# 1. Find the area of the shaded region.

```1. Find the area of the shaded region.
(a) y = x2 + 1, y = x, x = −1, x = 2
Z
2
A=
(x2 + 1) − x dx
−1
2
Z
=
x2 − x + 1 dx
−1
2
x3 x2
=
−
+ x 3
2
3
−1
2
22
(−1)3 (−1)2
=
−
+2 −
−
+ (−1)
3
2
3
2
8
1 1
=
−2+2 − − − −1
3
3 2
8 1 1
= + + +1
3 3 2
9 3
3
6 3
9
= + =3+ = + =
3 2
2
2 2
2
1
(b) x =
1
,
y2
x = y, y = 2
Z
2
(y − y −2 ) dy
1
2
2
y
y −1 =
−
2
−1 1
2
1 2 1 y +
=
2
y 1
1 2 1
1 2
=
(2) +
(1) +
−
2
2
2
4 1 1
= + − −1=2−1=
2 2 2
A=
2
1
1
1
(c) x = sin(y), x = 0, y = π4 , y =
Z
3π
4
3π/4
(sin(y) − 0) dy
A=
π/4
Z 3π/4
=
sin(y) dy
π/4
3π/4
= (− cos(y))|π/4
= − (cos(3π/4) − cos(π/4))
√
√ !
2
2
−
=− −
2
2
!
√
√
2 2
= 2
=− −
2
3
(d) y = 2 + |x − 1|, y = − 15 x + 7
(
2 + −(x − 1) = −x + 3 x ≤ 1
First note that 2 + |x + 1| =
2 + (x − 1) = x + 1
x≥1
4
Now to find the intersection points of the lines we set them equal to each
other.
1
−x + 3 = − x + 7
5
−5x + 15 = −x + 35
−20 = 4x
−5 = x
→ y = −(−5) + 3 = 5 + 3 = 8
1
x+1=− x+7
5
5x + 5 = −x + 35
6x = 30
x=5
→y =5+1=6
So we have intersections at the points (−5, 8) and (5, 6).
Looking at the graph, we see that we will have to integrate over two
regions to find the total area enclosed by the lines.
Z 6 Z 1 1
1
− x + 7 − (x + 1) dx
A=
− x + 7 − (−x + 3) dx +
5
5
1
−5
Z
1
Z 6
1
1
=
− x + x + 7 − 3 dx +
− x − x + 7 − 1 dx
5
5
−5
1
Z
1
=
−5
4
x+4
5
Z
dx +
1
5
6
6 − x dx = A1 + A2
5
To keep things in order, Integrate one piece at a time.
5
1
4
x + 4 dx
A1 =
5
−5
2
1
4 x
=
+ 4x 5 2
−5
1 !
1 2
=4
x + x
10
−5
1
1
2
2
(1) + 1 −
(−5) + (−5)
=4
10
10
1
25
=4
+1−
+5
10
10
24
=4 − +6
10
−24 + 60
36
72
=4
=2
=
10
5
5
Z
5
6
A2 =
6 − x dx
5
1
2 5
6 x = 6x −
5 2 1
5 !
1 2 =6 x− x 10 1
1
1
2
2
= 6 5 − (5) − 1 − (1)
10
10
1
25
−1+
=6 5−
10
10
24
=6 4−
10
40 − 24
16
48
=6
=3
=
10
5
5
Z
Total area is then A = A1 + A2 =
6
72
5
+
48
5
=
120
5
= 24
(e) y = sin(x), y = cos(x), x = 0, x = 2π
Here is a graph of the indicated region:
The total area would be set up like this:
Z
0
π/4
Z
(cos(x) − sin(x)) dx+
5π/4
π/4
Z
(sin(x) − cos(x)) dx+
2π
(cos(x) − sin(x)) dx
5π/4
However, we can make integrating easier if we change the region from
π
to 9π
and note the symmetry of this graph around the line x = 5π
.
4
4
4
Because of symmetry, we can integrate over half of the region and double
its value.
7
This will be same area as before except we can now write the area as
Z 9π/4
(cos(x) − sin(x)) dx
A=2
5π/4
9π/4
= 2 sin(x) + cos(x)|5π/4
= 2 [(sin(9π/4) + cos(9π/4)) − (sin(5π/4) + cos(5π/4))]
&quot;√
√
√
√ !#
2
2
2
2
=2
+
− −
−
2
2
2
2
h√
√ i
√
√
= 2 2 − (− 2) = 2(2 2) = 4 2
8
2. Find a horizontal line y = k that divides the area between y = x2 and y = 9
into two equal parts.
From the graph, we can see that it is better to Integrate with respect to y
√
than it is with x. Use the functions, x = y and x = 0. One area will be from
0 to k, the other will be from k to 9. We want them equal, so set the integrals
equal to each other, and solve for k.
Z 9
Z k
√
√
y − 0 dy = 2
y − 0 dy
2
k
0
Z 9
Z k
√
√
y dy =
y dy
k
0
2 3/2 k 2 3/2 9 y |0 =
y |k
3
3
k 3/2 − 0 = 93/2 − k 3/2
k 3/2 = 27 − k 3/2
2k 3/2 = 27
27
k 3/2 =
2
2/3
27
9
k=
= √
3
2
4
9
3. Find the Value of m so that the line y = mx divides the region enclosed by the
parabola y = 2x − x2 and the x-axis into two regions of equal area.
First need to value of the total area under the parabola.
2 Z 2
x3 8
12 − 8
4
2
2
(2x − x ) dx = x −
= 4−
− (0) =
=
3 0
3
3
3
0
Here is one way to sketch what we want.
We see that if we integrate the area between the parabola and the line and set
it equal to half of the total area, then we can find our value m. We need to
find the intersection point of the parabola and the line.
mx = 2x − x2
mx = x(2 − x)
m=2−x
m − 2 = −x
x=2−m
10
Now the Area under the parabola and the line is given by:
Z 2−m
(2x − x2 ) − (mx) dx
0
Z 2−m
x(2 − m) − x2 dx
=
0
2
2−m
x
x3 =
(2 − m) −
2
3 0
(2 − m)3
(2 − m)2
(2 − m) −
− (0)
=
2
3
1 1
3
= (2 − m)
−
2 3
=
(2 − m)3
6
Now we want this to be equal to half of the total area under the parabola.
1 4
(2 − m)3
=
6
2 3
3
(2 − m) = 4
√
3
2−m= 4
√
3
m=2− 4
11
```