MODULE 6

APPLICATIONS OF INTEGRALS

Definite Integral:

read as “integral of f(x) from a to b”, where F(x) is the antiderivative of f(x)

Examples:

Note: This is the area of the triangle bounded by y = x , x = 0 and x = 3.

Fundamental Theorem of Calculus:

If f is continuous on [a,b], then the function

has a derivative at every point x in the interval [a,b], and

Area under the curve:

Let y = f(x) be a nonnegative and integrable function over the closed interval [a,b].Then the area under the curve

y = f(x) from a to b bounded below by the x-axis is

Approximating the area under a curve y = f(x):

Divide the interval [a,b] in n equal parts each of length h = (b –a)/n.

The following algorithms can be used to approximate the area.

Left end points:

Right end points:

Trapezoidal Rule:

Simpson’s Rule:

Examples:

Approximate the area under y = sqrt (1 + x) on the interval [1,5] with n = 10.

Here n= (5 – 1)/10 = 0.4

Properties of Definite Integrals:

Note:

Note:

Evaluate the following definite integrals:

(Try checking your answers with a graphing by using : fnInt( f(x), x, a, b ) )

Evaluate it.

Definition: Let f be an integrable function on [a,b].The average mean value of f is defined by

Examples:

1.Find the average value of f(x) = 3x + 7 on the interval [1,4]

2. Find the average value of f(x) = x^2 +1 on the interval [0,2].

Does f assume its average value in the interval?

The function assumes this value when

One of these values lies in the interval [0,2].

3. Use fnInt( f(x), x, a, b) operation on your graphing calculator to find the average value

of:

sqrt(sin(x)) on the interval [0 , 2].
sqrt(4 – x^2) on the interval [-1 , 1].

Answer: a. 0.810368b. 1.913223

Area between curves:

Let f and g be continuous functions on [a,b] such that f(x) is greater than of equal to g(x) for all x in the interval. Then the area of the region enclosed between the curves y = f(x) and y = g(x) from x = a to x = b is given by

Examples:

Note: Use the function fnInt(f(x) - g(x), x, a, b) on a TI graphing calculator to verify

your answer.

Area under a Curve:

Find the area of the region enclosed by y = x^2 and y = x + 2.

The two curves intersect at x = -1 and x = 2

by solving:

2. Find the area of the region enclosed between the curves y = x^2 and y = 8 – x^2.

The curves intersect in x = -2 and x = 2.

3.Find the area of the region enclosed

between y = 1 and y = (sinx)^2
The curves intersect in x =0 and x = pi/2

4. Find the area enclosed in the region between y = x and y = sqrt(x).

Evaluate the integral:

The answer is 1/6.

5. Find the area of the region enclosed between y = x^2 and y = x^4 – 3x^2.

The curves intersect in x = -2, x = 0 and x = 2 and the first curve lies above the second.

Differential Equations and Initial Value Problems: Variables Separable

Procedure:

Step1: Separate the variables.

Step 2: Integrate both sides and add a constant of integration on the right.

Step 3: Evaluate the constant using the initial values.

Solve the following initial value problems:

Using y = 1 when x = 3, we get 1 = (27/2 – 12)+c, c = -1/2.

The solution is:

is the general solution.

is the general solution

y = 2 when x = 0, gives 2 = C

6.The velocity v of a particle is given below:

Find the particles position s at time t if s(0) = 12.

Here ds = (9.8t + 45)dt

Integrating both sides, we get

s = 4.9 t^2 + 45 t + c

Using s = 12 when t = 0, we get 12 = c.

7.Given the acceleration a(t) = sint, find velocity v and particle’s position so that

s(0) = -1 and v(0) = 1.

Note that dv/dt = a(t) = sint

dv = sint dt

Integrating both sides we get, v = -cost +c1

Now v = -1 when t = 0.

-1 = -cos0 +c1

c1 = 0

Thus v = -cost

ds = -cost dt

Integrating both sides, s = -sint +c2

s = 1 when t = 0.

1 = -sin0 +c2

c2 = 1

Thus s = -sint + 1.

Volume of a Solid of a Solid of Revolution:

Let A(x) be an integrable cross section of a solid form x =a to x = b.Its volume is:

Circular Cross Sections:

Examples:

The region in the first quadrant enclosed by the x-axis, y = x^2, and x = 2 is revolved about the x –axis. Find the volume.

Solution:

Find the volume of the solid obtained by rotating about the y-axis the region in the first quadrant bounded by the y-axis, y = 4 and y = x^2.

Solution:

Note here the radius is x = sqrt (y)

Find the volume of the solid obtained by rotating about the x-axis the region bounded by

y = sqrt (16 – x^2), x = 0 and y = 0

Solution:

Note: the solid is a hemisphere of radius 4 and so its area is one half the area of a sphere which is:

Washer Cross Section:

Find the volume of a solid obtained by revolving the region bounded by y = f(x), y = g(x), on the interval [a,b]

so that the curve y = f(x) lies above y = g(x) and both curves lie above the y-axis.

In this case the cross sections perpendicular to the axis of revolution are washers instead of disks.

Example: Take y = f(x) = 2 – x^2 and y = g(x) = x^2.The curves intersect at x = -1 and x = 1 and so a = -1 and b =1.

Volume of a solid using cylindrical method:

Shell Method:

Find the volume of the solid obtained by rotating the region bounded by y = x^2, x = 2 about the x-axis.

Solution::
Note here the shell radius is y and the shell
height is 2 - x = 2 – sqrt (y)

Find the volume of the solid obtained by rotating the region bounded by y = x^2,

x = 2, y = 0 about the y-axis.
Solution:
Here the shell radius is x and the shell height is y.