Section 7.1 Areas Between Curves
2010 Kiryl Tsishchanka
Areas Between Curves
We define the area A of S as the limiting value of the sum of the areas of the above approximating
rectangles:
n
X
A = lim
[f (x∗i ) − g(x∗i )] ∆x
n→∞
i=1
From this it follows that the area A of the region bounded by the curves y = f (x), y = g(x), and the lines
x = a, x = b, where f and g are continuous and f (x) ≥ g(x) for all x in [a, b], is
A=
Zb
a
[f (x) − g(x)]dx
(1)
1
EXAMPLE: Find the area of the region bounded above by y = , bounded below by the x-axis, and
x
bounded on the sides by x = 1 and x = 2.
Solution 1: Since y =
A=
Z2
1
1
is positive on [1, 2], we have
x
i2
1
dx = ln |x| = ln 2 − ln 1 = ln 2 − 0 = ln 2
x
1
Solution 2: We can use (1) with f (x) =
since f (x) > g(x) on [1, 2]. We have
A =
Zb
a
[f (x) − g(x)]dx =
= ln |x|
Z2 1
i2
1
1
and g(x) = 0,
x
Z2
1
1
− 0 dx =
dx
x
x
1
= ln 2 − ln 1 = ln 2 − 0 = ln 2
EXAMPLE: Find the area of the region bounded above by the x-axis, bounded below by y = x3 , and
bounded on the sides by x = −1 and x = 0.
1
Section 7.1 Areas Between Curves
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of the region bounded above by the x-axis, bounded below by y = x3 , and
bounded on the sides by x = −1 and x = 0.
Solution 1: Since x3 ≤ 0 on [−1, 0], we have
A=−
Z0
x4
x dx = −
4
3
0
−1
04 (−1)4
1
=− +
=
4
4
4
−1
Solution 2: We can use (1) with f (x) = 0 and g(x) = x3 ,
since f (x) ≥ g(x) on [−1, 0]. We have
A =
Zb
a
[f (x) − g(x)]dx =
Z0
−1
0 − x3 dx = −
Z0
x3 dx
−1
0
x4
04 (−1)4
1
=−
=− +
=
4 −1
4
4
4
EXAMPLE: Find the area of the region bounded by y = x, y = cos x, x = 0, and x = π/6.
Solution: We can use (1) with f (x) = cos x and g(x) =
x, since f (x) > g(x) on [0, π/6]. We have
A =
Zb
Zπ/6
(cos x − x) dx
[f (x) − g(x)]dx =
a
π/6 0 x2
(π/6)2
02
π
= sin x −
−
− sin 0 −
= sin
2 0
6
2
2
1 π2
= −
2 72
EXAMPLE: Find the area of the region bounded by y = 9 − x2 , y = x + 1, x = −1, and x = 2.
2
Section 7.1 Areas Between Curves
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of the region bounded by y = 9 − x2 , y = x + 1, x = −1, and x = 2.
Solution: We can use (1) with f (x) = 9 − x2 and g(x) =
x + 1, since f (x) > g(x) on [−1, 2]. We have
A =
Zb
[f (x) − g(x)]dx =
a
Z2
−1
Z2
(9 − x2 ) − (x + 1) dx
2
x2 x3
−
=
8 − x − x dx = 8x −
2
3 −1
−1
8
1 1
1
39
= 16 − 2 −
− −8 − +
= 22 − 3 + =
3
2 3
2
2
2
EXAMPLE: Find the area of the region bounded by y = x2 − 2x and y = x + 4.
Solution: To find the points of intersection we solve the following equation
x2 −2x = x+4
⇐⇒
x2 −3x−4 = 0
⇐⇒
(x+1)(x−4) = 0
therefore x = −1, 4. We can use (1) with f (x) = x + 4 and
g(x) = x2 − 2x, since f (x) ≥ g(x) on [−1, 4]. We have
A =
Zb
a
Z4
[f (x) − g(x)]dx =
Z4
−1
(x + 4) − (x2 − 2x) dx
4
x3 3x2
+ 4x
=
−x + 3x + 4 dx = − +
3
2
−1
−1
125
64
1 3
+ −4 =
= − + 24 + 16 −
3
3 2
6
2
EXAMPLE: Find the area of the region bounded by y =
3
√
x + 3 and y =
x+3
.
2
Section 7.1 Areas Between Curves
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of the region bounded by y =
√
x + 3 and y =
x+3
.
2
Solution: To find the points of intersection we solve the following equation
2 x + 3 2
√
√
x+3
1
x+3=
x+3 =
⇐⇒
⇐⇒ x + 3 = (x + 3)2
2
2
4
which can be rewritten as
4(x + 3) − (x + 3)2 = 0
(x + 3)(1 − x) = 0
√
therefore x = −3, 1. We can use (1) with f (x) = x + 3 and
x+3
, since f (x) ≥ g(x) on [−3, 1]. We have
g(x) =
2
A=
Zb
a
⇐⇒
[f (x) − g(x)]dx =
Z1 √
x+3
x+3−
2
dx
−3
1
(x + 3)2
2
3/2
= (x + 3) −
3
4
−3
4
16
=
− 4 − (0 − 0) =
3
3
EXAMPLE: Find the area of the region bounded by y =
4
√
x, y =
x
, and x = 5.
2
Section 7.1 Areas Between Curves
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of the region bounded by y =
√
x, y =
x
, and x = 5.
2
Solution: To find the points of intersection we solve the
following equation
√ 2 x 2
√
1
x
⇐⇒
⇐⇒ x = x2
x=
x =
2
2
4
which can be rewritten as
x2 − 4x = 0
⇐⇒
x(x − 4) = 0
therefore x = 0, 4. Note that
x √
≥ x on [4, 9]. Therefore
2
A =
Z4 0
√
x
dx +
x−
2
2
2
x
= x3/2 −
3
4
4
Z5 4
√
x ≥
x
on [0, 4] and
2
x √ − x dx
2
x2 2 3/2
+
− x
4
3
0
5
4
√
107 − 40 5
=
12
EXAMPLE: Find the area of the region bounded by y = sin x, y = cos x, x = 0, and x = π/2.
Solution: The points of intersection occur when sin x =
cos x, that is, when x = π/4 (since 0 ≤ x ≤ π/2).
Observe that cos x ≥ sin x when 0 ≤ x ≤ π/4 but
sin x ≥ cos x when π/4 ≤ x ≤ π/2. Therefore the required area is
Zπ/2
| cos x − sin x|dx
A =
0
Zπ/2
Zπ/4
π/2
+ [− cos x − sin x]π/4
= (cos x − sin x)dx + (sin x − cos x)dx = [sin x + cos x]π/4
0
0
=
π/4
√
1
1
1
1
√ + √ − 0 − 1 + −0 − 1 + √ + √
= 2 2−2
2
2
2
2
EXAMPLE: Find the area enclosed by the line y = x − 1 and the parabola y 2 = 2x + 6.
5
Section 7.1 Areas Between Curves
2010 Kiryl Tsishchanka
EXAMPLE: Find the area enclosed by the line y = x − 1 and the parabola y 2 = 2x + 6.
Solution 1: By solving the equation
y2 − 6
y+1=
2
we find that the points of intersection are (−1, −2) and (5, 4). We solve the equation of the parabola for
x and notice from the picture that the left and right boundary curves are
1
xL = y 2 − 3 and xR = y + 1
2
We must integrate between the appropriate y-values, y = −2 and y = 4. Thus
A =
Z4
Z4 Z4 1 2
1 2
(xR − xL )dy =
(y + 1) −
y − 3 dy =
− y + y + 4 dy
2
2
−2
−2
3
2
y
1 y
+
+ 4y
= − ·
2 3
2
−2
4
=
−2
1
4
− · 64 + 8 + 16 −
+ 2 − 8 = 18
6
3
Solution 2: We can find the area by integrating with respect to x instead of y, but the calculation is much
more involved:
Z5
Z−1
√
√
√
A = [ 2x + 6 − (− 2x + 6)]dx + [ 2x + 6 − (x − 1)]dx
−1
−3
=2
Z−1
√
2x + 6dx +
−3
1
= 2 (2x + 6)3/2
3
Z5
√
[ 2x + 6 − x + 1]dx
−1
5
1
x2
3/2
+ (2x + 6) −
+x
= 18
3
2
−3
−1
−1
6