Names
Date
Mαth 10550 Worksheet 9
1. The following shows the graph of a function f (t):
2
1
2
4
6
8
-1
-2
(a) Give a rough sketch of the graph of the function g(x) =
FTC.
Rx
0
f (t) dt. Fix g(0) = 0 and use the
Solution: We expect to have a graph that looks more or less like
2
4
6
8
-0.5
-1.0
-1.5
-2.0
(b) Calculate the derivative of the following function:
Z
22
g(x) =
cos(11x)
√
t
dt
2016 − sin t
Solution: Notice that
d
d
g(x) = −
dx
dx
√
Z
cos(11x)
22
√
t
d
dt = − (G ◦ h)(x) = −G0 (h(x)) · h0 (x)
2016 − sin t
dx
Rx
Rx
t
Where G(x) := 22 2016−sin
dt = 22 f (t) dt and h(x) := cos(11x). We apply the FTC to obtain
t
G0 (x) = f (x). Hence,
p
cos(11x)
d
0
g(x) = −f (h(x)) · h (x) = −
· (−11 sin(11x))
dx
2016 − sin(cos(11x))
2. Sketch the region enclosed by the curves |3x| = y1 and y2 + x2 = 4, on the interval [−2, 2] and find
the area of the region. You may use the symmetry of this two functions.
Solution:
We have to look first for points (if any) where the graphs intersect and we should pay attention
to which function is above in the region of integration.
Let us solve the equation |3x| = 4 − x2 . If x > 0 we have
x2 + 3x − 4 = 0 =⇒ (x + 4)(x − 1) = 0 =⇒ x = 1
since x > 0. Since this two functions are symmetric with respect to the y-axis we know that
the other intersection point is x = −1. Note that the integral is then divided into four intervals,
[−2, −1], [−1, 0], [0, 1], [1, 2]. Next we sketch the graphs and look for the region enclosed between
this functions:
6
5
4
3
2
1
-2
1
-1
2
We see that
Z
−1
Z
0
−2
−1
1
Z
0
2
(y1 − y2 ) dx
(y2 − y1 ) dx +
(y2 − y1 ) dx +
(y1 − y2 ) dx +
A=
Z
1
=: A1 + A2 + A3 + A4 = 2(A3 + A4 )
We see by inspection that this is the case. We compute
Z 1
Z 2
(y2 − y1 ) dx +
(y1 − y2 ) dx
A/2 =
0
1
Z 1
Z 2
2
=
(−x + 4 − 3x) dx +
(3x − 4 + x2 ) dx
0
1
3
1 2
2
2
−x
3x 3x
x3 =
+ 4x −
+
− 4x +
3
2 0
2
3 1
−1
3
3(2)2
(2)3
3
1
=
+4−
+
− 4(2) +
−
−4+
3
2
2
3
2
3
13 2
15
=
+ =
=5
3
3
3
=⇒ A = 10
3. (a) A particle traveling on a line accelerates at a rate of 2t − 1 f t/sec2 . If the initial velocity is
known to be 6 f t/sec, find the displacement of the particle int he first two seconds.
Solution: Since the acceleration is 2t − 1 f t/sec2 . We know that
Z
Z
v(t) = a(t) dt = 2t − 1 dt = t2 − t + K =⇒ v(t) = t2 − t + 6
Applying the FTC we get (with ∆ = Displacemetn),
Z
2
2
t − t + 6 dt =
∆=
0
2
t3 t2
38
(2)3 (2)2
− + 6t =
−
+ 6(2) =
3
2
3
2
3
0
(b) If a(t) = −6 f t/sec2 and the initial velocity is not known, but the particle is know to travel
20 f t in the first 3 seconds, then find the possible value(s) for the initial velocity.
Solution: Let us first find the velocity function. Recall that dtd v(t) = a(t), so that
Z
Z
v(t) = a(t) dt = −6 dt = −6t + K =⇒ v(t) = v0 − 6t.
R3
The second part of the statement tells us that 20 = 0 |v0 − 6t| dt . We then look for t∗ so
that v(t∗ ) = 0, thus t∗ = v60 . Let us assume that 0 ≤ t∗ < 3 and compute
Z 3
Z t∗
Z 3
|v0 − 6t| dt
|v0 − 6t| dt +
|v0 − 6t| dt =
20 =
t∗
0
0
t
t
Z 3
Z t∗
(v0 − 6t)2 ∗ (v0 − 6t)2 ∗
v0 − 6t dt =
v0 − 6t dt −
=
+
−12 0
−12 3
t∗
0
−1
1 2
=
2(v0 − 6t∗ ) − v02 + (v0 − 18)2 =
v0 + (v0 − 18)2
12
12
2
2
2
=⇒ 0 = v0 + (v0 − 18) − 224 = 2[(v0 − 9) − 39]
√
=⇒ v0 = 9 ± 39 ≥ 0
If it were the case that t∗ ≥ 3 > 0, since v0 − 6t decreases from v0 ≥ 0 the line would have
always been above the real axis on [0, 3] and then
Z
3
Z
|v0 − 6t| dt =
20 =
0
0
3
3
47
v0 − 6t dt = (v0 t − 3t ) = 3v0 − 27 =⇒ v0 =
3
0
2
Finally looking for the possibility of having t∗ < 0 we get that t6∗ = v0 < 0 and this would
be the case if during the interval [0, 3] the velocity had always been negative. Notice then
that |v0 − 6t| = −(v0 − 6t) the we compute
3
Z 3
Z 3
2 20 =
|v0 − 6t| dt = −
v0 − 6t dt = −(v0 t − 3t )
0
0
7
= −3v0 + 27 =⇒ v0 = ≥ 0 =⇒⇐=
3
So that we obtain nothing from this case.
0
4. Compute the following integrals:
(a)
1
Z
1 − 2x2 dx
0
Solution: Recall that |x| = x if x ≥ 0 and −x otherwise. Notice that 1 − 2x2 defines
a negative parabola, so that before x = √12 it is positive and negative afterwards. We
compute
1
Z
1 − 2x2 dx =
1
√
2
Z
0
2
Z
1
− 1 − 2x2 dx
1 − 2x dx +
√1
2
0
√
√1
√1
2 3 2
2 3 2
2 2−1
= x− x + x− x =
3
3
3
0
1
(b)
Z
π/2
sin3 x dx
0
Solution: Set u(x) = u = cos(x). We compute u(0) = 1 and u(π/2) = 0. Also du/dx =
− sin(x) =⇒ −du = sin(x)dx = u0 (x)dx. So that with f (t) = 1 − t2 , we have as before
Z
π/2
Z
π/2
Z
π/2
(1 − (cos(x))2 ) sin(x) dx
0
0
1
Z π/2
Z u(π/2)
Z 1
u3 2
0
2
=
f (u(x))u (x) dx = −
f (u)du =
(1 − u )du = u −
=
3
3
0
u(0)
0
0
3
sin x dx =
0
2
sin x · sin(x) dx =
(c)
Z
9
q
√
4 − x dx
0
√
Solution: We use the change√of variable u(x) = u = 4 − x. Then we have u(0) = 4 and
u(9) = 1; also, du/dx = −1/(2 x) which implies 2(u − 4)du = dx thus we have
Z
0
9
Z
q
√
4 − x dx = 2
4
1
1/2
u
(u − 4)du = 2
u5/2
u3/2
−4
5/2
3/2
1
= 188
15
4
5. Compute the following definite integral using right endpoint approximation and the limit definition
of the definite integral. (You should compute the limit, and you should show all of your work.)
Verify your answer using the FTC.
Z
3
x2 − x dx
0
#
"
Note:
Pn
i=1
i=
n(n+1)
2
and
Pn
i=1
i2 =
n(n+1)(2n+1)
6
.
Solution:
Let us recall that the definite integral of a continuous function on an interval can be computed
as follows
Z
b
f (t) dt = lim
a
where ∆x =
case we have
• ∆x =
b−a
n
b−a
n
=
n→∞
n
X
f (xi )∆x
i=1
and xi = a + i · ∆x are the right-hand end points of the partition. Thus in our
3−0
n
=
3
n
• xi = a + i · ∆x = 0 + i ·
3
n
=
3i
n
) = ( 3i
)2 − ( 3i
)
• f (xi ) = f ( 3i
n
n
n
Thus using the formulas above our integral is given by
" #
Z 3
n
2
X
3i
3i
3
2
x − x dx = lim
−
·
n→∞
n
n
n
0
i=1
"
#
n
n
3 9 X 2 3X
= lim
i −
i
n→∞ n n2
n i=1
i=1
3 9 n(n + 1)(2n + 1) 3 n(n + 1)
·
= lim
− ·
n→∞ n n2
6
n
2
3·9·2 3·3
9
=
−
= .
6
2
2
Now we solve it by using the FTC, note that
3
3
Z 3
Z 3
d x3 x2
x
x2 33 32
9
2
x − x dx =
−
dx =
−
=
−
= .
dx 3
2
3
2
3
2
2
0
0
0