Answers & Hints for Selected Exercises 683EXERCISE SET 7.6- If a= b, then J: f + J: = J: f + fca f = J: f - fca = 0 = J: f = J: f.
If b = c, then J: f + J: = J: f + lbb f = J: f + 0 = l: f = J: f ·
If a = c, then l: f + J: = J: f + J: f = 0 + J: = J: f · - Sgn(x) is piecewise continuous on [-1, 1], so it is integrable there. Sgn(x)
does not have the intermediate value property, so by Thm. 6.3.7, it cannot be
a derivative on [-1, 1]; that is, it does not have an antiderivative there. - In Exercise 6.2.17 we showed that f'(O) = 0. For x-=/= 0, f'(x) = 2xsin ~ -
cos.!.. Note that f' is not continuous at 0 since lim f'(x) does not exist.
x x-o
For 0 < € < ~' f' is continuous on [e, ~] so by FTC-I, J} f' = [f(x)J! =
2 ~
[x^2 sin~];'= !>-e^2 sin:. By Thm. 7.4.7, f' is integrable on [O, ~]and f 02 f' =
e-O lim J.¥: e f' = e-0 lim [4 7r -e^2 sin l] e = 4. 7r
- (a) Suppose x < 0.
If x < -1, J~ 1 sgn = - fx-
1
(-1) = fx-
1
l = -l -x = lxl -1;
If x = -1, f~ 1 sgn = f~ 1
1
sgn = 0 =Ix! - 1;
If -1 < x < 0, J~ 1 sgn = J~ 1 (-1) = -(x + 1) = - x - 1 = lxl - l.(b) For x = 0, f~ 1 sgn = lim J~ 1 (-1) = - ( lim € + 1) = -1 =Ix! -1.
e-o- e-o-
(c) Suppose x > 0. Then f~ 1 sgn = f~ 1 sgn +fox sgn = -1 + lim+ J"x 1 =
e-o
-1 + lim (x - e) = x - 1 = Jxl - 1.
e-o+
- Suppose f is integrable on [a, b] and define F(x) =fax f on [a, b].
(a) Suppose f ~ 0 on [a, b]. If x 1 < x2 in [a, b], then by Exercise 7.2.12 (a),
J: 12 f ~ 0, SO F(x2) = fax
2
f = J:^1 f + fxx 1
2
f ~ J:
1
f = F(x1).
(b) Redo (a), changing inequalities appropriately. - (a) -f(x) (b) (! o g)(x)g'(x) (c) -(! o g)(x)g'(x)
(d) (! o h)(x)h'(x) -(! o g)(x)g'(x) - (a) 3/4 (b) 8/15
- (a) ~(x^2 + 2).Jx2 - 1 + C
(g) sin(ln 2) - cos(ln 2) + 1 /2
(c) 6ln2 - 2 (e) 4e^3 + 2
(i) x tan-^1 x - ~ ln(l + x^2 ) + C- (a) [(x -t)f'(t) - J f'(t)(-dt)J: = [O - (x - a)f'(a)] + J: f'(t)dt
= -(x - a)f'(a) + f(x) - f(a) = f(x) - [f(a) + f'(a)(x -a)]= f(x) -T1(x).