# PRACTICE PROBLEMS: SET 2 1. Problems , ```PRACTICE PROBLEMS: SET 2
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. A particle travels on the x-axis with velocity given by the function
v(t) = 2t − 5.
(a) Find the displacement of the particle in the time interval [1, 4].
(b) Find the total distance traveled by the particle in the same time interval.
Problem 2. Evaluate
∫
Problem 3. Let
−3 + x2
√
dx .
5
x
∫
x4
f (x) =
√
t &middot; cos(t2 ) dt .
x
Compute f ′ (x).
Problem 4. Compute
∫
1
2
ex
dx .
x2
1
Problem 5. Let
∫
2
sin3 (t) dt .
f (x) =
ex
Find f ′ (x).
Problem 6. Compute
∫
6
1
2x − 1
dx .
(x + 2)3
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1.
(a) The displacement s(4) − s(1) is the net change in the position function s(t)
for the particle in the time interval [1, 4], and it equals (by the Net Change
Theorem) the integral of the velocity:
∫
∫
4
s(4) − s(1) =
4
(2t − 5) dt .
v(t) dt =
1
1
The antiderivative for 2t − 5 is t2 − 5t and so, by the Fundamental Theorem
of Calculus, we obtain that the displacement equals
(t2 − 5t) |41 = (16 − 20) − (1 − 5) = −4 + 4 = 0.
Hence, the particle ended exactly where it started the movement in the
time interval [1, 4].
(b) The total distance traveled by the particle is the integral of the absolute
value of the velocity:
∫
∫
4
|v(t)| dt =
1
4
|2t − 5| dt .
1
Now, we split the above integral on the intervals [1, 2.5] and [2.5, 4] since at
t = 2.5, the function v(t) = 2t − 5 changes from negative to positive, and
so the definition of its absolute value changes at this point. Therefore, the
total distance traveled by the particle is
∫
∫
5
2
∫
|2t − 5| dt +
1
=
5
2
∫
(5 − 2t) dt +
1
4
5
2
|2t − 5| dt
4
5
2
(2t − 5) dt
5
= (5t − t2 ) |12 +(t2 − 5t) |45
2
(
)
(
)
25 25
25 25
=
−
− (5 − 1) + (16 − 20) −
−
2
4
4
2
25
25
=
−4−4+
4
4
9
= .
2
∫
Problem 2. In general, the indefinite integral ∫f (x) dx represents the most
general antiderivative of the function f (x); i.e., f (x) dx = F (x) + C where
F ′ (x) = f (x).
PRACTICE PROBLEMS: SET 2
3
We evaluate the integral by first dividing both terms in the numerator by the
given denominator:
∫
−3 + x2
√
dx
5
x
∫
1
9
=
−3x− 5 + x 5 dx
∫
∫
9
− 15
= −3 x dx + x 5 dx
5 4
5 14
= −3 &middot; x 5 + x 5 + C
4
14
15 4
5 14
= − x5 + x 5 + C .
4
14
Problem 3. We define two new functions:
∫ x4 √
g(x) =
t cos(t2 ) dt
0
∫
and
x
h(x) =
√
t cos(t2 ) dt .
0
Then f (x) = g(x) − h(x), which yields that f ′ (x) = g ′ (x) − h′ (x). We compute
g ′ (x) and h′ (x) using the chain rule and the Fundamental Theorem of Calculus.
So,
g ′ (x)
=
√
x4 cos((x4 )2 ) &middot; 4x3
= x2 &middot; cos(x8 ) &middot; 4x3
= 4x5 cos(x8 ),
and
So,
h′ (x) =
√
x cos(x2 ).
f ′ (x) = 4x5 cos(x8 ) −
√
x cos(x2 ).
Problem 4. First we compute the antiderivative
∫ 1
ex
dx .
x2
For this we use the substitution u = x1 , which yields
du = −
So,
∫
1
dx .
x2
1
ex
dx
x2
∫
=
−eu du
= −eu + C
1
= −e x + C .
4
MATH 101: PROF. DRAGOS GHIOCA
Therefore, using the Fundamental Theorem of Calculus, we obtain
∫ 2 1
( 1)
√
1
ex
dx
=
−e x |21 = −e 2 − (−e1 ) = − e + e.
2
1 x
Problem 5. We have that
∫
f (x) = −
ex
sin3 (t) dt,
2
and so, using the Fundamental Theorem of Calculus and the chain rule, we obtain
f ′ (x) = − sin3 (ex ) &middot; (ex )′ = −ex sin3 (ex ).
Problem 6. We first evaluate the indefinite integral using the substitution u =
x + 2, and thus du = dx. Note that
2x − 1 = 2(x + 2) − 5 = 2u − 5.
So,
∫
2x − 1
dx
(x + 2)3
∫
=
∫
=
=
2u − 5
du
u3
2u−2 − 5u−3 du
∫
∫
2 u−2 du −5 u−3 du
5 −2
&middot;u +C
2
5
= −2(x + 2)−1 + (x + 2)−2 + C .
2
= −2u−1 +
So, we evaluate the integral
∫ 6
2x − 1
dx
3
1 (x + 2)
(
)
5
−1
−2
=
−2(x + 2) + (x + 2)
|61
2
5
5
= −2 &middot; 8−1 + 8−2 − (−2 &middot; 3−1 + 3−2 )
2
2
1
5
2
5
= − +
+ −
4 128 3 18
205
.
=
1152
```