Solutions For Calculus Quiz #3
1. (a) We can rewrite the equation as dy dx = ex
2 +e−x 2 .
Thus, we have the general solution by finding the antiderivative of the above function. Hence y= ex
2 − e−x
2 +c wherec is a constant. Substituting the values of y= 0 andx= 0, we get c= 0. Thus, y = ex
2 −e−x 2 .
(b) We have the general solution by finding the antiderivative of the function.
Hence,y=−1
πcos(πx) +cwherecis a constant. Substituting the values ofy= 0 and x= 0, we get c= 1
π. Thus, y=−1
πcos(πx) + 1 π. 2. (a) We use the right endpoints; thusc1 = 3π
10, c2 = 6π
10, c3 = 9π
10, c4 = 12π
10 , c5 = 15π We find 10
SP = 3π
10[sin(3π
10) + sin(6π
10) + sin(9π
10) + sin(12π
10 ) + sin(15π 10 )]
= 3π
10(0.809 + 0.9511 + 0.309−0.5878−1) = 0.4536 (b) We use the midpoints; thus c1 = 3π
20, c2 = 9π
20, c3 = 15π
20 , c4 = 21π
20 , c5 = 27π We find 20
SP = 3π
10[sin(3π
20) + sin(9π
20) + sin(15π
20 ) + sin(21π
20 ) + sin(27π 20 )]
= 3π
10(0.454 + 0.9877 + 0.7071−0.1564−0.891) = 1.038 (c)
Z 3π/2
0
sinx dx= [−cosx]3π/20 = [−cos(3π
2 )]−(−cos 0) = 0 + 1 = 1.
Comparing the result with those obtained in (a) and (b), we find that (b) is a better approximation than (a).
3. (a) lim
||P||→0 n
X
k=1
e−ck△k, where P is a partition of [1,3].
(b) lim
n→∞
n
X
k=1
sin(kπ n )π
n = Z π
0
sinx dx= [−cosx]π0 = 2.
4. (a) Integrating, we get Z
f(x) = ln|1 +x|+cwhere cis a constant.
(b) Integrating, we get Z
f(x) =−3 cos(x
3) +cwhere c is a constant.
5. (a) The Fundamental Theorem of Calculus (Part I)
If h(x) is a differentiable function and f(u) is continuous for u between a and h(x), then d
dx Z h(x)
a
f(u) d(u) =f(h(x))h′(x).
The Fundamental Theorem of Calculus (Part II) Assume thatf is continuous on [a, b], then
Z b
a
f(x) dx=F(b)−F(a) whereF(x) is an antiderivative off(x), that is, F′(x) =f(x).
(b) (i)
y =
Z 1−4x
0
(2t2+ 1) dt= [2
3t3 +t]1−4x0 = 2
3(1−4x)3+ (1−4x)
= 2
3(1−12x+ 48x2−64x3) + 1−4x=−128
3 x3+ 32x2−12x+5 3 dy
dx = −128x2+ 64x−12
(ii) Let f(t) = 2t2+ 1 and h(x) = 1−4x.
Then dy
dx = f(h(x))h′(x)
= [2(1−4x)2+ 1](−4) = [2(1−8x+ 16x2) + 1](−4)
= −128x2+ 64x−12, thanks to the FTC Part I 6. (a) (i)
1 4
Z 4
0
f(x) dx = 1 4[
Z 2
0
x2 dx+ Z 4
2
(x2+ 3) dx]
= 1 4[1
3x3]20+1 4[1
3x3+ 3x]42
= 1 4[8
3 + 64
3 + 12− 8
3−6] = 41 6
(ii) There does not exist xsuch that f(x) =f(x) becausef(x) is not continuous.
(b) (i) 1 2
Z 1
−1
g(x) dx= 1 2
Z 1
−1
e−x dx= 1
2[−e−x]1−1 = 1
2(−e−1+e) = e− 1e
2 . (ii)
e−x = e−1e
2 = e2−1 2e ex = 2e
e2−1 x = ln( 2e
e2−1) 2
7. (a) x=y−1, x=√ y+ 1 Z 3
0
(p
y+ 1−(y−1))dy = Z 1
0
py+ 1 dy+ Z 3
1
(p
y+ 1−(y−1))dy
= [2
3(y+ 1)3/2]10+ [2
3(y+ 1)3/2− 1
2y2+y]31
= 8 3.
0 0.5 1 1.5 2 2.5 3
0 0.5 1 1.5 2 2.5 3
(b)
0 0.5 1 1.5 2
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
A
A
0 0.5 1 1.5 2
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
B
AreaA = Z 1
1/2
(1
x −1dx+ Z 3/2
1
(1− 1 x) dx
= [lnx−x]11/2+ [x−lnx]3/21
= −1−(ln1 2− 1
2) + (3
2−ln3
2)−1 = ln 2−(ln 3−ln 2) = 2 ln 2−ln 3.
AreaB = Z 1
1/2
1dx+ Z 3/2
1
1 x dx
= [x]11/2+ [lnx]3/21
= −1−1
2 + ln3 2 = 1
2+ ln3 2.
4
8. (a) To find the equilibrium, we need to solvex= 5x2
4 +x2 ⇒ 4x+x3 = 5x2 ⇒x(x2−5x+ 4) = 0 Therefore 0,4,and 1 are equilibria.
Now,f′(x) = 40x (4 +x2)2.
And, |f′(0)|= 0<1, |f′(4)|= 2/5<1, |f′(1)|= 8/5>1. We conclude that 0 and 4 are stable and 1 is unstable, thanks to the stability criterion.
(b) The sequence xt with x0 = 2 converges to 4.
0 0.5 1 1.5 2 2.5 3 3.5 4
0 0.5 1 1.5 2 2.5 3 3.5 4
