Download MATH141 – Tutorial 2

MATH141 – Tutorial 2 - solutions
1. Use the Fundamental Theorem of Calculus to evaluate the integral, or explain why it does
not exist.
(a)
(b)
R √3
1
Rπ
−π
6
dx
1+ x2
f ( x)dx where f ( x) =

x
if − π ≤ x ≤ 0
sin( x)
if 0 < x ≤ π .
Solution.
(a) The function is defined everywhere in the given interval, the antiderivative of (1 + x2 )−1
√
is arctan( x); we have 6 arctan( 3) − 6 arctan(1) = 6 π3 − π4 = π2 .
(b)
Z 0
−π
xdx +
Z π
0
x2
sin( x)dx =
2
=−
0
−π
π2
2
+ [− cos( x)]π0
+ (− cos(π ) + cos(0) = 2 −
π2
2
2. Find the derivative of the function
R 3x u2 −1
(a) g( x) = 2x
du
u2 +1
2
x
√1
(b) g( x) = tan
( x) 2+t4 dt
Solution.
(a) We use the Fundamental theorem of Calculus once again; as prescribed in the hint, break
the integral into two parts, and flip the sign of the first part, as to get
R
g( x) = −
Z 2x 2
u −1
0
u2 + 1
1
du +
Z 3x 2
u −1
0
u2 + 1
du
Using the chain rule, we get for final answer
−2 ·
4x2 − 1
9x2 − 1
+
3
·
4x2 + 1
9x2 + 1
(b) This time, we have a harder substitution, which involves the derivative of tan( x) = sec2 ( x).
I leave it to you to work out the details, the answer is
sec2 ( x)
2x
−√
+√
2 + t4
2 + x8
3. If F ( x) = 1x f (t)dt where f (t) =
Solution. Write the problem as
R
R t2 √1+u4
du, find F 00 (2)
1
u
d2
dx2
Z x Z t2 √
1 + u4
1
u
1
dudt.
Because the derivative with respect to x of the function will be by FTC f ( x), the second derivative involves thus
Z x2 √
1 + u4
d
du
dx 1
u
where now we make a substitution and use the chain rule
F 00 ( x) = √
and F 00 (2) =
√
2
1 + x8
x2
2
dx2
=√
x
dx
1 + x8
257.
4. Find the interval on which the curve y given below is concave upward
y=
Z x
0
1
dt
1 + t + t2
Solution. To get the curve to be concave upward, we require that y00 > 0 (recall the second
derivative test). Using FTC, we get y0 is of the form (1 + x + x2 )−1 , differentiating again, we
2
get
y00 = −
1 + 2x
(1 + x + x2 )2
which is positive if (1 + 2x) < 0, or equivalently if x < − 21 , so the interval in question is
(−∞, − 12 ).
5. Let
f ( x) =
and
g( x) =



0




x
if x < 0
if 0 ≤ x ≤ 1


2−x




0
Z x
0
if 1 < x ≤ 2
if x > 2
f (t)dt
(a) Find an expression for g( x) similar to the one for f ( x)
(b) Sketch the graphs of f and g
(c) Where is f differentiable? Where is g differentiable?
R
Solution. We apply FTC by intervals, with the first part being 0x 0dx = 0; next, the interval
Rx
x2
0 x1{ x∈(0,1]} dt = 2 . For the third part, we break the integral into two pieces, with
Z 1
0
f ( x)dx +
Z x
1
f ( x)1{x∈(1,2]} dx =
x
x2
x2
1
+ 2x −
= 2x −
−1
2
2 1
2
h
For the interval x > 2, we evaluate the endpoint of g(2), that is 2x −
g( x) =



0




 x2
−1
i
x=2
=1
if x < 0
if x ∈ [0, 1]
2


2x −





1
x2
2
x2
2
−1
if x ∈ (1, 2]
if x > 2
6. f is not differentiable at x = 0, 1, 2, so has derivative on the interval (−∞, 0) ∪ (0, 1) ∪
3
(1, 2) ∪ (2, ∞) while g is differentiable everywhere on R.
7. Find a function f and a number a such that for all x > 0
6+
Z x
f (t)
t2
a
√
dt = 2 x
3
Solution. Certainly f (t) has to be a polynomial; consider f (t) = −t 2 – using FTC, we
differentiate both sides, i.e.
1
f ( x)
= −√
2
x
x
3
which justifies the claim that f ( x) = − x 2 . Plugging back in the integral, integrating the
function, we obtain
√
√
√
6+2 x−2 a = 2 x
which after cancellation gives a = 9.
8. Prove that
Z b
a
x2 dx =
b3 − a3
3
Solution. Here, we apply the definition: we have ∆x =
( a + i∆x)2 . This leads to
Z b
a
2
n
x dx = lim
∑
n→∞
b−a
(b − a)2
a + 2ia
+ i2
n
n2
2
i =1
a2 (b −
n
2a(b − a)2
n→∞
n
n2
i =1
1
2
2
= (b − a) a + a(b − a) + (b − a)
3
1 2
= (b − a)
(b + ab + a2 )
3
= lim
a)
∑ 1 + nlim
→∞
4
b− a
n
and xi = a + i∆x, with f ( xi ) =
b−a
n
n
∑ i + nlim
→∞
i =1
(b − a)3
n3
n
∑ i2
i =1
=
1 3
(b − a3 )
3
by first expanding, regrouping terms together, using linearity and pulling out constants in
front of sums. We have formulas for ∑in=1 i, ∑in=1 i2 ; finally, taking limits and regrouping the
polynomial terms show that all but the cubic terms vanish. This concludes the proof.
Comparison properties
Theorem 1
Let a ≤ x ≤ b and f ( x), g( x) be integrable functions. Then
R
◦ If f ( x) ≥ 0 ∀ x then ab f ( x)dx ≥ 0
R
R
◦ If f ( x) ≥ g( x) ∀ x, then ab f ( x)dx ≥ ab g( x)dx
◦ If m ≤ f ( x) ≤ M for all x, then
m(b − a) ≤
Z b
a
f ( x)dx ≤ M(b − a)
9. Use the properties of integrals to verify the inequality without evaluation the integrals
2≤
Z 1 p
−1
√
1 + x2 dx ≤ 2 2
Solution. We know that −1 ≤ x ≤ 1, since this is the domain on which we integrate.
Squaring both sides gives 0 ≤ x2 ≤ 1, we get
0 ≤ x2 ≤ 1
⇔
1 ≤ x2 + 1 ≤ 2
⇔
1≤
p
x2 + 1 ≤
√
2
It suffices now to use the comparison properties and remember the formula for the integral of
a constant, which yields the inequality of the statement.
5
1. Fundamental theorem of calculus
Theorem 2 (Fundamental theorem of calculus - part 1)
If f is a continuous function on an interval I = [ a, b], then the function F ( x) defined as
F ( x) =
Z x
a
f (t)dt
for x ∈ I is continuous on the interval, differentiable on ( a, b) and its derivative F 0 ( x) = f ( x).
If we see instead ag(x) f (t)dt, then we can use the chain rule and make the substitution u =
g( x), du = g0 ( x)dx.
10. Verify by differentiation that the formula is correct
R
Z
√
√
1
x2 x2 + a2
dx = −
x2 + a2
+C
a2 x
Solution. On the left hand side, the FTC allows us to get precisely the integrand. On the
right, we differentiate using the product of the quotient rule to get
2x
1
− √
+
2 x2 + a2 a2 x
√
x2 + a2
a2 x2
√
and rearranging gives indeed ( x2 x2 + a2 )−1 .
Find the general indefinite integral
(a)
Z
sin( x)
dx
1 − sin2 ( x)
Solution. Make the substitution 1 − sin2 ( x) = cos2 ( x), so that cos( x) = u, du = − sin( x)dx.
We then have the simpler equivalent integral
−
Z
1
1
1
du = + C =
+C
u2
u
cos( x)
6
provided that {0} is excluded from the domain so that the integral is well-defined.
(b)
sin(2x)
dx
sin( x)
Z
Solution. Use the double angle formula sin(2x) = 2 sin( x) cos( x), in which case we have
simply
2
Z
cos( x)dx = 2 sin( x) + C
Evaluate the integral
(a)
Z 2
−1
| x − x2 |dx
Solution. The domain on which the parabola is negative is (−∞, 0) ∪ (1, ∞); we use linearity to break the integral into three regions. Note that the integrand will be the same, although
the absolute value guarantees that the function stays positive.
Z 0
−1
( x2 − x)dx +
Z 1
0
( x − x2 )dx +
Z 2
1
( x2 − x)dx
Recall that the absolute value function is defined as

 f ( x)
if f ( x) ≥ 0
| f ( x)| =
− f ( x) otherwise
We then have our primitive is either ±
second to
1
2
− 13 , the third to
7
3
x3
3
−
x2
2
. The first integral evaluates to
− 32 . Final answer is thus
7
11
6 .
1
3
+ 12 , the
(b)
Z −2 4
x −1
−5
x2 + 1
dx
Solution. We have a perfect square, might as well break it. Surprise; you get ( x2 − 1)( x2 + 1)
i−2
3
in the numerator, are left with x2 − 1, which integrates to x3 − x
−5
(c)
π
4
Z
0
1 + cos2 (θ )
dθ
cos2 (θ )
Solution. Here, instead of making a substitution, we break the integral into two parts, one
of which is
π
4
Z
0
π
sec2 (θ )dθ = sec(θ ) tan(θ )]04
and the latter which is just the integral of a constant, that is π /4.
Substitution rule If u = g( x) is a differentiable function whose range is an interval I and f is
continuous on I, then
Z
f ( g( x)) g0 ( x)dx =
Z
f (u)du
If we have an indefinite integral on I = [ a, b] and g0 ( x) is continuous on [ a, b], f is continuous
on the range of u = g( x), then
Z b
a
f ( g( x)) g0 ( x)dx =
Z g(b)
g( a)
Evaluate the indefinite integral
(a)
Z
√
1 + 4x
1 + x + 2x2
8
dx
f (u)du
Solution.
Let u = 1 + x + 2x2 , so that du = dx · (1 + 4x). The integral is then after
substitution equal to
Z
1
√ du
u
which gives 1 + x + 2x2 + C.
(b)
Z
cos4 ( x) sin( x)dx
Solution. Take cos( x) = u, − sin( x)dx = du. The integral becomes − u4 du, so the answer
5
is − cos5(x) + C.
R
(c)
Z
sec3 ( x) tan( x)dx
Solution. It is perhaps more obvious to see which substitution is appropriate if we write
R
sec2 ( x) × sec( x) tan( x). We take u = sec( x), get u2 du and the answer is sec3 ( x)/3 + C. 4.
Z
1+x
dx
1 + x2
Solution. Break into two integrals, the first has antiderivative arctan, the second can be
dealt with using substitution 1 + x2 = u, du = 2xdx, so as to get
1
2
Z
1
1
du = log(|1 + x2 |) + C
u
2
9
5.
Z
√
x2
dx
1−x
√
Solution. Here, no choice but to get ride of the annoyance 1 − x; take the change of
variable u = 1 − x, x = 1 − u and the differentials are du = −dx. This gives us something
slightly easier to work with:
Z
√
1 − 2u + u2
2 3
2 5
√
du = 2 u − u 2 + u 2 + C
3
5
u
and the answer in terms of x is then obtained by replacing back u = 1 − x.
Evaluate the definite integral if it exists.
(a)
Z 2
0
dx
(2x − 3)2
Solution. This integral does not exist, since 2x − 3 = 0 ⇔ x =
3
2
∈ [0, 2].
(b)
Z 3
0
dx
2x + 3
Solution. Make a change of variable 2x + 3 = u, du = 2dx. Answer is log(3)/2.
(c)
π
3
Z
0
sin(θ )
dθ
cos2 (θ )
Solution. Let cos( x) = u; the definite integral is 1.
10
4.
π
3
Z
− π3
sin5 (θ )dθ
Solution. Odd function on a symmetric interval: the integral is zero.
5.
1
2
Z
0
Solution. Let u = arcsin( x), du =
approximately 0.137078.
arcsin( x)
√
dx
1 − x2
to get arcsin2 ( x)/2. The numerical answer is
√ 1
1− x2
6.
Z a
−a
x
p
x2 + a2 dx
Solution. Let u = x2 + a2 , du = 2xdx
11. If a and b are positive numbers, show that
Z 1
0
x a (1 − x)b dx =
Z 1
0
xb (1 − x) a dx
Solution. Making the change of variable x 7→ 1 − x, that corresponds to u = 1 − x, we get
[0, 1] → [1, 0], and
Z 1
0
x a (1 − x)b dx =
Z 0
1
(1 − u)a ub (−du) =
Z 1
0
(1 − u)a ub du =
Z 1
0
(1 − x)a xb dx
The integrand corresponds to the probability distribution function Beta random variable. 11