MATH 101 MIDTERM 1, OCTOBER 6, 2004
n Z j
X
1. Evaluate in closed form
(x2 + x)dx.
Z
2
Z
2
+
0
j−1
j=1
Z
n
+...+
1
Z
n
(x2 + x)dx = (
=
n−1
0
x3
n3
x2
n2
+ )|n0 =
+
.
3
2
3
2
Z
4
(2 − x2 )dx,
2. Write the upper and lower Riemann sums approximating the integral
1
corresponding to the partition of [1, 4] into n intervals of equal length. (Indicate which is
which. Do not try to evaluate the sums.)
For a partition as indicated, we take ∆xj = 3/n, xj = 1 + (3j/n). The function 2 − x2 is
decreasing for x > 0, so on each [xj−1 , xj ] the maximal and minimal values are attained
at the left and right endpoints, respectively. Thus the upper Riemann sum is
n
X
∆xj · (2 −
x2j−1 )
j=1
n
X
3
3(j − 1) 2
· (2 − (1 +
) ),
=
n
n
j=1
and the lower Riemann sum is
n
X
∆xj · (2 − x2j ) =
j=1
Z
x2 −2x−1
3. Let F (x) =
p
n
X
3j
3
· (2 − (1 + )2 ).
n
n
j=1
t3 + 1dt. Find the equation of the line tangent to the graph of
2
F (x) at x = 3.
Z
2
We have 3 −2·3−1 = 9−6−1 = 2, so F (3) =
2
t3 + 1dt = 0. Using the Fundamental
2
Theorem of Calculus we get
F 0 (x) =
p
p
(x2 − 2x − 1)3 + 1 · (2x − 2),
√
√
so F 0 (3) = 23 + 1 · (2 · 3 − 2) = 9 · 4 = 12. Thus the tangent line has the equation
y = 12(x − 3) = 12x − 36.
4. Evaluate the integrals:
Z
Z
Z
3
2
2
2
(a) sin x cos x dx = (1 − cos x) cos x sin x dx = − (1 − u2 )u2 du
Z
=
(u4 − u2 )du =
u5
u3
cos5 x cos3 x
−
+C =
−
+ C.
5
3
5
3
1
We used the substitution u = cos x, du = − sin x dx.
(b) We substitute u = ln x, du = dx/x:
Z
tan2 (ln x)
dx =
x
Z
Z
2
tan u du =
(sec2 u − 1)du = tan u − u + C = tan(ln x) − ln x + C.
√
5. Find the average value of f (x) = 9 − x2 on [0, 3]. (Hint: you may interpret the
integral as an area of a planar region.)
Z 3p
The integral
9 − x2 dx is equal to the area of one quarter of the disc of radius 3,
0
centered at 0. The area of the entire disc is π · 32 = 9π. Thus the average value is
1
3
Z
3
p
9 − x2 dx =
0
3π
1 9π
·
=
.
3 4
4
6. There are 2 finite planar regions bounded by the graphs of y = x3 − x2 + x and y = 3x.
Find the area of each of them.
We first find the x-coordinates of the intersection points: x3 −x2 +x = 3x, x3 −x2 −2x = 0,
x = 0 or x2 − x − 2 = 0, i.e. x = 0, 2, −1. The first area is
Z
0
−1
(x3 − x2 + x − 3x)dx = (
x4
1 1
x3
5
−
− x2 )|0−1 = − − + 1 =
,
4
3
4 3
12
and the second one is
Z
2
(3x − x3 + x2 − x)dx = (−
0
x3
8
x4
16 8
+
+ x2 )|20 = − + + 4 = .
4
3
4
3
3
2