PRACTICE PROBLEMS: SET 3
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1.
(a) Find a function f (x) such that
∫ x
f (t) dt = x2 − 5 sin(x).
0
(b) Find a function f (x) and a number a such that
∫ x
f (t) dt = x5 − 2.
a
(c) Find a function f (x) and a strictly positive number a such that
∫ x√
4
f (t) dt = x5 − x.
a
Problem 2. Compute
∫
1
0
Problem 3. Compute
∫
3
−3
3 − 4x
dx .
1 + x2
x2 sin(x)
dx .
x6 + 1
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Solution.
(a) By the Fundamental Theorem of Calculus, we know that f (x) is the derivative for x2 − 5 sin(x), and thus
f (x) = 2x − 5 cos(x).
(b) Again using FTC we conclude that f (x) is the derivative of x5 − 2, and
therefore f (x) = 5x4 . In order to find the number a, we note that letting
x = a in the formula:
∫ x
f (t) dt = x5 − 2,
a
yields 0 = a −2 since the left side becomes an integral with both endpoints
being the same number a when we
√ let x = a. So, from the easy equation
a5 − 2 = 0 we conclude that a = 5 2.
√
(c) Again FTC yields that, this time, the function 4 f (x) is the derivative of
x5 − x, and so,
√
4
f (x) = 5x4 − 1.
5
So, f (x) = (5x4 − 1)4 . Now, for finding the number a, again we let x = a
in the formula:
∫ x√
4
f (t) dt = x5 − x
a
and obtain 0 = a − a. We are asked to find a solution a which is strictly
positive; so a ̸= 0 is not allowed and thus we may divide by a the above
equation and obtain a4 − 1 = 0. This last equation factors as
5
(a2 − 1)(a2 + 1) = 0,
and since a2 + 1 is always strictly positive, we conclude that we must have
a2 − 1 = 0 which yields two possible solutions: a = −1 and a = 1. Since we
are asked for a positive solution, we arrive at the conclusion that a must
be equal to 1.
Solution. We first split the integral as follows:
∫ 1
3 − 4x
dx
2
0 1+x
∫ 1
∫ 1
4x
1
dx
−
dx
= 3·
2
2
0 x +1
0 x +1
∫ 1
4x
= 3 · tan−1 (x) |10 −
dx
2+1
x
0
(π
) ∫ 1 4x
= 3
−0 −
dx
2
4
0 x +1
∫ 1
3π
4x
=
−
dx .
2+1
4
x
0
PRACTICE PROBLEMS: SET 3
3
For the second integral we use the substitution: u = x2 + 1, and thus du = 2x dx.
Therefore, we compute the antiderivative first:
∫
4x
dx
2
x +1
∫
2
du
=
u
= 2 ln |u| + C
= 2 ln(x2 + 1) + C .
Hence
∫
1
4x
dx
x2 + 1
0
We conclude that
∫
0
1
=
2 ln(x2 + 1) |10
=
2 ln(2).
3 − 4x
3π
dx =
− 2 ln(2).
2
x +1
4
Solution. The main observation here is that the function
x2 sin(x)
f (x) = 6
x +1
is an odd function, and we integrate over an interval [−3, 3] which is symmetric
around the origin. In such cases, the integral always equals 0. Now, the fact that
the interval [−3, 3] is symmetric with respect to the origin is obvious; all we need
to show is that the function f (x) is indeed odd, i.e.,
f (−x) = −f (x).
For proving that f (x) is odd we use the fact that sin(x) is an odd function and
thus,
f (−x)
(−x)2 sin(−x)
(−x)6 + 1
x2 · (− sin(x))
=
x6 + 1
2
x sin(x)
= − 6
x +1
= −f (x).
=
So, f (x) is indeed an odd function and this concludes the proof that
∫ 3 2
x sin(x)
dx = 0.
6
−3 x + 1