A. Area Under a Curve

A. AREA UNDER A CURVE:

If y = f (x) is continuous and non-negative on [a, b], then the area under the curve of f from a to b is: A = ∫ f (x) dx

If y = f (x) is continuous and f (x) < 0 on [a, b], then the area under the curve from a to b is: A = - ∫ f (x) dx

If x = g (y) is continuous an non-negative on [c, d], then the area under the curve of g from c to d is: A = ∫ g (y) dy

EXAMPLES:

Find the area under the curve of f (x) = (x – 1) 3 from x = 0 to x = 2.
Find the area of the region bounded by the graph of f(x) = x 2 – 1, the lines x = -2 and

x = 2 and the x-axis.

Find the area of the region bounded by x = y 2, y = -1 and y = 3.
Using a calculator, find the area bounded by f (x) = x 3 + x 2 – 6x and the x-axis.
The area under the curve y = e x from x = 0 to x = k is 1. Find the value of k.
The region bounder by the x-axis, and the graph of y = sin x between x = 0 and x = π is divided into 2 regions by the line x = k. The area of the region 0 ≤ x ≤ π is twice the area of the region k ≤ x ≤ π. Find k.

B. AREA BETWEEN TWO CURVES:

a. If f and g are continuous functions on the interval [a, b], and if f (x) ≥ g (x) for x in

[a, b], then the area of the region bounded above by y = f (x), below by y = g (x), on the left by the line x = a, and on the right by the line x = b is: A = ∫ [f (x) – g (x)] dx

b. If w and v are continuous functions and if w (y) ≥ v (y) for all y in [c, d], then the area of the region bounded on the left by x = v (y), on the right by x = w (y), below by y = c, and above by y = d is: A = ∫ [w (y) – v (y)] dy

NOTE: The choice between the formulas is generally dictated by the shape of the region, and one would usually choose the formula that requires the least amount of splitting. However, if the integrals resulting by one method are difficult to evaluate, then the other method might be preferable, even if it requires more splitting.

EXAMPLES:

1. Find the area of the region bounded by the graphs of f (x) = x 3 and g (x) = x.

(Sketch the graph, find the points of intersection, set up integrals.)

2. Find the area of the region bounded by the curve y = e x, the y – axis and the line

y = e 2.

3. Using a calculator, find the area of the region bounded by y = sin x and y = x / 2

between 0 ≤ x ≤ π.

4. Find the area of the region bounded by the curve xy = 1 and the lines y = -5, x = e and

x = e 3.

VOLUMES AND DEFINITE INTEGRALS

C. SOLIDS WITH KNOWN CROSS SECTIONS:

If A (x) is the area of a cross section of a solid and A (x) is continuous on [a, b], then the volume of the solid from x = a to x = b is: V = ∫ A (x) dx

NOTE: A cross section of a solid is perpendicular to the height of the solid.

EXAMPLES:

1.The base of a solid is the region enclosed by the ellipse (x 2 / 4) + (y 2 / 25) = 1. The cross sections are perpendicular to the x-axis and are isosceles triangles whose hypotenuses are on the ellipse. Find the volume of the solid.

2. Find the volume of a pyramid whose base is a square with a side of 6 feet long, and a height 10 feet.

D. THE DISC METHOD: The volume of a solid of revolution using discs:

a. Revolving about the x-axis:V = π ∫ (f (x)) 2 d x; f (x) represents the radius.

b. Revolving about the y-axis:V = π ∫ (g (y)) 2 d y; g (y) represents the radius.

c.Revolving about a line y = k: V = π ∫ (f (x) – k) 2 dx; | f (x) – k | represents the radius.

d.Revolving about a line x = h:V = π ∫ (g (y) – h) 2 dy; | g (y) – h | represents the radius.

EXAMPLES:

1. Find the volume of the solid generated by revolving the region bounded by the graph of

f (x) = √ x –1, the x – axis and the line x = 5 about the x – axis.

2. Find the volume of the solid generated by revolving the region bounded by the graph of

y = √ cos x where 0 ≤ x ≤ π / 2, and the x – axis about the x – axis.

3. Find the volume of the solid generated by revolving the region bounded by the graph of

y = x 2, the y – axis, and the line y = 6 about the y – axis.

4. Using a calculator, find the volume of the solid generated by revolving the region bounded

by the graph of y = x 2 + 4, the line y = 8 about the line y = 8.

E. THE WASHER METHOD: The volume of a solid (with a hole in the middle) generated by revolving a region bounded by 2 curves:

About the x – axis: V = π ⌠[g (x) 2 – g (x) 2] dx; where f (x) = outer radius and

g (x) = inner radius.

About the y– axis: V = π ⌠[p (y) 2 – q (y) 2] dy; where p (y) = outer radius and

q (y) = inner radius.

About a line x = h: V = π ⌠[R (x) 2 – r (x) 2] dx; where R (x) = outer radius and

r (x) = inner radius.

d. About a line y= k: V = π ⌠[R (y) 2 – r (y) 2] dy; where R (y) = outer radius and

r (y) = inner radius.

EXAMPLES:

1. Using the washer method, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x in the first quadrant about the x – axis.

2. Using the washer method and a calculator, find the volume of the solid generated by revolving the region in example 1 about the line y = 2.

3. Using the washer method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x 2 and x = y 2 about the y – axis.

F. THE CYLINDRICAL SHELLS METHOD: Method for finding volumes that may be applicable when the cross-sectional areas cannot be found or the integration is too difficult.

Note: A cylindrical shell is a solid enclosed by two concentric right circular cylinders. The volume of a cylindrical shell with inner radius r 1, and outer radius r 2, and height h can be written as V = 2 π (average radius) (height) (thickness)

About the y-axis: V = 2 π ∫ x f(x) dx; where f (x) is the height

b. About the x-axis: V = 2 π ∫ y g(y) dy; where f (y) is the height

Notes: 1. This method is applicable in a variety of situations such as: the region may be

enclosed between two curves, or the axis of revolution may be some line other

than the axis.

2. While in the disk and washer method the cross sections are perpendicular to the

axis of revolution, in the cylindrical shell method, the cross sections are parallel

to the axis of revolution.

EXAMPLES:

Use cylindrical shells to find the volume of the solid generated when the region R in the first quadrant enclosed between y = x and y = x 2 is revolved about the

y-axis.

Use cylindrical shells to find the volume of the solid generated when the region R under y = x 2 over the interval [0, 2] is revolved about the x-axis.

PRACTICE PROBLEMS:

Section 1: The use of a calculator is not allowed in this section.

1. Let f (x) = ⌠f (t) dt where the graph of f is given in the figure below:

Evaluate F(0), F(3), and F(5).
On what interval(s) is F decreasing?
At what value of t does F have a maximum?
On what interval is F concave up?

2. Find the area of the region(s) enclosed by the curve f (x) = x 3, the x – axis, and the lines

x = -1 and x = 2.

3. Find the area of the region(s) enclosed by the curve y = │2x – 6 │, the x – axis, and the lines

x = 0 and x = 4.

4. Find the approximate area under the curve f (x) = 1 / x from x = 1 to x = 5, using four

right-endpoint rectangles of equal lengths.

5. Find the approximate area under the curve y = x 2 + 1 from x = 0 to x = 3 using the

Trapezoidal Rule with n = 3.

6. Find the area of the region bounded by the graphs of y = √ x, y = - x, and x = 4.

7. Find the area of the region bounded by the curves x = y 2 and x = 4.

8. Find the area of the region bounded by the graphs of all four equations: f(x) = sin (x / 2);

x – axis, and the lines x = π / 2 and x = π.

9. Find the volume of the solid obtained by revolving about the x – axis, the region bounded by

the graph of y = x 2 + 4, the x – axis, the y – axis, and the line x = 3.

10. The area under the curve y = 1 / x from x = 1 to x = k is 1. Find the value of k.

11. Find the volume of the solid obtained by revolving about the y – axis the region bounded by

x = y 2 + 1, x = 0, y = -1 and y = 1.

12. Let R be the region enclosed by the graph y = 3x, the x – axis and the line x = 4. The line

x = a divides the region R into two regions such that when the regions are revolved about the

x – axis, the resulting solids have equal volume. Find a.

Section 2: The use of a calculator is allowed in this section.

13. Find the volume of the solid obtained by revolving about the x – axis the region bounded by

the graphs of f (x) = x 3 and g (x) = x 2.

14. The base of a solid region is a region bounded by the circle x 2 + y 2 = 4. The cross of the

solid sections are perpendicular to the x – axis and are equilateral triangles. Find the volume

of the solid.

15. Find the volume of the solid obtained by revolving about the y – axis, the region bounded by

the curves x = y 2 and y = x – 2.

For problems 16 – 19, find the volume of the solid obtained by revolving the region as described. Use the figure below.

16. R 1 about the x – axis.

17. R 2 about the y – axis.

18. R 1 about the line BC.

19. R 2 about the line AB.

20. The function f (x) is continuous on [0, 12] and the selected values of f (x) are shown below.

x / 0 / 2 / 4 / 6 / 8 / 10 / 12
f(x) / 1 / 2.24 / 3 / 3.61 / 4.12 / 4.58 / 5

Find the approximate area under the curve of f from 0 to 12 using three midpoint rectangles.