### Ch. 5 Review: Integrals

```Ch. 5 Review:
Integrals
{
AP Calculus
Ch. 5 Test Topics
5.2: The Differential dy
5.2: Linear Approximation
5.3: Indefinite Integrals
5.4: Riemann Sums (Definite Integrals)
5.5: Mean Value Theorem/Rolle’s Theorem
Tangent line
dx & dy: change in x and y for tangent (derivative)
∆  ∆: ℎ
The Differential dy

=  ′  ,   = ′() ∙

Find the differential dy:
y = 3 2 − 4 + 2
dy = (6x – 4) dx
Write the equation of the line that best
fits  = 3 2 − 7 at x = 2. Then find ∆, ∆,
dx, and dy if f(2.01) is approximated.
∆
Equation:
∆
dx
dy
Linear Approximation
Write the equation of the line that best
fits  = 3 2 − 7 at x = 2. Then find ∆, ∆,
dx, and dy if f(2.01) is approximated.
Point of tangency: f(2) = -2
Slope of tangent (deriv):
y’ = 6x – 7 when x = 2  5
Sub into pt-slope equation:
y – 1 = ( − 1 )
y + 2 = 5(x – 2)  y = 5x – 12
If x = 2.01, y = -1.95
∆: Function change in x: 2.01 − 2 = .01
∆ : Function change in y: f(2.01) – f(2) = .0503
dx: Tangent line change in x -- 2.01 – 2 = .01
dy: Tangent line change in y for x = 2 to 2.01: -1.95 - -2 = .05
or dy = f’(x) dx at x = 2  (6(2) – 7)(.01) = .05
Linear Approximation
If a function is
continuous and
differentiable on the
interval [a, b], then there
is at least one point x = c
at which the slope of the
tangent equals the slope
of the secant connecting
f(a) and f(b)
Mean Value Theorem
If a function f is:
1)
Differentiable for all values of x in the
open interval (a, b) and
2)
Continuous for all values of x in the
closed interval [a, b]
Then there is at least one number x = c in
(a, b) such that
f’(c) =
−()
−
Mean Value Theorem (MVT)
If a function is
differentiable and
continuous on the
interval [a, b], and
f(a) = f(b) = 0, then
there is at least one
value x = c such
that f’(c) = 0.
Rolle’s Theorem
Remember – Function must be
CONTINUOUS and DIFFERENTIABLE on
interval! Otherwise, conclusion of MVT may
not be met.
Mean Value Theorem
8 1/3  =
5 4 + 1  =
(7 + 3)8  =
52  =
5 tan 5  =
Integrals Self-Quiz
8 1/3  = 6 4/3 +
5 4 + 1  =  5 +  +
(7 +
3)8
1
=
(7 + 3)9 +
63
5
52  = − cos 2 +
2
1
5 tan 5  = sec 5 +
5
Integrals Self-Quiz
sin    =
( 2 − 3)5  =
4    =
2  3 − 7  =
Integrals Self-Quiz
sin    =   +
( 2
− 3)5
1 2
=
( − 3)6 +
12
1
= −  5  +
5
4
2
3
2 5
− 7  =  − 7 2 +
5
Integrals Self-Quiz
R Problems, pg. 260:
R1 –R5 ab
```