The Fundamental Theorem of Calculus

We’ve learned two different branches of calculus so far: differentiation and integration. Finding slopes of tangent lines and finding areas under curves seem unrelated, but in fact, they are very closely related. It was Isaac Newton’s teacher at Cambridge University, a man name Isaac Barrow (1630 – 1677), who discovered that these two processes are actually inverse operations of each other in much the same way division and multiplication are. It was Newton and Leibniz who exploited this idea and developed the calculus into it current form.The Theorem Barrow discovered that states this inverse relation between differentiation and integration is called The Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus 1 (FToC1)

.

1.∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥=𝐹 (𝑏)−𝐹 (𝑎)

h𝑤 𝑒𝑟𝑒𝐹 𝑖𝑠 𝑎𝑛𝑦 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 (𝑖𝑛𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 )𝑜𝑓 𝑓 , h𝑠𝑢𝑐 h𝑡 𝑎𝑡 𝐹 ′= 𝑓The integral gives the NET change

𝐹 (𝑥 )=∫𝑎

𝑥

𝑓 (𝑡 ) 𝑑𝑡 ,𝑎≤ 𝑥≤𝑏2. h𝑇 𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝐹 ′ (𝑥 )= 𝑑𝑑𝑥∫𝑎

𝑥

𝑓 (𝑡 ) 𝑑𝑡= 𝑓 (𝑥)

Fundamental Theorem of Calculus 2 (FToC2)

This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.

FToC1 bridges the antiderivative concept with the area problem.

To evaluate an integral, take the antiderivatives and subtract.

It can be proved and the proof can be found elsewhere (WEB)

It is a “shortcut” rule for integration: an easier way (from Riemann sums and other methods) to calculate definite integrals.

𝑭𝑻𝒐𝑪𝟏∫𝒂

𝒃

𝒇 (𝒙 ) 𝒅𝒙=𝑭 (𝒃 )−𝑭 (𝒂)

𝑭𝑻𝒐𝑪𝟐 (𝒑𝒂𝒓𝒕𝟏)𝑭 (𝒙)=∫𝒂

𝒙

𝒇 (𝒕 )𝒅𝒕where is a continuous function on, and varies between and . Notice that this integral equation is a function of , which appears as the upper limit of integration. If happens to be positive, and we let then we can define as the area under the curve from to

𝐹 ′ (𝑥 )=cos𝑥

Lower limit of integration is a constant.

20

1

1+t

xddt

dx∫ 2

1

1 x

𝑭𝑻𝒐𝑪 2 (𝒑𝒂𝒓𝒕 2)𝑭 ′ (𝒙 )= 𝒅𝒅𝒙∫𝒂

𝒙

¿¿

𝐹 (𝑥 )=∫−𝜋

𝑥

cos𝑡 𝑑𝑡 𝐹 ′ (𝑥 )=?

𝐹 ′ (𝑥 )= 𝑑𝑑𝑥 ∫

−𝜋

𝑥

cos𝑡 𝑑𝑡= 𝑑𝑑𝑥

¿¿¿

Or, longer way:

Directly:

¿𝑑𝑑𝑥

(sin 𝑥−sin (−𝜋 ))=𝑐𝑜𝑠𝑥

53 sin

x

dt t dt

dx∫

53 sin xdt t dt

dx ∫

3 sinx x

=

=

2

0cos

xdt dt

dx∫

2 2cosd

x xdx

2cos 2x x 22 cosx x

The upper limit of integration does not match the derivative, but we could use the chain rule.

= = =

The lower limit of integration is not a constant, but the upper limit is.

We can change the sign of the integral and reverse the limits.

What if upper limit is g(x) not x itself?Or upper limit is constant and lower limit is x?

FToC 2, the most general form

𝑑𝑑𝑥 [ ∫

h(𝑥)

𝑔(𝑥)

𝑓 (𝑡 ) 𝑑𝑡= 𝑓 [𝑔(𝑥 )] ∙𝑔 ′ (𝑥 )− 𝑓 [h (𝑥 ) ] ∙ h ′ (𝑥)]2

2

1

2

x

tx

ddt

dx e∫2

2

1

2

x

tx

ddt

dx e∫

2 2

1 12 2

22xx

xee

¿

Example 10:Evaluate the following using the FTOC2, then if feeling bored verify by doing in the Loooooooong way.

The Mean Value Theorem (for Integrals)

∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥= 𝑓 (𝑐) (𝑏−𝑎 )

Where is called the average value of the function on the interval . The above equation above can be explicitly solved for .

Geometric interpretation: For a positive function , there exists at least one number such that the rectangle with base and height has the same area as the region under the graph of from to .

Find the mean (average) value of the function over the interval

𝐴𝑉= 13−1

∫1

3

(𝑥2+2   )𝑑𝑥

𝐴𝑉=12 ( 𝑥

3

3+2𝑥)

𝐴𝑉=12 [( 27

3+6)−( 1

3+2)]

𝐴𝑉=12 [15−

73 ]

𝐴𝑉=193

=6.33

3

1

𝐴𝑣𝑔𝑣𝑎𝑙𝑢𝑒=∫

0

12

[50+14𝑠𝑖𝑛( 𝜋𝑡12 )]𝑑𝑡12−0

= 112 [50 𝑡−14

12𝜋𝑐𝑜𝑠( 𝜋𝑡12 )]|0

12

=50+214𝜋≈58.913

𝑓 (𝑐 )=∫−1

2

[1+𝑥2 ]𝑑𝑥

2−(−1)=1

3 [𝑥+𝑥3

3 ]|−1

2

=13 [2+ 8

3−(−1)−(− 1

3 )]=2

1 + x2 = 2 for x = ⟹

For positive function , there exist two numbers such that the rectangle with base and height has the same area as the region under the graph of from -1 to 2.

The average rate of change of a car’s position over an interval is represented graphically as the slope of the secant line to the graph position/time over the interval.

𝑣𝑎𝑣𝑔=𝑥(𝑡2)−𝑥 (𝑡1)

𝑡 2−𝑡 1

Average value of velocity over the same interval is:

𝑣𝑎𝑣𝑔=1

𝑡2−𝑡 1∫𝑡 1

𝑡 2

𝑣 (𝑡)𝑑𝑡

𝑝𝑟𝑜𝑜𝑓 :1

𝑡2−𝑡1∫𝑡 1

𝑡 2

𝑣 (𝑡)𝑑𝑡=1

𝑡2−𝑡1∫𝑡 1

𝑡 2

𝑥 ′(𝑡 )𝑑𝑡=1

𝑡2−𝑡1[𝑥 (𝑡 2)−𝑥(𝑡1) ]𝑏𝑦 𝐹𝑇𝑜𝐶 1

𝑓 𝑎𝑣𝑔=∑ 𝑓 (𝑥 )

7≈40.429

Area

area = 1160 average=1160/30 38.667

trapezoidal approximation is more accurate that LRAM (basically arithmetic mean)

Sometimes we might have to solve an integral equation! Being able to simplify definite integrals with variables in the interval of integration is important. Here are a couple of examples showing an important application that is important.

1𝑏∫0

𝑏

( 2+6 𝑥−3𝑥2 )𝑑𝑥=3

2+3𝑏−𝑏2=3→𝑏2−3𝑏+1=0→𝑏=3±√52

(𝑏) 𝑓 (𝑥)= 𝑓 (0)+∫0

𝑥

𝑓 ′ (𝑥 )𝑑𝑥 𝑓 (2)=5+∫0

2

(2𝑥2−2 )𝑑𝑥=193

(𝑎 ) 𝑓 (𝑥 )=∫ ( 2𝑥2−2 )𝑑𝑥=23𝑥3−2𝑥+𝐶

𝑓 (0 )=5=𝐶

In the previous example, the first method relied heavily upon our ability to find the antiderivative of the integrand. This is not always easy, possible, or prudent! Being able to express a particular value of a particular solution to a derivative as a definite integral is of paramount importance, especially when we don’t know how to find a general antiderivative. (calculator can do easily definite integrals – see problem 20) Hard Facts To Refute:A. Where you are at any given time is a function of 1) where you started and 2) where you’ve gone from your starting point (displacement).B. What you have at any given moment is a function of 1) what you started with plus 2) what you’ve accumulated since then.When you accumulate at a variable rate, you can to use the definite integral to find your net accumulation.Important Idea of Accumulation***************************(* means VERY IMPORTANT)

What one has now = What one started with + What one has accumulated since one started. This can be expressed mathematically as

Example 19:

If and find

(a) and integral equation for f(x) (b) f(3) (c) f(-2)

𝑎 ¿ 𝑓 (𝑥 )= 𝑓 (2 )+∫2

𝑥

4 𝑠𝑖𝑛2 (2 𝑥 )𝑑𝑥= 𝑓 (2 )+2∫2

𝑥

(1− cos 4 𝑥 )𝑑𝑥=¿¿

¿ 𝑓 (2 )+2 𝑥|2𝑥−

12

sin 4 𝑥|2

𝑥

=−2+2𝑥−4−sin 4 𝑥+ 12

sin(8)

𝑓 (𝑥)=2 𝑥−6− sin 4 𝑥+12

sin (8)

cos 2𝑥=𝑐𝑜𝑠2𝑥−𝑠𝑖𝑛2𝑥

¿1−2𝑠𝑖𝑛2𝑥→𝑠𝑖𝑛2 𝑥=1− cos2𝑥

2

¿1−2𝑠𝑖𝑛2𝑥→𝑐𝑜𝑠2𝑥=1+cos2 𝑥

2

𝑏¿ 𝑓 (3 )= 𝑓 (2 )+∫2

3

4𝑠𝑖𝑛2 (2𝑥 )𝑑𝑥= 𝑓 (2 )+2∫2

3

(1− cos4 𝑥 )𝑑𝑥=¿¿

Example 19 continues:

¿−2+2𝑥|23−

12

sin 4 𝑥|2

3

=−2+6−4−12

sin 12+ 12

sin (8)

𝑓 (3 )=12

sin 8−12

sin12

b

¿−2+2𝑥|2− 2−

12

sin 4 𝑥|2

−2

=−2−4−4−12

sin(−8)+ 12

sin (8)

𝑓 (−2 )=sin (8 )−10

(𝑎)𝑔(2)=−𝜋

(𝑏)𝑔 (𝜋 )=−𝜋+∫2

𝜋

𝑒𝑠𝑖𝑛 𝑥𝑑𝑥≈−1.169

𝑔 (𝑥)=𝑔(2)+∫2

𝑥

𝑔 ′ (𝑥)𝑑𝑥=−𝜋+∫2

𝑥

𝑒𝑠𝑖𝑛 𝑥𝑑𝑥

(𝑐 )𝑔 (𝜋)=−𝜋+∫2

0

𝑒𝑠𝑖𝑛 𝑥𝑑𝑥 ≈−7.378

𝑊 (𝑡)=𝑊 (0)+∫0

𝑡

𝑊 ′ (𝑡)𝑑𝑡⟹𝑊 (14 )=180+𝑊 (𝑡)|0

14

𝑊 (14)=180+10 𝑠𝑖𝑛7𝜋

4≈172.930

𝑑𝑢𝑟𝑖𝑛𝑔 h𝐶 𝑟𝑖𝑠𝑡𝑚𝑎𝑠 𝑡𝑖𝑚𝑒? h𝑤 𝑜𝑐𝑜𝑜𝑘𝑒𝑑?