finding definite integrals by substitution and solving separable differential equations

9
Finding Definite Integrals by Substitution and Solving Separable Differential Equations

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Example 8: Wrong! The limits don’t match! Using the original limits: Leave the limits out until you substitute back. This is usually more work than finding new limits

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Page 1: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Finding Definite Integrals by Substitution and

Solving Separable Differential Equations

Page 2: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Example 8:

240

tan sec x x dx

The technique is a little different for definite integrals.

Let tanu x2sec du x dx

0 tan 0 0u

tan 14 4

u

1

0 u du

We can find new limits, and then we don’t have to substitute back.

new limit

new limit

12

0

12

u

12

We could have substituted back and used the original limits.

Page 3: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Example 8:

240

tan sec x x dx

Let tanu x

2sec du x dx40

u du

Wrong!The limits don’t match!

42

0

1 tan2

x

2

21 1tan tan 02 4 2

2 21 11 02 2

u du21

2u

12

Using the original limits:

Leave the limits out until you substitute back.

This is usually more work than finding new limits

Page 4: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Example: (Exploration 2 in the book)

1 2 3

13 x 1 x dx

3Let 1u x

23 du x dx 1 0u

1 2u 122

0 u du

232

0

23

u Don’t forget to use the new limits.

322 2

3

2 2 23

4 23

Page 5: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Separable Differential Equations

A separable differential equation can be expressed as the product of a function of x and a function of y.

dy g x h ydx

Example:

22dy xydx

Multiply both sides by dx and divide both sides by y2 to separate the variables. (Assume y2 is never zero.)

2 2 dy x dxy

2 2 y dy x dx

0h y

Page 6: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Separable Differential Equations

A separable differential equation can be expressed as the product of a function of x and a function of y.

dy g x h ydx

Example:

22dy xydx

2 2 dy x dxy

2 2 y dy x dx

2 2 y dy x dx 1 2

1 2y C x C

21 x Cy

2

1 yx C

2

1yx C

0h y

Combined constants of integration

Page 7: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Example 9:

222 1 xdy x y edx

2

2

1 2 1

xdy x e dxy

Separable differential equation

2

2

1 2 1

xdy x e dxy

2u x2 du x dx

2

11

udy e duy

11 2tan uy C e C

211 2tan xy C e C

21tan xy e C Combined constants of integration

Page 8: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Example 9:

222 1 xdy x y edx

21tan xy e C We now have y as an implicit function of x.

We can find y as an explicit function of x by taking the tangent of both sides.

21tan tan tan xy e C

2

tan xy e C

Notice that we can not factor out the constant C, because the distributive property does not work with tangent.

Page 9: Finding Definite Integrals by Substitution and Solving Separable Differential Equations

Until then, remember that half the AP exam and half the nation’s college professors do not allow calculators.

In another generation or so, we might be able to use the calculator to find all integrals.

You must practice finding integrals by hand until you are good at it!