MATH 101 MIDTERM 1 – SOLUTIONS
Wednesday, October 1, 2003
1. Evaluate
100
X
j2 −
j=1
100
X
j=1
j −
2
100
X
100
X
(100 − j)2 .
j=1
(100 − j)2 = (12 + 22 + . . . + 992 + 1002 ) − (992 + 982 + . . . + 12 + 02 )
j=1
= 1002 = 1000.
2. WriteZ (but do not try to evaluate) the lower Riemann sum approx4
imating
(2x − x2 ) dx, corresponding to the partition of the interval [0, 4]
0
into 8 subintervals of equal length.
We have n = 8, xi = i/2, ∆xi = 1/2. The function is increasing on [0, 1]
and decreasing on [1, 4]. Hence its minimum on each of the intervals [0, 1/2]
and [1/2, 1] is attained at the left endpoints, and its minimum on each of the
intervals [(i − 1)/2, 1/2], i ≥ 3, is attained at the right endpoints. We thus
have
2 8 X
i − 1 i − 1 2 X
i i L=
−
) +
2
2 − )2 .
2
2
2
2
i=1
i=3
3. Evaluate the following integrals
(a) Let u = e2x , then du = 2e2x dx,
Z
e2x
dx =
e4x + 1
Z
du
= tan−1 u + C = tan−1 (e2x ) + C.
+1
u2
(b) Let u = cos x, du = − sin x,
Z
sin3 x
dx =
cos4 x
Z
(1 − cos2 x) sin x
dx = −
cos4 x
= −u−1 +
Z
(1 − u2 ) du
=
u4
Z
1
u−3
+ C = sec3 x − sec x + C.
3
3
(u−2 − u−4) du
Z
Z
Z
Z
1
1
1
(c) x cos x dx =
x(cos(2x) + 1) dx =
x cos(2x) dx +
x dx.. In2
2
2
1
tegrating by parts with f = x, f 0 = 1, g 0 = cos(2x), g = 2 sin(2x), we get
2
Z
1
x cos(2x) dx = x sin(2x) −
2
Z
Z
1
1
1
sin(2x) dx = x sin(2x) + cos(2x) + C,
2
2
4
1
11
x2
x sin(2x) + cos(2x) +
+C
x cos x dx =
2 2
4
4
1
1
x2
= x sin(2x) + cos(2x) +
+ C.
4
8
4
2
4. Find F 0 (x), if F (x) =
R
2
R
2
Write x4x
= a4x −
2
the Chain Rule:
F 0 (x) = ln
R x2
a
√
Z
4x2
x2
√
ln( t) dt, x > 0.
, and use the Fundamental Theorem of Calculus and
4x2 · (4x2 )0 − ln
√
x2 · (x2 )0 = 8x ln(2x) − 2x ln x.
5. Find the area of the finite planar region bounded by the parabola y =
x2 − 5x + 4 and the line y = −x + 1.
We first find the intersection points: x2 − 5x + 4 = −x + 1, x2 − 4x + 3 = 0,
x = 1, 3. Thus the area in question is
Z
A=
1
3
(−x + 1) − (x − 5x + 4) dx =
Z
2
1
3
(−x2 + 4x − 3) dx = −
4
1
= −9 + 18 − 9 − (− + 2 − 3) = .
3
3
3
x3
+ 2x2 − 3x
1
3