MAT1320
answers/solutions for practice sheet for second midterm
It is hoped that the solutions do not contain too many typos. . .
Please let me know if you find any errors.
Note that sometimes only the answer is given, but on the test you should show your work!
1. Implicit differentiation, find y 0 (3.5, 3.6)
1 − ey
4. y =
x · ey + 1
p
x − y x2 + y 2
0
p
5. y =
x · x2 + y 2 − y
2y − x
1. y =
y − 2x
0
0
2x(x + y) − x2
2y(x + y)2 + x2
2x − y 2 − 1
or y 0 =
. see note1
2y(x + y) + y 2 + 1
2. y 0 =
3. y 0 =
x − xy y/x
1 + ln(x)xy )
6. y 0 =
4x + y
2y − x
7. y 0 =
1 + y 2 + x4 y 2 + x4 y 4 − 2xy
x2 − 2xy − 2x5 y 3
2. Related Rates
Exercises in Chapter 3.9: 3–7, 9–11, 37–40.
3. Find the linearization L of each function f at the given point a, then use this to
give a linear approximation of the given number u. (3.10)
1
1. Write f (x) = x 5 . Then the linearization is given by L(x) =√f (a) + f 0 (a)(x − a). We have
4
1
f 0 (x) = x− 5 · 15 and so L(x) = 2+ 80
(x−32). To approximate 5 30 we evaluate L(30) = 1.975.
2. We get f 0 (x) = 1/x. So f 0 (a) = 1/1 = 1 and f (a) = ln(1) = 0. This gives L(x) =
f 0 (a)(x − a) + f (a) = 1(x − 1) + 0 = x − 1 Then L(0.9) = −0.1.
3. First we have f 0 (x) = 3(x + 3)2 so f 0 (a) = 3(0 + 3)2 = 27 and f (a) = (0 + 3)3 = 27. Then
L(a) = f 0 (a)(x − a) + f (a) = 27(x − 0) + 27 = 27 + 27x. This gives L(3.01) = 108.27
4. Write f (x) = xx = eln(x)·x so f 0 (x) = xx (1 + ln(x)). Thus f (a) = 11 = 1 and f 0 (a) =
11 (1 + ln(1)) = 1 and L(x) = f 0 (a)(x − a) + f (a) = 1(x − 1) + 1 = x. This gives L(0.9) = 0.9.
4. Riemann Sums (5.1)
Set up the Riemann Sums for the following areas. In particular, give the approximation with n = 6
rectangles using right-hand endpoints; the same using the left-hand endpoints; the same using the
midpoints. Furthermore, write down the general expression in terms of n rectangles. You do not
need to evaluate any of these.
2
x
These two expressions are exactly equal (using x+y
= y 2 + 1). The first is what you might have obtained from
the quotient rule; the second from cross-multiplying and product rule.
1
1. For some reason this grew from an answer to a solution to a review of the topic. . .
We have ∆x = (3−1)/n = (3−1)/6 = 1/3. The first rectangle goes from a = 1 to a = 1+∆x
so from 1 to 1 + 1/3 = 4/3. It’s right-hand endpoint is at 4/, it’s left-hand endpoint is at 1
and its midpoint is at 1 + ∆x/2 = 7/6. To get the next rectangle we just add ∆x to these,
and so on.
For the right-hand endpoints we evaluate the function at the points 1 + 13 , 1 + 2 13 , 1 + 3 13 , 1 +
4 13 , 1 + 5 13 , 1 + 6 31 , which is the same thing as 43 , 53 , 36 , 73 , 38 , 93 . This gives the approximation as
the following.
2( 34 )
2( 53 )
2( 36 )
2( 73 )
2( 83 )
2( 39 )
(1/3) + 5 2
(1/3) + 6 2
(1/3) + 7 2
(1/3) + 8 2
(1/3) + 9 2
(1/3)
2( 43 )2 + 1
2( 3 ) + 1
2( 3 ) + 1
2( 3 ) + 1
2( 3 ) + 1
2( 3 ) + 1
Of course we can pull out the ∆x as a common factor.
2( 43 )
2( 53 )
2( 63 )
2( 73 )
2( 83 )
2( 93 )
(1/3)
+
+
+
+
+
2( 43 )2 + 1 2( 53 )2 + 1 2( 63 )2 + 1 2( 73 )2 + 1 2( 38 )2 + 1 2( 93 )2 + 1
For the left-hand endpoints we evaluate it at the left-hand of each rectangle. So instead of
starting at a + ∆x = 1 + 1/3 and going to a + n∆x = 1 + 6/3 = b = 3 we start at a and go to
a + (n − 1)∆x. So we evaluate the function at 33 , 34 , 53 , 36 , 73 , 83 . This gives the approximation
as the following.
2( 43 )
2( 53 )
2( 63 )
2( 73 )
2( 83 )
2( 33 )
+
+
+
+
+
(1/3)
2( 33 )2 + 1 2( 43 )2 + 1 2( 53 )2 + 1 2( 63 )2 + 1 2( 37 )2 + 1 2( 83 )2 + 1
For the midpoints we evaluate at the middle of each rectangle, namely 76 , 69 , 11
, 13 , 15 , 17 . This
6 6 6 6
gives the approximation as the following.
2( 96 )
2( 11
)
2( 13
)
2( 15
)
2( 17
)
2( 76 )
6
6
6
6
+
+
+ 13 2
+ 15 2
+ 17 2
(1/3)
2+1
)
)
+
1
)
+
1
)
+
1
2( 76 )2 + 1 2( 96 )2 + 1 2( 11
2(
2(
2(
6
6
6
6
P
We could have written these in -notation as well. Number the rectangles 1, 2, 3, 4, 5, 6. For
the right-hand endpoint of rectangle i we add i∆x, so we evaluate it at 1 + i/3 = (3 − i)/3.
For the left-hand endpoint of rectangle i we add (i−1)∆x, so we evaluate it at 1+(i−1)/3 =
(2 − i)/3. For the midpoint of rectangle i we add (i − 1)∆x + ∆x/2, so we evaluate it at
(i − 1)/3 + 1/6 = (2i − 1)/6. So we could have written the following.
6
right
)
1 X 2( 3−i
3
3−i 2
3 i=1 2( 3 ) + 1
left
)
1 X 2( 2−i
3
3 i=1 2( 2−i
)2 + 1
3
mid
)
1 X 2( 2i−1
6
3 i=1 2( 2i−1
)2 + 1
6
6
6
The general expression in terms of n rectangles is the same idea. For the right-hand evaluation
we would have ∆x = (3 − 1)/n = 2/n. The right-hand endpoint of the i-th rectangle is at
a + i∆x = 1 + 2i/n. This gives the following
n
1 X 2(1 + 2i/n)
n i=1 2(1 + 2i/n)2 + 1
In fact we’re -close to the definition of the defininte integral, so let’s just give it.
n
1 X 2(1 + 2i/n)
n→∞ n
2(1 + 2i/n)2 + 1
i=1
lim
Which
P is easier, writing out in full an expression for a particular value of n, writing it down
in -notation, or just writing the general expression in terms of n.
5. Use the given information to evaluate the integral (5.2)
R4
f
(x)
dx
so
f (x) dx = 3 + 3 = 6.
0
−4
−4
R4
R0
R4
2. Since f is an odd function 0 f (x) dx = − −4 f (x) dx so −4 f (x) dx = 0. Then
1. Since f is an even function
R4
f (x) dx =
4
Z
Z
4
(f (x) + 5) dx =
−4
Z
4
3.
Z
0
f (x) dx =
−4
Z
4
f (x) dx +
−4
Z
5 dx = 0 + 5(8) = 40
−4
4
f (x) dx = (−3) + (−12) = −15.
f (x) dx +
−4
R0
0
4. We subtract integrals to get the range of integration we want.
Z 4
Z 4
Z 2
f (x) dx =
f (x) dx −
f (x) dx = 243 − 1 = 242
2
−4
−4
Z 2
Z 4
Z 6
f (x) dx = 99 − 243 = −144
f (x) dx −
f (x) dx =
−4
−4
−6
5. Split up the integrand.
Z 4
Z
(5f (x) − 2g(x) + 3) dx = 5
−4
4
−4
Z
4
f (x) dx − 2
1.
(3x)2 − 1
(3x)2 − 1
0
(3x)
=
3
(3x)2 + 1
(3x)2 + 1
4
2. e( x2 )2 (x2 )0 = 2xex .
4
g(x) dx + 3
−4
= 5(−2) − 2(7) + 3(8)
=0
6. Find the derivatives (5.3)
Z
dx
−4
√
2
√
x +1
x+1
1
3. √ 4
· √
( x)0 = 2
√ 2
2x + 5x + 1 2 x
2 x +5 x +1
√
√ √ 0
arctan( x)
√
4. arctan(2x)(2x) − arctan( x)( x) = 2 arctan(2x) −
2 x
0
5. sin((x2 )3 + 1)(x2 )0 = 2x sin(x6 + 1)
7. Compute the following integrals (4.9, 5.5, 7.1, 7.2 and 7.3)
8. More integrals.
1. We might be tempted to try the substitution u = cos(5x) but this won’t help since we
need an x outside the cos. Looks like a job for integration by parts. We set u = x and
dv = cos(5x)dx, giving du = dx and v = sin(5x)/5. Notice the 5 in the denominator, quick
check that the derivative of v gives the desired dv. . . all done? Perfect, on with the solution.
Z
Z
sin(5x)
sin(5x)
x · cos(5x)dx = x
−
dx
5
5
Now we have a small substitution, namely w = 5x (for sanity, let’s not use u since we already
did!). This gives dw = 5dx.
Z
Z
Z
sin(w) 1
1
cos(5x)
cos(w)
sin(5x)
dx =
× dw =
+C =−
+C
sin(w)dw = −
5
5
5
25
25
25
So our final answer is
Z
x · cos(5x)dx =
x sin(5x) cos(5x)
−
+C
5
25
2. Integration by parts looks good, since we would like the x2 to go away (which it will once we
differentiate it a couple of times). We set u = x2 and dv = e−x dx, which gives x = 2x and
v = −e−x .
Z
Z
Z
2 −x
2 −x
−x
2 −x
x e dx = −x e − 2x(−e )dx = −x e + 2 xe−x dx
This is another integration by parts, we set u = x (the polynomial again) and dv = e−x dx
to get du = dx and v = −e−x . We work out the second integral by itself.
Z
Z
Z
−x
−x
−x
−x
xe dx = −xe − (−e )dx = −xe + e−x dx = −xe−x + e−x + C
This gives the final answer.
Z
x2 e−x dx = −x2 e−x + 2 −xe−x + e−x + C = −x2 e−x − 2xe−x + 2e−x + C
3. Surprisingly enough this is an integration by parts. There are a few ways to do this.
R
We set u = ln(x) and dv = ln(x)dx.
This
gives
du
=
(1/x)dx
and
v
=
ln(x)dx. If this is
R
going to workR we need to know ln(x)dx, which is itself an integration by parts. So in order
to work out ln(x)dx we set w = ln(x) and dz = dx, giving dw = (1/x)dx and z = x.
Z
Z
Z
ln(x)dx = x ln(x) − x(1/x)dx = x ln(x) − dx = x ln(x) − x + C
Now we can go back to our first
R integration by parts, with u = ln(x) and dv = ln()x)dx,
giving du = (1/x)dx and v = dv = x ln(x) − x
Z
Z
2
(ln(x)) dx = ln(x) (x ln(x) − x) − (x ln(x) − x) (1/x)dx
Z
Z
= ln(x) (x ln(x) − x) − ln(x)dx + dx
= ln(x) (x ln(x) − x) − (x ln(x) − x) + x + C
= x(ln(x))2 − 2x ln(x) + 2x + C
Alternatively, we could have started with a different split into “parts”. Set u = (ln(x))2 and
dv = dx. This gives something initially simpler, namely du = (2 ln(x)/x)dx and v = x.
Z
Z
Z
2
2
2
(ln(x)) dx = (ln(x)) x − x(2 ln(x)/x)dx = (ln(x)) x − 2 ln(x)dx
We are back with the integral of ln(x). So we still have another integration by parts. Since
we’ve done it already a few lines up we’ll just reuse that result.
Z
Z
2
2
(ln(x)) dx = (ln(x)) x − 2 ln(x)dx
= (ln(x))2 x − 2 (x ln(x) − x) + C
= x(ln(x))2 − 2x ln(x) + 2x + C
As yet a third option, we might hope to avoid the integration by parts altogether. This is
essentially wishful thinking, but let’s try it. It would be nice to get rid of the ln(x). . . We’re
running out of letters, so let a = ln(x), which makes da = (1/x)dx. Note that this also
means that a = ex .
Z
Z
Z
2
2
(ln(x)) dx = (a) xda = a2 ea da
Now we’re back to a (double) integration by parts, which is very close to a previous question.
As an exercise, finish this integration by parts, substitue back in tersm of x and verify that
we get the same thing.
4
4. We’d like to get rid of the polynomial, so an
R integration by parts is in order. Set u = t and
3
dv = ln(x). This gives du = 4t dt and v = ln(x)dx = . . .. If you read the previous solution
you know what v is but it looks scary. Let’s put that on hold for a moment and give a quick
look around for alternatives.
Notice that the real issue is the product of a polynomial and a logarithm, and if we differentiate a logarithm then. . . So let’s try another strategy: set w = ln(t) and dz = t4 dt. This
gives dw = (1/t)dt and z = t5 /5.
Z 5
Z 4
Z
t
t
1
t5 ln(t)
t5 ln(t)
t5
t5 ln(t)
4
−
× dt =
−
dt =
−
+C
t · ln(t)dt =
5
5
t
5
5
5
25
This worked out better than expected, since the ln(t) went away in one step. What happens
if you follow through with the first attempt?
5. This is not an obvious one. Based on√the fact that it seems
√ like a chain rule kind of situation,
one might try a substitution of w = x and dw = (1/(2 x))dx. This gives dw = (1/(2w))dx
meaning dx = 2w dw, so we get the following.
Z
Z
√
ln( x)dx = 2 ln(w)w dw
It would be nice to differentiate the ln(w), so this suggests u = ln(w) and dv = w dw, which
gives du = (1/w)dw and v = w2 /2.
Z
Z 2
Z
√
w2 ln(w)
w
1
−2
× dw
ln( x)dx = 2 ln(w)w dw = 2
2
2
w
Z
= w2 ln(w) − w dw
= w2 ln(w) −
w2
+C
2
Of course we need to rewrite this in terms of x, which gives the following.
√
Z
√
√ 2 √
√
( x)2
x
ln( x)dx = ( x) ln( x) −
+ C = x ln( x) − + C
2
2
Alternatively,
we might have tried an integration
√
√ by parts
√ right away. We could start with
a = ln( x) and db = dx. This gives da = (1/ x)(1/(2 x))dx = (1/(2x))dx and b = x. We
used a and b instead of u and v, but in fact they are very similar. . . In any case we get the
following.
Z
Z
Z
Z
Z
Z
√
√
√
√
1
x
dx = x ln( x) − + C
ln( x)dx = x ln( x) − x(1/(2x))dx = x ln( x) −
2
2
In the first method, the initial substitution might have made things more clear, but in the
end we can always include this in the integration by parts.
6. Integration by parts will make the x2 go away after two steps. Since it looks like we will
have two rounds of integration by parts, let’s compute the indefinite integral first, and then
evaluate it. Start with u = x2 and dv = sin(x)dx, which gives du = 2x dx and v = − cos(x).
Z
Z
Z
2
2
2
x · sin(x)dx = −x cos(x) − (− cos(x))2x dx = −x cos(x) + 2 x cos(x) dx
Now we do another integration by parts with u = x and dv = cos(x) dx, which gives du = dx
and v = sin(x).
Z
Z
Z
2
2
2
x · sin(x)dx = −x cos(x) + 2 x cos(x) dx = −x cos(x) + 2 x sin(x) − sin(x)dx
= −x2 cos(x) + 2 (x sin(x) + cos(x)) + C
= −x2 cos(x) + 2x sin(x) + 2 cos(x) + C
Now we evaluate this at the limits of integration.
Z 2π
2π
x2 · sin(x)dx = −x2 cos(x) + 2x sin(x) + 2 cos(x) 0
0
= −(2π)2 cos(2π) + 2(2π) sin(2π) + 2 cos(2π)
− −(0)2 cos(0) + 2(0) sin(0) + 2 cos(0)
= −4π 2
Does itRlook weird that putting an x2 in the integrand would make it negative? Quick sanity
2π
check: 0 sin(x) = 0, right? But x2 sin(x) will be smaller where sin(x) > 0 and larger where
sin(x) < 0, so it makes sense that this integral should be negative.
R5
R5
7. The eM in the denominator is really just a negative power, 1 eMM dM = 1 M e−M dM . So this
is an integration by parts. Set u = M and dv = e−M , which gives du = dM and v = −e−M .
Z 5
Z 5
−M
−M 5
M e dM = −M e
−
(−e−M ) dM
1
1
Z1 5
5
= −M e−M 1 +
e−M dM
1
−M 5
−M 5
= −M e
+
−e
1
1
= −5e−5 + 1e−1 − e−5 + e−1
2
6
=− 5 +
e
e
8. Integration by parts, although it looks a bit unclear. Try u = (ln(x))2 and dv = x4 dx, to
get du = 2 ln(x)(1/x) dx and v = (x5 /5)dx.
Z
1
2
2
Z
x5 (ln(x))2 2 2 4
x · (ln(x)) dx =
x ln(x) dx
−5
5
1
1
Z
25 (ln(2))2 − 15 (ln(1))2 2 2 4
=
x ln(x) dx
−
5
5 1
Z
32(ln(2))2 2 2 4
=
−
x ln(x) dx
5
5 1
4
2
Now we integrate the seccond integral, also by parts. This time u = ln(x) and dv = x4 dx,
giving du = (1/x) dx an v = (x5 /5) dx.
2 Z 2 4
5
Z 2
x ln(x) x
4
dx
x ln(x) dx =
−
5
1
1 5
1
5
2 5 2
x ln(x) x =
−
5
25 1
1
32 ln(2) 1 ln(1) 25
15
=
−
−
+
5
5
25 25
32 ln(2) 31
=
−
5
25
Putting it all together we get the following:
Z 2
32(ln(2))2 64 ln(2)
62
32(ln(2))2 2 32 ln(2) 31
4
2
−
−
=
−
+
x · (ln(x)) dx =
5
5
5
25
5
25
125
1
9. This one looks like we want to set u = cos(t) to get du = − sin(t)dt. But there is a problem,
it’s sin(2t) not sin(t). So we have to fix that, using sin(2t) = 2 sin(t) cos(t). Then we set
u = cos(t) to get du = − sin(t)dt.
Z
Z
Z
cos(t)
cos(t)
e
· sin(2t)dt = e
· 2 sin(t) cos(t)dt = −2 eu u du
This is a classic integration by parts to get rid of the factor of u leaving a pure exponental.
Set w = u and dz = eu , which gives dw = du and z = eu .
Z
Z
u
u
u
−2 e u du = −2 ue − e du = −2 (ueu − eu ) + C = −2 cos(t)ecos(t) + 2ecos(t) + C
10. Integration by parts. If the ln were a cos or some such, we would want to differentiate the
polynomial, but here we’d rather differentiate the ln. Set u = ln(x + 1) and dv = x dx, which
gives du = 1/(x + 1) dx and v = x2 /2.
Z
Z
x2
x2 ln(x)
−
dx
x · ln(1 + x)dx =
2
2(x + 1)
The last integral is slightly annoying; it’s a matter of substituting w = x + 1 and dw = dx
to make the denominator be a single term.
Z
Z
x2 ln(x)
x2
−
dx
x · ln(1 + x)dx =
2
2(x + 1)
Z
(w − 1)2
x2 ln(x) 1
=
−
dw
2
2
w
Z 2
x2 ln(x) 1
w − 2w + 1
=
−
dw
2
2
w
Z
x2 ln(x) 1
1
=
−
w − 2 + dw
2
2
w
2
2
x ln(x) 1 w
=
−
− 2w + ln(w) + C
2
2 2
x2 ln(x) 1 (x + 1)2
=
−
− 2(x + 1) + ln(x + 1) + C
2
2
2
11. This is a substitution of u = ln(x) giving du = (1/x) dx. Notice how the denominator fits
perfectly as part of du.
Z
e2
e
5(ln(x))1/5
dx =
x
Z
e2
5(u)
e
1/5
x=e2
x=e2
5u6/5 25 ln(x)6/5 du =
=
6/5 x=e
6
x=e
25 ln(e2 )6/5 25 ln(e)6/5
−
6
6
6/5
25(2)
25
=
−
6
6
25 6/5
= (2 − 1)
6
=
9. And more integrals.
1. Since (cos x)0 = − sin x the antiderivative is − cos x. This gives the following.
Z 8
sin(x)dx = − cos(x)|80 = −(cos 8 − cos 0) = 1 − cos 8
0
9
Z
2.
√
1
Z
xdx =
9
x
1/2
1
9
x3/2 2 2
4
dx =
= − =−
3/2 1 9 3
9
3. One of these is a power and the other is an exponential, careful!
1 Z 1
1 e+1
1 e+1
1 e+1
e2
e
x
x
1
0
(x + e )dx =
x +e =
1 +e −
0 +e =
e+1
e+1
e+1
e+1
0
0
4. The substitution u = z 4 + 1 looks reasonable, giving du = 4z 3 dz. Then we get the following.
√
Z
1
x
z3
dz =
z4 + 1
Z
√
z= x
z=1
z=√x
z=√x
1
1
1
1
1
du = ln(u)
= ln(z 4 + 1)
= ln(x2 + 1) − ln(2)
4u
4
4
4
4
z=1
z=1
Note that we could have changed the limits
instead√instead of rewriting the
√ of integration
4
result in terms of the variable z. If z = x then u = z + 1 = ( x)4 + 1 = x2 + 1 and if
z = 1 then u = z 4 + 1 = 14 + 1 = 2.
x2 +1
Z √x
Z x2 +1
z3
1
1
1
1
dz
=
du
=
ln(u)
= ln(x2 + 1) − ln(2)
4
z +1
4u
4
4
4
1
2
2
What’s definitely not allowed is to leave the integral in terms of u, but substitute the limits
for z!
5. Notice that this function has a discontinuity. . . but it’s outside our range of integration so
we’re ok.
8
Z 8
x1/3 −2/3
1/3
1/3
x
dx =
=3×8 −3×1 =3
1/3
1
1
Z
4
4
2s 20
15
24
2 ds =
−
=
=
ln 2 0 ln 2 ln 2
ln 2
s
6.
0
7. This is essentially the integral of z −1/2 , so power. There
is a discontinuity but it is safely
√ R 18 −1/2
√ z1/2 18
R 18 q 3
tucked out of the way of our range of integration. 1
dz = 3 1 z
dz = 3 1/2 =
z
1
√
√
√ √
2 3 18 − sqrt1 = 2 3 3 2 − 1
8. A substitution of w = 1 + r with dw = dr turns this into a power.
r=1
r=1
Z r=1
Z 1
w4 15
(1 + r)4 16 1
3
3
(1 + r) dr =
w dw =
− =
=
=
4 r=0
4
4
4
4
0
r=0
r=0
We can also change the limits of integration to w. If r = 0 then w = 1 and if r = 1 then
w = 2.
2
Z 1
Z 2
w4 16 1
15
3
3
(1 + r) dr =
w dw =
− =
=
4 1
4
4
4
0
1
9. Trig substitution with x = sin θ, giving dx = cos θdθ.
Z
√
1/ 2
4
√
dx =
1 − x2
1/2
Z
√
x=1/ 2
Z
4
p
cos θdθ =
1 − sin2 θ
x=1/2
√
x=1/ 2
x=1/2
4
cos θdθ = 4
cos θ
Z
√
x=1/ 2
dθ
x=1/2
This is easily integrated, giving the following.
√
√
√
x=1/ 2
x=1/ 2
4 θ|x=1/2 = 4 arcsin x|x=1/2 = 4 arcsin(1/ 2) − arcsin(1/2) = 4 (π/4 − π6) = π/3
Alternatively we can work out the limits of integration in terms of θ. For the same√substitution, we notice
√ that when x = 1/2 we get θ = arcsin(1/2) = π/6 and when x = 1/ 2 we get
θ = arcsin(1/ 2) = π/4.
√
1/ 2
Z
1/2
4
√
dx =
1 − x2
Z
π/4
π/6
Z
4
p
cos θdθ =
1 − sin2 θ
π/4
4
cos θdθ = 4
cos θ
Z
8
sec2 θdθ = 8
sec2 θ
Z
π/6
π/4
dθ
π/6
Again this is now an easy integral.
π/4
4 θ|π/6 = 4 (π/4 − π6) = π/3
10. Trig substitution with x = tan θ, giving dx = sec2 θdθ.
√
Z
3
√
1/ 3
8
dx =
1 + x2
Z
√
x= 3
√
x=1/ 3
8
sec2 θdθ =
1 + tan2 θ
Z
√
x= 3
√
x=1/ 3
√
x= 3
√
x=1/ 3
dθ
This is now an easy integral so we get the following.
√
√
√
√ x= 3√
x= 3√
8 θ|x=1/ 3 = 8 arctan x|x=1/ 3 = 8 arctan( 3) − arctan(1/ 3) = 8 (π/3 − π/6) = 4π/3
Alternatively
√ of integration in terms of√θ. Using the same substitution,
√
√ we write the limits
when x = 3 then θ = arctan( 3) = π/3 and when x = 1/ 3 then θ = arctan(1/ 3) = π/6.
√
Z
3
√
1/ 3
8
dx =
1 + x2
Z
π/3
π/6
8
sec2 θdθ =
1 + tan2 θ
Z
π/3
π/6
8
sec2 θdθ = 8
sec2 θ
Z
π/3
dθ
π/6
This is now an easy integral so we get the following.
π/3
8 θ|π/6 = 8 (π/3 − π/6) = 4π/3
10. Given the clues, find f .
1. Since f 00 (x) is the derivative of f 0 (x), then f 0 (x) is an anti-deivative of f 00 (x) and so f 0 (x) =
− cos(2x)/2 + C. But f 0 (0) = 4 so − cos(2 × 0)/2 + C = 4 so C = 9/2. Thus f 0 (x) =
− cos(2x)/2+9/2. Then f (x) is an anti-derivative of f 0 (x) so f (x) = − sin(2x)/4+(9/2)x+C
(not necessarily the same constant C of course). This time we have f (0) = 1 so sin(2×0)/4+
(9/2) × 0 + C = 1 so C = 1. Thus f (x) = − sin(2x)/4 + (9/2)x + 1.
√
2. We have f 00 (x) = 3x1/2 . Then f 0 (x) will be an antiderivative of this, and f (x) will be an
anti-derivative of f 0 (x). We give the general anti-derivatives, notice that it is really important
to not use the same letter for the two constants.
√
3/2
√
2
3x3
x
+A=
+A
f 0 (x) = 3
3/2
3
√
√
4 3x5
x5/2
f (x) = 3
+ Ax + B =
+ Ax + B
(3/2)(5/2)
15
Now f (0) = 0 so B = 0. Then f (3) = 4 so A = −8/5.