MATH 106 MIDTERM EXAM

SOLUTION
1) Evaluate
Z 2 the2 integrals (5 points each).
x + sin x
(a) dx
x2 sin2 x

x2 + sin2 x x2 sin2 x
Z Z   Z  
1 1 1
dx = + dx = + 2 dx = − cot x − + c.
x2 sin2 x x2 sin2 x x2 sin2 x 2
sin x x x
Z 3
(b) |4 − 2t | dt
0

On the interval [0, 3] we have 2t ≥ 4 for t ∈ [2, 3] and 2t ≤ 4 for t ∈ [0, 2] . So we split the
integral into two parts:
Z 3 Z 2 Z 3 Z 2 Z 3
t t t t
|4 − 2 | dt = |4 − 2 | dt + |4 − 2 | dt = 4 − 2 dt + 2t − 4 dt
0 0 2 0 2

t=2 t=3 
2t 2t
       
4 1 8 4
= 4t − + − 4t = 8− + + − 12 − −8
ln 2 t=0 ln 2 t=2 ln 2 ln 2 ln 2 ln 2

1
=4+ .
ln 2
Z 1
dy
(c) √
0 1+ y
√ √
We use the substitution u = 1 + y, which gives du = 2√1 y dy, so dy = 2 y du = (2u − 2) du
√
(Since y = u − 1). The integral becomes
Z 2 Z 2 
2u − 2 1
du = 2 1− du = 2(u − ln u)|u=2
u=1 = 2 − ln 4.
1 u 1 u

√
Z 2 2
zdz
(d) √
3
.
0 7z 2 + 8

Let u = 7z 2 + 8, then du = 14zdz which means zdz = 1

du. For the limits, z = 0 ⇒ u = 8 and
√ 14
z = 2 2 ⇒ u = 64. Thus, the integral becomes
64 u=64
1 u2/3
Z
1 du 3 9
√ = · = (16 − 4) = .
14 8
3
u 14 2/3 u=8 28 7
 √
Z 2 √ Z π
dx
2) (a) (8 points) Show that arctan 1 + x2 dx < p
0 0 4 − sin2 x
√ √
Let f (x) = arctan 1 + x2 and g(x) = √ 1
. Since 1 + x2 and arctan(·) are increasing
4−sin2 x
functions
√ for x √ > 0, their superposition,
√ that is f is also increasing. Hence the maximum of f
on [0, 2] is f ( 2) = arctan 3 = π/3. This means
√ √ √
Z 2 √ Z 2 √ √ √ 2π
arctan 1 + x2 dx ≤ arctan 3 dx = 2 arctan 3 = .
0 0 3

On the other hand, it is clear that the minimum value of g(x) = √ 1 2 on the interval [0, π]
4−sin (x)
p 2
is precisely when the denominator 4 − sin (x) is maximum, which is when sin x = 0, hence
x = 0. Thus the minimum of g on [0, π] is g(0) = g(π) = √14 = 21 . This gives
Z π Z π
dx dx π
p ≥ = .
0 4 − sin2 x 0 2 2
Finally, we have
√ √
Z 2 √ 2π π
Z π
dx √
arctan 1 + x2 dx ≤ < ≤ p as 2 2 < 3.
0 3 2 0 4 − sin2 x

4x2
(b) (8 points) Find the area of the region enclosed by the curves y = sin x and y = .
π2
2
The two curves intersect at x = 0 and x = π/2. In the interval (0, π/2), we have sin x > 4x
π2
,
which can be simply observed without drawing the graphs, by√evaluating the functions at one
point from (0, π/2) (For example, evaluating at π/4 we get 1/ 2 > 1/4).

Thus, the area of the mentioned region is given by the integral

π/2 x=π/2
4x2 4x3
Z   
π
sin x − 2 dx = − cos x − 2 =1− .
0 π 3π x=0 6
3) (16 points) Solve the differential equation (1 + sin2 x) y ′ = sin 2x (1 + y 2 ) subject to the
condition y(0) = 1.

dy
(1 + sin2 x) = sin 2x (1 + y 2 )
dx
dy sin 2x
⇒ = dx
1+y 2 (1 + sin2 x)
dy 2 sin x cos x
(Since sin 2x = 2 sin x cos x) ⇒ = dx
1+y 2 (1 + sin2 x)
Z Z
dy 2 sin x cos x
⇒ = dx
1+y 2 (1 + sin2 x)
⇒ arctan y = ln (1 + sin2 x) + c
⇒ y = tan (ln (1 + sin2 x) + c)
(y(0) = 1) ⇒ 1 = tan (ln (1 + sin2 0) + c) ⇒ c = π/4
⇒ y = tan (ln (1 + sin2 x) + π/4).
4) Find f ′ (x0 ) if p
1 − sin3 x · (4 + sin2 x)2
(a) (8 points) f (x) = √3
, x0 = 0.
8+x

First of all, f (x0 ) = f (0) = 8. Now, taking logarithm of both sides, we get
1 1
ln f (x) = ln(1 − sin3 x) + 2 ln(4 + sin2 x) − ln(8 + x).
2 3
Differentiating both sides, we obtain

f ′ (x) 1 3 sin2 x cos x 2 sin x cos x 1 1

= · 3 +2· 2 − · .
f (x) 2 1 − sin x 4 + sin x 3 8+x

Evaluating at x = x0 = 0 we get

f ′ (0) 1 1 −1
=− · = .
f (0) 3 8 24
−1 −1 −1
Then f ′ (0) = · f (0) = ·8= .
24 24 3
x
(b) (8 points) f (x) = (ln x)e , x0 = e.

Again, we use logarithmic differentiation.

x
ln f (x) = ln[(ln x)e ] = ex ln ln x.
Differentiating both sides gives

f ′ (x) ex
= ex ln ln x +
f (x) x ln x
Evaluating at x0 = e we obtain

f ′ (e) ee
= ee ln ln e + = ee−1 .
f (e) e ln e
e e
Since f (e) = f (x) = (ln e)e = 1e = 1, we get

f ′ (e) = ee−1 .
 sin x R 1/√2
Z
2 2
5) (16 points) Let f (x) = e−t dt for 0 < x < π/2 and y0 = 0
e−t dt. Prove that f
0
has an inverse function f −1 . Find (f −1 )′ (y0 ).

We want to show that f is one-to-one by showing that its derivative on (0, π/2) is positive. If
f is one-to-one, then it has an inverse defined on the range of f . By FTC 1, we have
2
f ′ (x) = e− sin x
cos x > 0, x ∈ (0, π/2),
since cosine is positive at that interval and exponential is a positive function. Now we know
that f −1 exists. By the definition of an inverse we have x0 = f −1 (y0 ) ⇐⇒ f (x0 ) = y0 , that is
√
Z sin x0 Z 1/ 2
−t2 2
f (x0 ) = e dt = y0 = e−t dt.
0 0
√
Now it is clear that y0 corresponds to x0 such that sin x0 = 1/ 2, i.e. x0 = π/4.
As for (f −1 )′ (y0 ) we have
1
(f −1 )′ (y0 ) = .
f ′ (x 0)

Finally we obtain
1 1 √
(f −1 )′ (y0 ) = = − sin2
= 2e.
f ′ (π/4) e π/4 cos π/4
6) Let f (x) = arcsin x. Find

(a) (3 points) f (− 12 )

arcsin(−1/2) = −π/6.

(b) (3 points) f ′ (− 21 )

1 1 1 2
f ′ (x) = √ ⇒ f ′ (− ) = p =√ .
1−x 2 2 3/4 3

(c) (3 points) f (sin 5π

4
)

5π −1 −π
arcsin(sin ) = arcsin( √ ) = .
4 2 4

(d) (3 points) f (cos 1)

π
arcsin(cos 1) = arcsin(sin(π/2 − 1)) = − 1.
2

Z 1/2
f (t) dt
(e) (4 points) √ .
0 1 − t2

1/2 1/2 t=1/2

1  π 2 π 2
Z Z
f (t) dt 1 1
√ = arcsin t d(arcsin t) = arcsin2 t = arcsin2 (1/2) = = .
0 1 − t2 0 2 t=0 2 2 6 72