Math 1310 Lab 11. (Sec 5.4 - Sec 5.6)
Name/Unid:
Lab section:
1. (Play with fundamental theorem of calculus.)
R x2
(a) Let f (x) = 2x · π/2 cos y dy. What is f 0 (x)? (2 pts)
(b) Consider the function f given in part (a). What is
(
1 if x ≥ 0
(c) Let g(x) =
−1 if x < 0
. What is G(x) =
Rx
0
R
f (x) dx? (2 pts)
g(y) dy? (2 pts)
(d) Consider the function g given in part (c). If fundamental theorem of calculus (FTC)
works, then G0 (0) exists and equals g(0). But now G0 (0) does not exist, that means
some assumption for FTC does not hold. What’s going wrong? (2 pts)
(e) What if we instead consider the following function;
(
1 if t ≤ 1
h(t) =
t if t > 1
Rx
What is H(x) = 0 h(t)dt? Sketch a graph of H(x). Is this an antiderivative? If
so, what is H 0 (1)? (2 pts)
Page 2
(f) Suppose we have a continuous function f (x). Consider the two quantities;
Z
1
Z
f (x)dx
2
f
0
0
1
x dx
2
Sketch the graph of an example function and compare the two quantities. How are
they related? (2 pts)
Solution:
(a) f 0 (x) = 2x · 2x · cos(x2 ) + 2 ·
(b)
R
f (x) dx =
R
R x2
π/2
cos y dy = 4x2 cos(x2 ) + 2(sin x2 − 1).
2x(sin x2 − 1) dx = − cos x2 − x2 + C.
(c) It is the absolute value of x.
(d) f (x) is not continuous at 0, so FTC doesn’t work.
(
x
x<1
(e) G(x) =
The linear and quadratic chunks line up, so
2
1/2 + 1/2(x) x ≥ 1
there is no corner this time. G’(1) = 1.
(f) You should sketch some function, and then the same function stretched twice
as far and note that the area is twice as big.
Page 3
2. (Integration by parts: I wake up LATE.)
Rb
The integration by parts formula for definite integral is a f (x)g(x) dx = (F (b)g(b) −
Rb
formula
F (a)g(a)) − a F (x)g 0 (x) dx, where F (x)Ris any antiderivative for f (x),
R and the
0
for indefinite integral (anti-derivative) is f (x)g(x) dx = F (x)g(x)− F (x)g (x) dx+C.
This means you may divide your integrand into two different parts, integrate one of them,
and differentiate the other one. There’s a rule to show the priority of which should be
differentiated first: I wake up LATE rule.
I: Inverse trigonometric function. arccos x, arcsin x, · · ·
L: Logarithms. ln x, ln 3x, · · ·
1
A: Arithmetic functions. 1, x2 , 1+x
2,···
T: Trigonometric functions. sin x, tan x, · · ·
E: Exponential functions. ex , 2x , · · ·
Functions in the class I has the top priority to be differentiated.
R2
Example: Compute 1 x ln x dx. Since x ∈ class A, ln x ∈ class L, x should be integrated
while ln x has to be differentiated.
2 R Therefore, we set f (x) = x, andR g(x) = ln x, and we
R2
2
2
have 1 x ln x dx = 12 x2 ln x − 1 12 x2 · x1 dx = 12 22 ln 2 − 21 12 ln 1 − 1 21 x dx = 2 ln 2 − 34 .
1
R1
(a) Find 0 arccos x dx. (2 pts)
(b) Find
R
x2 ln x dx. (2 pts)
Solution:
1 R
R1
1
(a) 0 1 · arccos x dx = x arccos x − 0 x ·
0
(b)
R
x2 ln x dx = 13 x3 ln x −
R
1 2
x
3
√ −1
1−x2
dx = 0 −
√
dx = 13 x3 ln x − 19 x3 + C.
Page 4
1
1 − x2 = 1.
0
3. (Miscellaneous integration problems.)
(a) Find
R3
(b) Find
R1
(c) Find
R 2015
0
0
|(x − 1)(x − 2)| dx (2 pts)
cos (1 + x2 ) · x(1 + x2 ) dx (2 pts)
−2015
cos x2 ·
x2015
1+x2
dx (2 pts)
Solution:
R3
R1
R2
R3
(a) 0 |(x − 1)(x − 2)| dx = 0 (x − 1)(x − 2) dx − 1 (x − 1)(x − 2) dx + 2 (x − 1)(x −
2) dx = 5/6 − (−1/6) + 5/6 = 11/6.
(b)
R1
0
cos (1 + x2 )·x(1+x2 ) dx =
1
2
R2
1
u cos u· du = 21 (− cos 1−sin 1+cos 2+2 sin 2).
Page 5
(c) 0, since the integrand is an odd function.
Page 6
4. (More integration techniques) Solve the following integrations.
R 2π
(a) 0 cos(x)ex dx (2 pts)
(b)
R2
(c)
R1
1
0
ln(x)dx (2 pts)
x
dx (2 pts)
1 + x2
Page 7
(d)
R4 x+3
dx (2 pts)
2
x2 − 1
Solution:
(a) We use integration by parts. We will take an antiderivative of ex (which is ex
again) and take the derivative of cos(x). A term of the form ex sin(x) will pop
out, and we will use integration by parts on that again. Thus we get
Z 2π
Z 2π
x
x
2π
cos(x)e dx = [e cos(x)]0 +
ex sin(x)dx
0
0
Z 2π
x
2π
x
2π
(1)
= [e cos(x)]0 + [e sin(x)]0 −
ex cos(x)dx
0
Z 2π
2π
=e −1+0−
ex cos(x)dx.
0
Thus we get
R 2π
0
ex cos(x)dx =
e2π −1
.
2
(b) We know thatR an antiderivative of ln(x) is x ln(x) − x. Thus we can use the
2
TFC and get 1 ln(x)dx = [x ln(x) − x]21 = 2 ln(2) − 1.
(c) We can use u-substitution. Setting u = 1 + x2 , we get
R du = 2xdx. When x = 0
1 2 1
then u = 1, when x = 1 then u = 2. Thus we get 2 1 u du. We know that ln(u)
R2
is an antiderivative for u1 , thus TFC gives us 12 1 u1 du = 12 (ln(2) − ln(1)) = ln(2)
.
2
(d) We can break the integral as
Z 4
Z 4
Z 4
x+3
x
3
dx =
+
.
2
2
2
2 x −1
2 x −1
2 x −1
Page 8
(2)
We use u-substitution for the fisrt summand with u = x2 − 1: when x = 2 then
u = 3 and when x = 4 then u = 15, also du = 2xdx. Thus the fisrt of the two
integrals gives us
Z
Z 4
1 15 1
x
dx =
du
2
2 3 u
2 x −1
1
(3)
= (ln(15) − ln(3))
2
ln(5)
=
.
2
For the second part, we will factor x2 − 1 = (x + 1)(x − 1) and solve the equation
A
B
3
+
= 2
.
x+1 x−1
x −1
(4)
After bringing the left hand side to common denominator, we get the equality
between the two numerators Ax − A + Bx + B = 3. Thus we get B = −A and
A = − 23 . Thus we are left with
3
−
2
Z
2
4
3
1
dx +
x+1
2
We know ln(x − 1) is an antiderivative for
1
for x+1
. Thus TFC gives us
Z
2
4
x2
Z
2
1
x−1
4
1
dx.
x−1
(5)
and ln(x + 1) is an antiderivative
3
3
3
3
= − (ln(5) − ln(3)) + (ln(3) − 0) = − ln(5) + 3 ln(3).
−1
2
2
2
(6)
Thus,the integral we were looking for is − ln(5) + 3 ln(3) = − ln(5) + ln(27) =
ln 27
.
5
Page 9