Question

To evaluate 43/2 find
A) the square of 4 and then take the cube root.
B) the cube of 4 and then take the square root.
C) the square root of 2 times the cube root of 4.
D) the cube root of 4 and then take the square root.

Answer 1

Are you sure it's 43/2?

Answer 2

Problem 26 from Section 7.3 in Stewart:
Evaluate
Z
x
2
(3 + 4x − 4x
2
)
3/2
dx.
Solution:
The denominator is a mess; we need to complete the square to get any closer
to an integral that we might know how to evaluate. To find the appropriate
constants, we collect terms according to the power of x:
3 + 4x − 4x
2 = A
2 − (Bx + C)
2
= (A
2 − C
2
) − 2BCx − B
2x
2
.
This means, in particular, that B = 2, since the coefficient of x
2 must agree
on both sides of the equation. Furthermore,
2BC = 4C = −4,
by equating coefficients of the linear terms and plugging in B = 2. Solving
for C, this gives C = −1. By comparing the constant terms, we have
A
2 − C
2 = A
2 − 1 = 3,
since we know that C = −1. Then A = 2, and we can rewrite the integral as
Z
x
2
(4 − (2x − 1)2
)
3/2
dx.
Now, we make a substitution: 2x − 1 = A sin θ = 2 sin θ. Solving for x gives
x = sin θ +
1
2
Differentiating the substitution equation yields 2 dx = 2 cos θ dθ, i.e., dx =
cos θ dθ. Now, we may plug it in:
Z
x
2
(4 − (2x − 1)2
)
3/2
dx =
Z
(sin θ +
1
2
)
2
(4 − 4 sin2
θ)
3/2
cos θ dθ
=
Z
(sin2
θ + sin θ +
1
4
) cos θ
(4 − 4 sin2
θ)
3/2
dθ
=
Z
(sin2
θ + sin θ +
1
4
) cos θ
4
3/2
(1 − sin2
θ)
3/2
dθ
1
=
Z
(sin2
θ + sin θ +
1
4
) cos θ
8(cos2 θ)
3/2
dθ
=
Z
(sin2
θ + sin θ +
1
4
) cos θ
8 cos3 θ
dθ.
Now we may cancel a single factor of cos θ from the numerator and denominator,
and split up the integral:
Z
sin2
θ + sin θ +
1
4
8 cos2 θ
dθ =
Z
sin2
θ
8 cos2 θ
dθ +
Z
sin θ
8 cos2 θ
dθ +
Z
1
32 cos2 θ
dθ
=
1
8
Z
tan2
θ dθ +
1
8
Z
sec θ tan θ dθ +
1
32 Z
sec2
θ dθ.
The last two integrals come straight from our integral tables. The first one
requires a touch more work, namely, using the trigonometric identity tan2
θ =
sec2
θ − 1. This gives
1
8
Z
(sec2
θ − 1) dθ +
1
8
sec θ +
1
32 Z
sec2
θ dθ
= −
θ
8
+
1
8
sec θ +

1
8
+
1
32 Z
sec2
θ dθ
= −
θ
8
+
1
8
sec θ +
5
32
tan θ + C.
Now we need to convert this back to an expression in terms of x. First of all,
since 2x − 1 = 2 sin θ, we can solve for θ to get
θ = sin−1
(x −
1
2
).
In order to write sec θ and tan θ in terms of x, we need a right triangle. To
have sin θ = x − 1/2, we make the length of the side opposite the angle
θ equal to x − 1/2; the hypotenuse we just set to 1. (Recall that sin =
opposite/hypotenuse.) By the Pythagorean Theorem, this makes the side
adjacent to the angle θ have length
s
1
2 −

x −
1
2
2
=
s
1 −

x
2 − x +
1
4

=
r
3
4
+ x − x
2
.
Then
sec θ =
hypotenuse
adjacent =
1
q
3
4 + x − x
2
,
2
and
tan θ =
opposite
adjacent =
x −
1
q
2
3
4 + x − x
2
=
2x − 1
√
3 + 4x − 4x
2
.
Putting the pieces together, we have the answer
Z
x
2
(3 + 4x − 4x
2
)
3/2
dx = −
sin−1

x −
1
2

8
+
1
8
q
3
4 + x − x
2
+
10x − 5
32√
3 + 4x − 4x
2
+ C
= −
sin−1

x −
1
2

8
+
1
4
√
3 + 4x − 4x
2
+
10x − 5
32√
3 + 4x − 4x
2
+ C
= −
sin−1

x −
1
2

8
+
8
32√
3 + 4x − 4x
2
+
10x − 5
32√
3 + 4x − 4x
2
+ C
= −
sin−1

x −
1
2

8
+
10x + 3
32√
3 + 4x − 4
got it