AP Calculus
Name
CHAPTER 5 WORKSHEET
INTEGRALS
Seat #
Date
Fundamental Theorem of Calculus (Part 1)
ESSAY FROM 1995 AB EXAM:
1.
The graph of a differentiable function f on the closed interval [1, 7] is shown above.
x
Let hx    f t dt for 1  x  7.
1
(a)
(b)
(c)
(d)
Find h1 .
Find h' 4 .
On what interval or intervals is the graph of h concave upward? Justify your answer.
Find the value of x at which h has its minimum on the closed interval [1, 7]. Justify
your answer.

ESSAY FROM 1997 AB EXAM:












(3, –1)

2.
The graph of a function f consists of a semicircle and two line segments as shown above. Let g
x
be the function given by g x    f t  dt .

0
a) Find g 3 and g  2 .
b) Find all values of x on the open interval (–2, 5) at which g has a relative maximum.
Justify your answer.
c) Write an equation for the line tangent to the graph of g at x = 3.
d) Find the x-coordinate of each point of inflection of the graph of g on the open interval
(–2, 5). Justify your answer.
SEE OTHER SIDE
ESSAY FROM 2002 AB EXAM:
3.
The graph of a differentiable function f on the closed interval [–3, 15] is shown in the figure
x
above. The graph of f has a horizontal tangent at x = 6. Let g x   5   f t  dt for
6
 3  x  15 .
a) Find g 6 , g ' 6 and g" 6 .
b) On what intervals is g decreasing? Justify your answer.
c) On what intervals is the graph of g concave down? Justify your answer.
15
d) Find a trapezoidal approximation for
 f t  dt
3
using six subintervals of length t  3 .
1995 CALCULUS AB EXAM
AB-6
(a) h1 
1
1 f t dt  0
(b) h' 4  f 4  2
(c) 1  x  3 and 6  x  7
h is concave up when:
 h’ is increasing, or
 f is increasing, or
 h" x   0
1 : answer
1 : h'  x   f  x 
2
1 : answer
2 : 1 for each interval
3
1 : justificat ion
(d) minimum at x  1 because:
1 : answer
 h increases on [1, 5] and
2 : justificat ion
decreases on [5, 7], so

3

minimum is at an endpoint,
1 : eliminates interior points

and
 1 : eliminates x  7
 h7  area R1  area R 2  0 ,
and
h1  0
1997 CALCULUS AB EXAM
AB-5, BC-5
3
(a)
g 3   f t  dt
0
1
1
1
   22    
4
2
2
g  2 
2 : answer

2   -1  each incorrect area
  -1  error in summing

2
 f t  dt  
0
(b) g  x  has relative max at x = 2
because g '  x   f  x  changes from
positive to negative at x = 2
(c)
1
2
g ' 3  f 3  1
1 :
1 :

3

1 :
relative max at x  2 only
g '  x   f  x  or interprets
g  x  as area acumulator
justificat ion
g 3   
1

y        x  3
2

1 : g ' 3  1
2
1 : equation using g 3 and g ' 3
(d) graph of g has points of inflection at
x = 0 and x = 3
1 : points of inflection with
because g" x   f '  x  changes from
 x - coordinate s 0 and 3 only

positive to negative at x = 0 and
2
from negative to positive at x = 3
1 : justificat ion
 (ignore discussion at x  2)
or
because g '  x   f  x  changes from
increasing to decreasing at x = 0 and
from decreasing to increasing at x = 3