Page 204 - Calculus Workbook For Dummies

188 Part IV: Integration and Infinite Series

Solutions to Reverse Differentiation Problems

a Where (from 0 to 8) does A g equal zero? At about x = 2 or 2 ⁄2 and about x = 6.
1
A g equals zero twice between 0 and 8. First at about x = 2 or 2 ⁄2 where the negative area begin-
1
ning at x = 1 cancels out the positive area between x = ⁄2 and x = 1. The second zero of A g is
1
roughly at 6 (see the dotted line in the figure). After the first zero at about 2, negative area is
added between 2 and 4. The positive area from 4 to 6 roughly cancels that out, so A g returns
to zero at about 6.
b Where (from x = 0 to x = 8) does A g reach
a. its maximum value? A g reaches its max at x = 8. After the zero at x = 6, A g grows by about
4 or 4 ⁄2 square units by the time it gets to 8.
1
b. its minimum value? The minimum value of A g is at x = 4 where it equals something like –2.

c In what intervals between 0 and 8 is A g
a. increasing? A g is increasing from 0 to 1 and from 4 to 8.

b. decreasing? A g is decreasing from 1 to 4.
d Approximate A 1 ^ h, A 3 ^ h, and A 5 ^ h.
g
g
g
A 1 ^ h is maybe a bit bigger than the right triangle with base from x = ⁄2 to x = 1 on the x-axis and
1
g
1
vertex at _ 1 2 , 4i. So that area is about 1 or 1 ⁄4.
There’s a zero at about x = 2 or 2 ⁄2. Between there and x = 3 there’s very roughly an area of –1,
1
so A 3h is about –1.
g^
In problem 2b, you estimate A 4 ^ h to be about –2. Between 4 and 5, there’s sort of a triangular
g
shape with a rough area of ⁄2. Thus A 5h equals about –2 + ⁄2 or –1 ⁄2.
1
1
1
g^
x
e a. If A x = # sint dt, d A x = sin x.
f ^ h
f ^ h
dx
0 x
b. If A x = # sint dt, d A x = sin x.
g^ h
g^ h
dx
/ π 4
cos x
*f Given that A x = # sint dt, find d A xh. The answer is sin x- $ sin cos xh.
^
f ^
f ^ h
dx
- / π 4 x
This is a chain rule problem. Because the derivative of # sint dt is sinx, the
stuff - / π 4 cos x
derivative of # sint dt is sin stuff $ h stuff'. Thus the derivative of # sint dt is
^
- / π 4 - / π 4
l
^
sin cosx $ ^h cosx = - sinx $ sin cosxh.
^
h
d d d
g For A xh from problem 5a, where does A f equal zero? A f = sin x, so A f is zero at
f ^
dx dx dx
all the zeros of sin x, namely at all multiples of π: kπ (for any integer, k).
π
*h For A xh from problem 6, evaluate A f ' c m. In problem 6, you found that
f ^
4
π
π
A f ' x = - sin x $ sin cos xh, so A f ' c m = - sin $ sin cos π m = - 2 $ sin 2 . - .459.
c
^ h
^
4 4 4 2 2
i What’s the area under y = sinx from 0 to π? The area is 2. The derivative of cosx
-
is sinx, so cosx- is an antiderivative of sinx. Thus, by the Fundamental Theorem,
π
# sinx dx = - cosx = - - - h ^ 2 = 2.
π
1
@
^
1 = - - h
0
0