Page 196 - Calculus Workbook For Dummies

180 Part IV: Integration and Infinite Series

The Fundamental Theorem of Calculus (the difficult, mostly useless version): Given an
x
f ^
area function A f that sweeps out area under f t ^ h, A x = # ^h f t dt, the rate at
h
s
which area is being swept out is equal to the height of the original function. So, because
the rate is the derivative, the derivative of the area function equals the original function:
x
d A x = ^h f xh. Because A x = # ^h f t dt, you can also write the previous equation
dx f ^ x f ^ h
as follows: d # ^ h = ^ s
f t dt f xh.
dx
s
The Fundamental Theorem of Calculus (the easy, useful version): Let F be any anti-
derivative of the function f ; then
b
f x dx F b -
^
# ^ h = ^ h F ah
a
Q. a. For the area function Q. What’s the area under x2 2 + 5 from 0 to 4?
x Note this is the same question you worked
2
t dt, what’s
f ^ h
A x = # _ t - 5 i d A xh?
f ^
dx on in Chapter 9 with the difficult, sigma-
10
3 x 2 sum-rectangle method.
b. For the area function, B x = # sint dt,
f ^ h
d - 4 A. 188
what’s B xh? 3
f ^
dx
A. a. No work needed here. The answer is Using the second version of the
2
simply x - 5 x Fundamental Theorem,
4
b. x6 sin x 2 # _ 2 x + 5i dx F 4 - F 0 ^ h where F
3
2
=
^ h
0
The argument of an area function is the is any antiderivative of x2 2 + 5.
expression at the top of the integral
symbol — not the integrand. Because By trial and error, you can find that the
2
the argument of this area function, x3 , 2 3
is something other than a plain old x, derivative of 3 x + 5 x is an antiderivative
this is a chain rule problem. Thus, of x2 2 + 5. Thus,
d sin x3 $ 2 4 4 V
2
3
f ^ h
dx B x = 6 , x or x6 sin x . # _ 2 x + 5i dx = 2 x + 5 x W
3
2
3 W W
0
X 0
2 2
3
3
5 4 - cm
= c 4 + $ $ 0 + $
5 0m
3 3
188
=
3
The same answer with much less work than
adding up all those rectangles!