Integration by Parts and Improper Integrals

MAT 1300 B

Fall 2011 1 UNDOING THE PRODUCT RULE 2

1 Undoing the Product Rule

Recall the product rule for derivatives:

d [f(x)g(x)] = f 0(x)g(x) + f(x)g0(x) dx

Alternatively, if we let u = f(x) and v = g(x),

d du dv [uv] = v + u dx dx dx dv d du u = [uv] − v dx dx dx Z dv Z d du u dx = [uv] − v dx dx dx dx Z dv Z d Z du u dx = [uv] dx − v dx dx dx dx Z dv Z du u dx = uv − v dx dx dx 2 INTEGRATION BY PARTS 3

2 Integration by Parts

Integration by Parts Formula:

For indeﬁnite integrals: Z dv Z du u dx = uv − v dx dx dx For deﬁnite integrals:

Z b Z b dv b du u dx = [uv]a − v dx a dx a dx

Rules of Thumb:

1. Choose u to be a function that gets no more complicated when you take its derivative.

dv 2. Choose dx to be a function that gets no more complicated when you take its integral.

Ex: Find Z xex dx. 2 INTEGRATION BY PARTS 4

Notes:

• It is crucial to make the right choice. Making the wrong choice for u and v0 will usually give you a harder integral than the one you started with. For example, say you were asked to solve Z xexdx

and chose u = ex v0 = x giving you x2 u0 = ex v = 2 Then integration by parts would give you Z x2 Z x2 xex = ex − exdx 2 2 | {z } even harder We ended up with a worse integral, so we know we made the wrong choice for u and v0.

• Sometimes you have to do integration by parts twice. For example, say you were asked to solve Z x2exdx

The reasonable thing to do would be to choose

u = x2 v0 = ex

giving you u0 = 2x v = ex Then integration by parts gives you Z Z x2exdx = x2ex − 2xexdx 2 INTEGRATION BY PARTS 5

The second integral we can now do, but it also requires parts. We take

u = 2x v0 = ex

giving us u0 = 2 v = ex So we have Z Z x2exdx = x2ex − 2xex − 2exdx = x2ex − 2xex + 2ex + C

In general, you need to do n integration by parts to evaluate R xnexdx.

• In the case of deﬁnite integrals, the integration by parts formula be- comes Z b b Z b 0 0 uv dx = uv − u vdx a a a

Ex: Find Z 2 4x ln(x) dx. 1 2 INTEGRATION BY PARTS 6

Ex: Find Z ln(x) dx. 2 INTEGRATION BY PARTS 7

Ex: Find Z 3 5x(x + 1)1/2 dx. 0 3 APPLICATIONS 8

3 Applications

Ex: The price per unit, p, in dollars for a product ﬂuctuates with time, t, in years according to p = t1/5 ln(t). Find the average price per unit during the second year. 4 IMPROPER INTEGRALS 9

4 Improper Integrals

Consider y = e−x, and the area under it and above [1, 7].

R 7 −x We know how to calculate this already - its just 1 e dx. But suppose that we wanted to know the area under y = e−x but above [1, ∞). Since e−x approaches the x-axis very quickly its possible that the total area is ﬁnite. R ∞ −x What we want is 1 e dx. But what does this mean? We can calculate R 100 −x R 10000000 −x 1 e dx, 1 e dx and so on. Each provides a better approximation R ∞ −x R ∞ −x to 1 e dx. So our intuition should be that 1 e dxis the limit of this.

If f(x) is continuous on [a, b) but not at b (b might be ∞) then Z b Z d f(x) dx = lim f(x) dx − a d→b a is an improper integral. If this limit exists then we say that the improper integral converges. If the improper integral does not exist then we say that the improper integral diverges. If f(x) is continuous on (a, b] but not at a (a might be −∞) then Z b Z b f(x) dx = lim f(x) dx + a d→a d 4 IMPROPER INTEGRALS 10 is also an improper integral. 4 IMPROPER INTEGRALS 11

R ∞ −x 1. Evaluate 1 e dx. solution:

Z ∞ Z b e−x dx = lim e−x dx 1 b→∞ 1  b  −x = lim −e  b→∞ 1 = lim −e−b + e−1 b→∞ 1 = e

−x 1 So the area under the curve y = e and above [1, ∞) is e .

∞ R √1 2. Calculate 2 x dx solution: The area in question looks like this:

√1 The function y = x approaches the x-axis as x goes to inﬁnity, so it 4 IMPROPER INTEGRALS 12

is possible that there will only be a ﬁnite area.

Z ∞ 1 Z b 1 √ dx = lim √ dx 2 x b→∞ 2 x  b  x1/2 = lim   b→∞ 1/2 2 h √ √ i = lim 2 b − 2 2 b→∞ = ∞

Thus the area under the curve is inﬁnite. In cases where the limit doesn’t exist (ie is inﬁnite) we say the improper integral diverges.

R 0 x 3. Calculate −∞ e dx. solution: Here our limit is negative inﬁnity, but we deal with it in the same way 4 IMPROPER INTEGRALS 13

Z 0 Z 0 ex dx = lim ex dx −∞ b→−∞ b  0 x = lim e  b→−∞ b = lim 1 − eb b→−∞ = 1

Ex: Compute Z ∞ 1 2 dx. 1 x

Ex: Compute Z 1 1 2 dx. 0 x 4 IMPROPER INTEGRALS 14

Ex: Compute Z 3 1 1/3 dx. −2 x 5 PRESENT VALUE 15

5 Present Value

For a continuous income function, c(t), in dollars per year at time t in years, the total actual income earned by time T is

Z T c(t) dt. 0

If the inﬂation rate is r then the present value of the total income earned by time T is Z T c(t) e−rt dt. 0

Ex: Compute the total income earned over a 4 year contract where the continuous income function is

c(t) = 90000 + 11000t.

Compute the present value of the income if the inﬂation rate is 1%. 5 PRESENT VALUE 16

Ex: A perpetuity pays out 120000 per year in a continuous manner. Compute the total income earned by the perpetuity from now until the end of time. If the inﬂation rate is constant at 1% during all that time, compute the present value of the perpetuity. 6 APPLICATION: CONSUMER AND PRODUCER SURPLUS 17

6 Application: Consumer and Producer Sur- plus

As discussed earlier a demand function p = D(x) relates the price of a product and the number of that product sold (or, looking at it a diﬀerent way, the number of units consumers are willing to buy at that price). On the ﬂip side of this is a supply function. A supply function gives price as a function of units, but instead relates the price to how many units a producer will make at that price. A supply function is usually given as p = S(x).

The point (x0, p0) at which a demand function p = D(x) and a supply function p = S(x) meet is called the equilibrium point.

The area below the demand function and above the equilibrium price is called the consumer surplus. It represents the total money saved by consumers who were otherwise willing to pay more than the equilibrium price. The area above the supply function but below the equilibrium price is called the 6 APPLICATION: CONSUMER AND PRODUCER SURPLUS 18 producer surplus. It represents the additional money made by producers who were otherwise willing to produce for less than the equilibrium price.

We can see that both the consumer surplus and producer surplus are just areas between curves. Therefore we have the formulas:

Z x0 CS = (D(x) − p0) dx 0

Z x0 PS = (p0 − S(x)) dx 0

Examples:

1. Suppose D(x) = −x2 − 10x + 1200 and S(x) = x2 + 40x. Find the consumer surplus and producer surplus. solution: To ﬁnd the equilibrium price we have to ﬁnd where the func- tions intersect, so we set D(x) = S(x):

D(x) = S(x)

−x2 − 10x + 1200 = x2 + 40x 2x2 + 50x − 1200 = 0 2(x + 40)(x − 15) = 0 So the curves intersect when x = 15 and x = −40. Since x represents the number of units produced we only consider the positive intersection

point. Thus x0 = 15. To ﬁnd p0 we can plug x0 into either equation:

2 p0 = S(x0) = 15 + 40(15) = 825 6 APPLICATION: CONSUMER AND PRODUCER SURPLUS 19

So our equlibrium point is (15, 825). We can now calculate our sur- pluses:

Z x0 CS = (D(x) − p0) dx 0 Z 15 = ((−x2 − 10x + 1200) − 825) dx 0 Z 15 = (−x2 − 10x + 375) dx 0 15 x3 = − − 5x2 + 375x 3 0 = $3375

Z x0 PS = (p0 − S(x)) dx 0 Z 15 = (825 − x2 − 40x) dx 0 15 x3 = − − 20x2 + 825x 3 0 = $6750

x 2. Suppose D(x) = 15 − x and S(x) = 3 + 2 . Find the consumer surplus and producer surplus. solution: We ﬁrst ﬁnd the equilibrium point as before:

D(x) = S(x) x 15 − x = 3 + 2 3x = 12 2 x = 8 6 APPLICATION: CONSUMER AND PRODUCER SURPLUS 20

So x0 = 8 and thus p0 = 7.

Z x0 CS = (D(x) − p0) dx 0 Z 8 = ((15 − x − 7) dx 0 Z 8 = (8 − x) dx 0 8 x2 = 8x − 2 0 = $32

Z x0 PS = (p0 − S(x)) dx 0 Z 8 x = (7 − 3 − ) dx 0 2 8 x2 = 4x − 4 0 = $16