AP Calc Notes: DI – 7 Functions Defined as Integrals/Derivatives of Integrals
Let f be continuous on the interval [a, b]. Then the area under the curve from a to x is given by
y
A(x) =
∫
x
a
f
f(t)dt
ΔA
A
dA
A(x + h) - A(x)
= lim
h→0
dx
h
x
x+h
a
ΔA
f(x)h
= lim
h→0 h
h →0
h
= lim
b
t
Note: ΔA ≈ f(x)h
dA
= f(x)
dx
Fundamental Theorem of Calculus Part I:
d x
f(t)dt = f(x)
dx ∫ a
Algebraically integrate, then differentiate. Compare your answer to the original integral
x
d
tdt
dx ∫1
d
dx
x
d
cos tdt
dx π∫
d
dx
5 x3
∫
d
dx
tdt
1
7 x2
sin x
∫
tdt
1
ex
d
cos tdt
dx 2∫π
∫ cos tdt
π
2
What’s a formula for
d g (x)
f(t)dt ?
dx ∫ a
f(g ( x ) )g'( x )
Where did the t go?
Recall:
b
Why doesn’t it seem to matter what value of the lower limit is?
∫ f ( t ) dt = F ( b ) − F ( a )
a
F ' (t ) = f (t )
Why is there sometimes another function multiplied and sometimes not?
Ex: Evaluate
a.
d 5x
cos t 2 dt
∫
dx −1
= cos (5x)2 (5) = 5cos(25x2 ) by the chain rule.
b.
d 3x2
cos t 2 dt
dx ∫ −1
(
= 6xcos 3x2
c.
2
d −1
cos t 2 dt
∫
dx x
=
d.
)
x
d
(-∫ cos t2 ) = - cos x2
−1
dx
d x2
cos t 2 dt
∫
dx x
a
d
=
( ∫ cos t2 +
dx x
∫
x
d
cos t ) =
(-∫ cos t2 +
dx
a
a
x
2
= cos (x ) (2x)- cos x
2 2
2
2
= 2x cos x4 – cos x2
e.
d 2
cos t 2 dt
∫
dx −1
= 0 because
∫
2
−1
cos t2dt is a constant.
∫
x
2
a
cos t2 )
x
Ex: Find the absolute extrema of f ( x ) = ∫ et cos 2tdt on the interval [0, π].
0
Find critical points:
f' = e cos2x = 0 when x = π/4 and x = 3π/4.
x
Check critical points and endpoints in table:
0
The absolute minimum is -4.4203 at x = 3π/4;
π
the absolute maximum is 4.4281 at x = π.
4
3π
4
x
See graph of f below.
x
f ( x ) = ∫ et cos 2tdt
0
0
0.6773
-4.4203
4.4281
π
y
5
4
3
2
1
x
π/4
−1
−2
−3
−4
5
π/2
3π/4
π