W03 Regular

Due date: Sunday 2/1, 11:59pm

Partial fractions

01

04

Partial fractions - irreducible quadratic

Compute the integral:

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Solution

Solutions - 0060-04

(1) Perform long division:

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(2) Use to integrate:

Recall formula:

Choose . Then:

The final answer is therefore:

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02

05

Partial fractions - long division

Compute the integral:

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Solution

Solutions - 0060-05

(1) Numerator degree is not smaller! Long division first:

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(2) Factor denominator:

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(3) Write the partial fractions general form equation (for the second term):

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(4) Solve for constants:

Cross multiply:

Plug in , obtain so .

Plug in , obtain so .

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(5) Integrate by terms:

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03

06

Partial fractions - big generic

Give the generic partial fraction decomposition (no need to solve for the constants):

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Solution

Solutions - 0060-06

(1) Observe that :

On the other hand, cannot be factored further. (Its zeros are imaginary.)

Now all denominator factors are either linear or irreducible quadratic.

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(2) Write the partial fractions general form equation:

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(3) Notice a few things:

• Quadratic acquires linear term on top
• Linear is to 3rd power so it has repetition up to 3rd power
• Linear and are only to 1st power.

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04

07

Partial fractions - linear and quadratic

Compute the integral:

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Solution

Solutions - 0060-07

(1) Denominator has degree 3, numerator has degree 2, therefore long division is not necessary.

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(2) Write the partial fractions general form equation:

Notice that is an irreducible quadratic (cannot be factored). So we have:

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(3) Solve for constants:

Cross multiply:

Plug in , obtain:

Expand RHS:

Comparing terms, obtain: and thus .

Comparing constant terms, .

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(4) Integrate by terms:

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Note A: For the last term, use the formula:

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05

08

Partial fractions - repeated factor

Compute the integral:

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Solution

Solutions - 0060-08

(1) Write the partial fractions general form equation:

Observe that appears in degree 3 in the integrand, so we have one term for each power up to 3 in the partial fraction decomposition.

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(2) Solve for constants:

Cross multiply:

Plug in , obtain so .

Plug in , obtain .

Plug in , obtain:

Plug in , obtain:

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(3) Integrate each term:

Optional simplification:

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06

09

Partial fractions - rationalize first

For each of these integrals, make a -substitution that changes the integrand into a rational function. Write the integral in terms of for your answer. You do not have to compute the -integral.

(a)

(b)

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Solution

Solutions - 0060-09

(a) (1) -substitution with and :

Notice that . Then:

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(2) Partial fractions:

Therefore:

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(b) (1) Set :

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(2) Partial fractions:

Therefore:

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Simpson’s Rule

07

02

Simpson’s Rule for volume by shells

Use Simpson’s Rule with to compute the volume of the solid obtained by revolving the pictured region about the -axis. Can you do it without using a calculator?

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Solution

Solutions - 0070-02

(1) Recall shells formula:

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(2) Interpret:

Bounded above by . Bounded below by -axis.

Bounded left by . Bounded right by .

Obtain:

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(3) Create table of values to apply Simpson’s Rule:

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(4) Recall Simpson’s Rule formula:

Here since in this formula represents the integrand values.

Note that .

Plug in:

Therefore:

Therefore:

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08

03

Area of a garden bed

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Solution

Solutions - 0070-03

Note: you can also start from and . This gives a different answer, .

(1) Set up integration:

Set at the left upper corner, with extending to the right, extending downwards. Then:

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(2) Create table of values:

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(3) Recall Simpson’s Rule formula:

Here and .

Thus:

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(4) Compute cubic yards from known surface area:

Mulch is deep, so the volume is:

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