2 + 1 = sec2 in mind, let What about dt ? Since we have that These substitutions transform the integral into This is a tricky integral we need to do by parts!

To integrate Let Then But tan2 = sec2 - 1 , so rewrite the last integral and get

Still going…. It’s remarkable that we’re almost done. The integral of secant is a known formula, and then you can add the integral of sec3 to both sides and get So we’ve got this so far for where

We need a triangle! So far, with From the triangle, So

Definite integral: So far, we have Therefore To get the answer Maple got before, we’d have to rationalize the numerator inside the logarithm.

Surface Area The area of a surface of revolution is calculated in a manner similar to the volume. The following illustration shows the paraboloid based on (for x=0..2) that we used before, together with one of the circular bands that sweep out its surface area.

To calculate the surface area, we first need to determine the area of the bands. The one centered at the point (x,0) has radius and width equal to . Since we will be integrating with respect to x (there is a band for each x), we'll factor the dx out of ds and write . So the area of the band centered at (x,0) is equal to: Thus, the total surface area is equal to the integral To calculate the surface area

The surface area turns out to be: 13 3 p

End of methods of integration!…. DON’T FORGET: 1. Send in your homework. 2. Meet TAs and me in office hours 3. Keep in touch by email! 4. Have a great weekend!

Calculus Weeks 3&4