Area

Back to Area:

We can calculate the area between the x-axis and a continuous function f on the interval [a,b] using the definite integral:

Where f(xi*) is the height of a rectangle and ∆x is the width of that rectangle. {(b-a)/n (n is the number of rectangles)}

Remember that the area above the axis is positive and the area below is negative.

f (x)dxa

b

∫ =limn→ ∞

f(xi* )

i=1

n

∑ Δx

Set up the integral needed to find the area of the region bounded by: and the x-axis.y =x2 −x−2

− (x2 − x − 2)dx−1

2

∫

Set up the integral needed to find the area of the region bounded by: , the x-axis on [0,2].y =1−x2

− (1− x2 )dx1

2

∫

(1−x2 )dx0

1

∫

Area bounded by two curves

Suppose you have 2 curves, y = f(x) and y = g(x)

Area under f is: f (x)dxa

b

∫ g(x)dxa

b

∫Area under g is:

Superimposing the graphs, we look at the area bounded by the two functions:

f (x)dxa

b

∫ − g(x)dxa

b

∫ = ( f (x) − g(x))dxa

b

∫

(top - bottom)*∆x

The area bounded by two functions can be found:

A =topfunction

⎛⎝⎜

⎞⎠⎟−

bottomfunction

⎛⎝⎜

⎞⎠⎟dx

a

b

∫             (a≤b)

Find the area of the region between the two functions: and

y =x2 −2x+ 2

y =−x2 + 6

Top Function?

Bottom Function?

Bounds?

y =−x2 + 6

[-1,2]

y =x2 −2x+ 2

Area? [(−x2 + 6)−(x2 −2x+ 2−1

2

∫ )]dx = (−2x2 + 2x + 4−1

2

∫ )dx = 9

Find the area bounded by the curves: and y =x2 −4x−5 y =x+1

Solve for bounds:

x +1=x2 −4x−50 =x2 −5x−6

0 =(x+1)(x−6)

x =6   x=−1

Find the area bounded by the curves: and y =x2 −4x−5 y =x+1

Sketch the graph:

[(x +1)−(−1

6

∫ x2 −4x−5)]dx

(top - bottom)*∆x

(−−1

6

∫ x2 + 5x+ 6)dx

=455

6

Find the area of the region determined by the curves:

andx =

12y2 −3 y =x−1

Bounds? In terms of y: [-2,4]

Points (-1,-2) & (5,4)

Graph?

Solve for y: x =12y2 −3

y =± 2x+ 6

Find the area of the region determined by the curves:

andx =

12y2 −3 y =x−1

Need 2 Integrals!

One from -3 to -1 and the other from -1 to 5.

Area?

[ 2x + 6−3

−1

∫ −(− 2x+ 6)]dx

+ 2x + 6 − (x −1)dx−1

5

∫

Horizontal Cut instead:

Bounds? In terms of y: [-2,4]

Right Function?

Left Function?

Area?

x =12y2 −3

x =y+1

[(y +1−2

4

∫ )−(12y2 −3)]dy = 18

In General:

Vertical Cut: Horizontal Cut:

A =topfunction

⎛⎝⎜

⎞⎠⎟−

bottomfunction

⎛⎝⎜

⎞⎠⎟dx

a

b

∫                   a≤b

A =rightfunction

⎛⎝⎜

⎞⎠⎟−

leftfunction

⎛⎝⎜

⎞⎠⎟dy

c

d

∫                   c≤d

Find the Area of the Region bounded by

andy = x y =x2

Bounds? [0,1]

Bottom Function?

y = xTop Function?

y =x2

Area?( x −x2 )dx

0

1

∫

Find the Area of the Region bounded by

, , and the y-axis y =cos(x) y =sin(x)

Bounds? [0,π/4] and [π/4, π/2]

Bottom Function?

y =cos(x)Top Function?

y =sin(x)Area?

(cos(x) −sin(x))dx0

π4

∫

x =π2

[0,π/4]

Find the Area of the Region bounded by

, , and the y-axis y =cos(x) y =sin(x)

Bounds? [0,π/4] and [π/4, π/2]

Bottom Function? y =cos(x)

Top Function? y =sin(x)

Area?

(cos(x) −sin(x))dx0

π4

∫

x =π2

+ (sin(x) − cos(x))dxπ

4

π

2

∫

Find the area of the Region bounded by y =2x2 +10,  y=4x+16,  x=−2  and  x=5

Bounds? Interval is from -2, 5

Functions intersect at x = -1 and x = 3

Graph?

Top function switches 3 times!

This calculation requires 3 integrals!

Find the area of the Region bounded by y =2x2 +10,  y=4x+16,  x=−2  and  x=5

Area? [(2x2 +10)−2

−1

∫ −(4x+16)]dx

+ [(4x +16)−1

3

∫ − (2x2 +10)]dx

+ [(2x2 +10)3

5

∫ − (4x +16)]dx

Find T so the area between y = x2 and y = T is 1/2.

Bounds? − T , T⎡⎣ ⎤⎦

Top Function?

Bottom Function?

Area?

y =T

y =x2

T −x2( )dx− T

T

∫

Taking advantage of Symmetry 2 T −x2( )dx0

T

∫

Area must equal 1/2: 2 T −x2( )dx0

T

∫ =12

T =93

4Ans: