18.01 Quiz 3
Fall 1999
Name: Solution
e-mail: mjg@math.mit.edu
1. (5 pts each) Evaluate the following definite and indefinite integrals.
Z
x
)
a)
cot x dx (remember cot x = cos
sin x
Z
cos x
dx =
sin x
Z
1
du
u
(u = sin x, du = cos x dx)
= ln u + c =
Z
√
b)
x2
dx
x3 + 1
Z
1
x2
√
dx =
3
3
x +1
=
Z
1
c)
−1
√
1
−1
1
√
dx =
1 − x2
Z
π/2
−π/2
Z
π/2
=
−π/2
9
d)
4
ex
Z
1
√ du
u
2√
u+c=
3
(u = x3 + 1, du = 3x2 )
2√ 3
x +1+c
3
1
dx
1 − x2
Z
Z
ln sin x + c
√
x
√
x dx
p
1
cos u du
1 − sin2 u
cos u
du = π
cos u
(x = sin u, dx = cos u du)
18.01 1999 Quiz 3
Name: Solution
Z
4
9
√ √
x x
e
2
x dx =
3
Z
27
8
√
3√
(u = x x, du =
x dx)
2
eu du
27
2 u = e =
3 8
2 27
(e − e8 )
3
2. (5 pts each).
a) Convert the expression
Z
π/3
0
sin x cos x e− cos x dx
into a definite integral which does not involve trigonometric functions. Do not evaluate!
Z
0
π/3
− cos x
sin x cos x e
Z
dx = −
1/2
1
Z
1
=
1/2
ue−u du
(u = cos x, du = − sin x dx)
ue−u du
b) Clifford and Emily are asked to evaluate the indefinite integral
Z
2 sin x cos x dx.
Clifford makes the substitution u = sin x and is led to the answer sin2 x. Emily makes
the substitution u = cos x and is led to the answer − cos2 x. Both check their answers
by differentiating. Frustrated, they turn to you for help. Who is right and who is wrong,
and what mistake was made?
0
0
Since sin2 x = 2 sin x cos x and (− cos2 x) = 2 sin x cos x, both are correct. The
results differ only by a constant, because sin2 x = 1 − cos2 x, that’s why they are
antiderivatives of the same function.
3. (5 pts each) Evaluate the following derivatives.
Z sin x
d
2
a)
et dt
dx 0
page 2 of 5
18.01 1999 Quiz 3
Name: Solution
d
dx
d
b)
dx
Z
Z
sin x
0
2
et dt =
2
esin
x
cos x
e2x
ex
ln t dt
d
dx
Z
e2x
ex
ln t dt = ln(e2x )2e2x − ln(ex )ex
=
4xe2x − xex
4. (10 pts each)
a) Sketch the graph of the function y = x3 − 9x (zeros, positive/negative sign).
10
pos.
−4
−3
−2
5
−1
−5
1
2
3
neg.
f (x) = x(x2 −9) = x(x−3)(x+3),
so the zeros are −3, 0, 3, and f is
positive between −3 and 0 and between 3 and ∞, and negative between −∞ and −3 and between 0
and 3.
−10
−15
b) Find the area of the region bounded by the curve y = x3 − 9x, the x-axis, and the lines
x = −1 and x = 3. (Take care of the signs you know from a)!)
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18.01 1999 Quiz 3
Name: Solution
R
An antiderivative of f is (x3 − 9x) dx = 14 x4 − 92 x2 . Since f is negative between 0
and 3 we have to split the integral and take care of the sign. So the area is:
Z 0
Z 3
3
A=
(x − 9x) dx −
(x3 − 9x) dx =
−1
9
1
= x4 − x2
4
2
81 17
+
=
4
4
=
0
0
9
1
− x4 − x2
4
2
−1
3
0
98
1
= 24
4
2
5. (10 pts each)
a) Find the general solution to the differential equation
1 y
dy
= − .
dx
x x
1
dy
= − (y − 1)
dx
x
dx
dy
=−
y−1
x
ln(y − 1) = − ln x + c
C
1
y = 1 + ec = 1 +
x
x
b) In part a), if y(1) = 2, what is y(2)?
2 = y(1) = 1 + C, so C = 1, and
y(2) = 1
1
2
6. (20 pts) In this problem we will examine exact and approximate integrals of f (x) =
3x2 + 2x + 1.
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18.01 1999 Quiz 3
Z
a) (4 pts) Compute
Name: Solution
3x2 + 2x + 1 dx exactly.
1
0
Z
1
0
Z
b) (6 pts) Approximate
1
0
1
3x2 + 2x + 1 dx = x3 + x2 + x 0 = 3
3x2 + 2x + 1 dx, using the trapezoidal rule for n = 2 (which
means evaluating f at 3 points).
Z
1
0
1
3x + 2x + 1 dx =
2
2
Z
c) (6 pts) Approximate
0
1
11 1
1
·1+1·
+ ·6
2
4
2
=
25
8
3x2 + 2x + 1 dx, using Simpson’s rule for n = 2 (which means
evaluating f at 3 points).
Z
1
0
11
1 1
1·1+4·
+1·6 = 3
3x + 2x + 1 dx = ·
2 3
4
2
d) (4 pts) Why does Simpson’s rule give the exact result?
Because Simpson’s rule uses quadratic approximation which is exact for a quadratic
polynomial.
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