Solving Definite Integrals

Lesson 13

Solving Definite Integrals

How to find antiderivatives We have three methods: 1. Basic formulas 2. Algebraic simplification 3. Substitution

Basic Formulas If f(x) is…

x

n

k cos(kx) sin(kx) ekx 1 x ax

…then an antiderivative is…

x n +1 except if n = –1 kx (assuming the variable is x!)

1 n +1

sin(kx)/k –cos(kx)/k ekx /k

ln x ax /ln(a)

Algebraic Simplification 2 1 3 ( x ! 1)( x + 1) dx = x ! 1 dx = 3 x ! x+C " "

! (3x + 1) dx = 2

2 3 2 9 3 6 2 9 x + 6 x + 1 dx = x + x + x + C = 3 x + 3 x + x+C 3 2 !

!

x + x2 dx = x

!

x (1 + x ) dx = ! 1 + x dx = x + 12 x 2 + C x

Integrals by Substitution Start with Let u = g(x).

!

x sin(x 2 )dx

du = g!(x) so du = g!(x) dx dx

Let u =?

du

u = x 2 ! du = 2xdx

1 1 1 2 2 x sin( x ) dx = sin( x ) 2 x dx = sin( u ) du = ! cos( x )+C " " " 2 2 2 2

Antiderivative Practice 4t t 2sin(3t) ! e + 4 dt Use basic formulas: "

Problem 1

4t t 2 1 4t 2 sin(3t) ! e + 4 dt = ! cos(3t) ! + 3 4 e "

1 ln(4 )

4t + C

z 2 + 2z dz Simplify algebraically first, then integrate. Problem 2 ! 2 z z2 + 2z z2 2z 2 ! z 2 dz = ! z 2 + z 2 dz = ! 1 + z dz = z + 2 ln z + C Problem 3

!

6t dt 2 4+t

Make a substitution: Let u=4+t2, so du=2tdt.

6t 2t 1 2 dt = 3 dt = 3 du = 3ln u + C = 3ln 4 + t +C ! 4 + t2 ! 4 + t2 !u

Antiderivative Practice Problem 4

!

!

2 dy Make a substitution: u=ln(ky), so du=dy/y. y ln(ky)

2 2 1 2 dy = ! dy = ! du = 2 ln u + C = 2 ln ln(ky ) + C y ln(ky ) ln(ky ) y u

Problem 5 Find the particular function F(x) such that F'(x) = x2 and the graph of F(x) passes through (1, 2). The general antiderivative is ! x 2 dx = 13 x 3 + C 3 Then to find C, we must have F(1) = 13 1 + C = 2 Thus, C = 5/3, and our function is F(x) = 13 x 3 + 53

Solving Definite Integrals Theorem: (Fundamental Theorem I)

Or: If F is an antiderivative for f, then Example We determined using Simpson’s Rule : Now use the fundamental theorem: An antiderivative for f(x) = 3x + 5 is So:

!

12

0

3x + 5 dx = F(12) - F(0) = 276 - 0 = 276

Example

!

3

2

x 3 dx

We have to • find an antiderivative; • evaluate at 3; • evaluate at 2; • subtract the results.

!2 x dx = x 3 3

1 4

4 3 2

= 14 3 " 14 2 = 4

This notation means: evaluate the function at 3 and 2, and subtract the results.

4

81 4

" 146 = 645 = 16.25

Don’t need to include “+ C” in our antiderivative, because any antiderivative will work.

Examples "

!

2sin( x ) + 3x dx

0

)

!

0

3x 2 2sin( x ) + 3x dx = $2 cos x + 2

Alternate notation

= –1 !

0

= 1

" 3! 2 # = % $2 cos ! + $ ($2 cos 0 + 0 ) & 2 ( ' " 3! 2 # 3! 2 = % $2( $1) + $ ($2(1) + 0 ) = 4 + & 2 ( 2 '

1 "!2 s ds !1

1 "!2 s ds = ln s !1

!1 !2

= ln1 ! ln 2 =!! ln 2

Practice Examples !

5

!

9

1

2

1 4 dx = e e

3 s ds =

!

9

2

3s ds = 2 s 1/ 2

3/ 2

9 2

()

=2 9

3/ 2

()

"2 2

3/ 2

= 54 " 4 2

ses + 1 "!2 s ds !1 !1 1 s s = " e + ds = (e + ln s ) !2 = e!1 + ln1 ! (e!2 + ln 2) !2 s 1 1 =!! ! 2 ! ln 2 e e !1

Substitution in Definite Integrals • We can use substitution in definite integrals. • However, the limits are in terms of the original variable. • We get two approaches:

– Solve an indefinite integral first – Change the limits Method I: First solve an indefinite integral to find an antiderivative. Then use that antiderivative to solve the definite integral. Note: Do not say that a definite and an indefinite integral are equal to each other! They can’t be.

Example First: Solve an indefinite integral.

!

3t 3 dt = 2 t +4 2

!

!

2t 3 dt = 2 t +4 2

u = t2 + 4

3t dt 2 t +4

!

du = 2t dt

1 du = 32 ln t 2 + 4 + C u Here’s an antiderivative!

2t dt becomes du

Pull out the 3, and put in a 2.

Second: Use the antiderivative to solve the definite integral.

!

2

1

3t dt = 2 t +4

(

3 2

ln t + 4 2

Here’s the antiderivative we just found.

) ( 2

1

=

3 2

) (

ln 8 "

3 2

)

ln 5 = 23 ln 85

Example When discussing population growth, we worked backwards to find out what we got from evaluating

P!(t ) Let’s find an antiderivative using substitution. " dt P (t )

"

P !(t) dt = P(t)

"

u = P(t) du = P'(t) dt

1 du = ln u + C = ln P(t) + C u

Of course, P(t) is always non-negative, so we don’t need absolute values… $ P(b) ' P !(t) b ) So we get: "a dt = ln(P(t)) a = ln (P(b)) # ln (P(a)) = ln& P(t) % P(a) ( b

Method II: Convert the Limits We start with x = a and x = b, and a substitution formula u = … Just put a and b into the substitution formula and get new limits. Note: You do not have to go back to x then! 2 3t dt Start with the same substitution Example !1 2 t +4

u = t2 + 4 du = 2t dt

8 3t 3 2 1 3 81 3 3 8 dt = 2 t dt = du = ln u = ln ( ) 2 2 5 !1 t 2 + 4 2 !1 t 2 + 4 5 2 !5 u When t = 1, u = 5. becomes du becomes u When t = 2, u = 8. What happens to t = 1? And when t = 2, 2 u = t2 + 4 = 1^2 + 4 = 5. u =  t  + 4 = 2^2 + 4 = 8.

2

Example !

8

1

)

8

1

y

y

( 5 y )dy 2

3

( 5 y )dy = ) 3

2

8

1

u = 5y2

y = 1: u = 5

du = 10y dy

y = 8: u = 320 4 3

1 1 320 1 8 1 1 u y (5 y ) dy = ) (5 y 2 )3 10 y dy = ) u 3 du = 10 1 10 5 10 4 3

Note that we can also do this problem without u-sub --try algebraic simplification

1 2 3

320

5

4 4 " 3 ! = % 320 3 # 5 3 & $ 163.5 40 ' (

!

8

1

y

(

3

5y

2

) dy = ! y (5y ) 8

1

1 2 3

1 5 " 1 2% 8 dy = ! $ 53 y 3 ' y dy = 53 ! y 3 dy 1 1 # &

8 1 3

5 +1 3

1 8 8 y 3 =5 = 53 (8 3 ( 13 ) ) 163.5 5 8 +1 3 1

8

!

1

e4 x

Practice Example

dx

1+ e Method I: Firstly compute 0

4x

u = 1 + e4 x du = 4e4 x dx

!

1

0

!

e4 x 1 + e4 x

e4 x

!

1+ e

4x

dx 1 " 2

1 " +1 2

1 4e4 x 1 1 1 1 u dx = ! dx = ! du = ! u du = +C 1 4 1 + e4 x 4 4 4 u " +1 2

1 12 1 + e4 x = u +C = +C 2 2

e4 x 1 + e4 x

dx =

1 1 1 1 1 1 + e4 x |10 = 1 + e4 " 1 + e0 = 1 + e4 " 2 2 2 2 2 2

u = 1 + e4 x u(0) = 1 + e4*0 = 2,u(1) = 1 + e4*1 = 1 + e4

Method II:

!

1

0

e

4x

1 + e4 x

dx =

4x

1 " 2

1 " +1 2

1 1 4e 1 1+ e 1 1 1+ e 1 u 1+ e4 dx = du = u du = | 2 4 !0 1 + e4 x 4 !2 4 !2 4 1 u " +1 2

1 12 1+ e4 1 + e4 2 = u |2 = " 2 2 2

4

4

Using Definite Integrals We can now evaluate many of the integrals that we have been able to set up. Example Find area between y = sin(x) and the x–axis from x = 0 to x = π, and from x = 0 to x = 2π. The area from 0 to π is clearly:

#

!

0

!

sin( x ) dx = " cos( x ) = 1 + 1 = 2 0

The area from 0 to 2π is more complicated. We note that But this is obviously not the area! The area from 0 to 2π can be found by:

#

!

0

2!

(

!

)(

2!

sin( x ) dx " # sin( x ) dx = " cos( x ) 0 + cos( x ) ! !

)= 4

"

2!

0

sin( x ) dx =0

Summary • Used the fundamental theorem to evaluate definite integrals. • Made substitutions in definite integrals – By solving an indefinite integral first – By changing the limits

• Used the fundamental theorem to evaluate integrals which come from applications.