unit #11 : integration by parts, average of a function

Unit #11 : Integration by Parts, Average of a Function

Goals:

• Learning integration by parts.

• Computing the average value of a function.

Integration Method - By Parts - 1

Integration by Parts

So far in studying integrals we have used

• direct anti-differentiation, for relatively simple functions, and

• integration by substitution, for some more complex integrals.

However, there are many integrals that can’t be evaluated with these techniques.

Try to find

∫xe4x dx.


This particular integral can be evaluated with a different integration technique,integration by parts. This rule is related to the product rule for derivatives.Expand

d

dx(uv) =

Integrate both sides with respect to x and simplify.

Express

∫udv

dxdx relative to the other terms.


Integration by PartsFor short, we can remember this formula as∫

udv = uv −∫vdu

Integration by parts:

• Choose a part of the integral to be u, and the remaining part to be dv.

• Differentiate u to get du.

• Integrate dv to get v.

• Replace

∫u dv with uv −

∫vdu.

• Hope/check that the new integral is easier to evaluate.

Integration By Parts - Example 1 - 1

∫udv = uv −

∫vdu

Example: Use integration by parts to evaluate

∫xe4x dx.


Verify that your anti-derivative is correct.


∫udv = uv −

∫vdu

Guidelines for selecting u and dv

• Try to select u and dv so that either

– u′ is simpler than u or

–∫dv is simpler than dv

• Ensure you can actually integrate the dv part by itself


Example: Consider the integral

∫x cosx dx.

Based on the guidelines, what choice of u and dv should you try first?

(a) u = x cos(x), dv = dx.

(b) u = 1, dv = x cos(x) dx.

(c) u = x, dv = cos(x) dx.

(d) u = cos(x), dv = x dx.


Evaluate the integral

∫x cosx dx.


Verify that your anti-derivative is correct.


Example: Evaluate the slightly more challenging integral∫x2 cosx dx

What choice of u and dv would be most likely to be helpful?

(a) u = 1, dv = x2 cos(x) dx.

(b) u = x, dv = x cos(x) dx.

(c) u = x2, dv = cos(x)dx.

(d) u = x2 cos(x), dv = dx.

Integration By Parts - Example 3 - 2∫x2 cosx dx

Integration By Parts - Example 3 - 3 ∫x2 cosx dx

Integration By Parts - Definite Integrals - 1

Integration By Parts - Definite Integrals

When using integration by parts to evaluate definite integrals, you need to applythe limits of integration to the entire anti-derivative that you find.

Example: Evaluate

∫ π

0

x sin(4x) dx

Integration By Parts - Definite Integrals - 2

Don’t forget that dv does not require any other factors besides dx. That can helpwhen there is only a single factor in the integrand.

Example: Find

∫ 2

1

lnx dx

Integration By Parts - Circular Case - 1

There are some classical problems that can be solved by integration by parts, butnot in the direct way we have seen so far.

Example: Integrate by parts twice to find

∫cos(x)ex dx.

Integration By Parts - Circular Case - 2∫cos(x)ex dx

Applications of Integrals - Average Value - 1

Applications of Integrals - Average Value

Example: If the following graph describes the level of CO2 in the air in agreenhouse over a week, estimate the average level of CO2 over that period.

0 1 2 3 4 5 6 7

350

400

450

Time (Days)

CO

2

(p

pm

)

Give the units of the average CO2 level.


Example: Sketch the graph of f (x) = x2 from x = −2 to 2, and estimatethe average value of f on that interval.


What makes the single “average” f value different or distinct from otherpossible f values?

Use this property to find a general expression for the average value of f (x)on the interval x ∈ [a, b].


Average Value of a Function on [a, b]The average value of a function f (x) on the interval [a, b] is given by

A =1

b− a

∫ b

a

f (x) dx

Interpret this form of the average value graphically.


Example: Find the exact average of f (x) = x2 on the interval x ∈ [−2, 2]using this formula.

Average Value - Examples 1 - 1

A =1

b− a

∫ b

a

f (x) dx

Example: The temperature in a house is given by H(t) = 18 + 4 sin(πt/12),where t is in hours and H is degrees C. Sketch the graph of H(t) from t = 0to t = 12, then find the average temperature between t = 0 and t = 12.


H(t) = 18 + 4 sin(πt/12)


A =1

b− a

∫ b

a

f (x) dx

Example: You are told that the average value of f (x) over the interval

x ∈ [0, 3] is 5. What is the value of

∫ 3

0

f (x) dx?


Example: Without computing any integrals, explain why the average valueof cos(x) on x ∈ [0, π/2] must be greater than 0.5.


Can you make a general statement about the average value of decreasing func-tions which are concave down?

unit #11 : integration by parts, average of a function

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