The Fundamental Theorem of
Calculus (Part 2)
Lesson 5.4
Mr. Peltier
The Fundamental Thm of Calculus Pt.
2
• Before we get to what part 2 actually is, let’s
talk about the area function of a function with
a lower limit of a:
x
A( x)   f (t )dt  signed area from a to x
a
The Fundamental Thm of Calculus Pt.
2
• EX: Find a formula for the area function
x
A( x)   t dt
2
3
x
t 
A( x)   
 3 3
3
3
x 3  1 3
A( x)      x  9
3 3 3
3
The Fundamental Thm of Calculus Pt.
2
• Now that we have our area function
• What is it’s derivative, A’(x) ?
A' ( x)  x
2
 
d x 2
2
So,
 t dt  x
dx 3
This must mean that…
1 3
A( x)  x  9
3
The Fundamental Thm of Calculus Pt.
2
• If f is continuous on an open interval I
containing a, then for every x in I
x


d
  f (t ) dt   f ( x)
dx  a

• Egads, this means integration and
differentiation are inverse operations !
The Fundamental Thm of Calculus Pt.
2
x
• EX: Find the derivative of A( x)  
calculate A’(2), A’(3), and A(2)
2
A' ( x)  1  x 3
A' (2)  1  23  9  3
A' (3)  1  3  28  2 7
3
2
A(2)   1  t dt  0
2
3
1  t 3 dt
and
The Fundamental Thm of Calculus Pt.
2
• EX: Let F(x) be the particular antiderivative of
f(x) = sin(x2) satisfying F (  )  0. Express F(x)
as an integral.
According to FTC II, the area function with lower limit a = 
derivative satisfying F (   )  0 :
F ( x) 
x

 
sin t dt
2

is an anti-
Assignment! 
Page 277
#7-12, 17, 18,
21-24
The Fundamental Thm of Calculus Pt.
2
• What to do when the upper limit of
integration is something other than “x”
• Chain Rule!!
x
d
• EX:
 sin t dt
dx 2
2
sin x   2 x 
2
2 x sin x 2 
Assignment 2! 
Page 277
#13-16, 19,
20, 29-34