# FTC 1

advertisement ```BC 1-2
FTC
Name:
The (First) Fundamental Theorem of Calculus
Let ƒ be a continuous function defined in an open interval I containing a.
&ograve;
Then F(x) =
x
f (t)dt is a differentiable function for all x in I and
x
d
f (t) dt = f (x) .
dx &ograve;a
(
a
)
Find F(x) = &ograve; (t 4 + 2t + 3) dt .
x
(1)
3
Then differentiate to find
(2)
Find
&ograve;
x
-2
(4)
Find
Find
4
3
d
dx
( &ograve; (sin 2t + cos3t) dt) .
x
-2
d&aelig; x
sin(t 2 )dt &ouml;&oslash; . In this case the
&ograve;
&egrave;
17
dx

1
x3
sec 2 t dt .
Then find
IMSA
(&ograve; ( t + 2t + 3) dt).
x
(sin 2t + cos3t ) dt .
Then find
(3)
dF d
=
dx dx
d  x 3 2 
sec t dt

dx 1
Euler5.1
S 12
BC 1-2
FTC
Name:
Note that problem (4) came out differently, due to the x3 as the upper limit of integration. The
chain rule comes into play as we take the derivative. Continue with the next problems.
(5)
Find
 t
sin x
3
2
Then find

 5t  2 dt .
d
dx

sin x
2
t
3


 5t  2 dt .


(6)
Find
d x4
10
2 sin t dt .
dx
(7)
Find
d  x cost 
e dt  .


dx  3
Check your answer on your TI-89 by entering: d(  (e^(cos(t)), t, 3,  (x)), x)
(8)
Find
d
dx

g(x)
a

f (t) dt .
Check your answer on your calculator.
IMSA
Euler5.2
S 12
```