EvaluatingIntegrals(Substitution RuleII)

EVALUATING INTEGRALS - SUBStitutiON EVALUATING INTEGRALS USING THE SUBSTITUTION RULE (PART II) This tutorial is continued from Part I. Previously, we examined the substitution rule rule and how it is used to evaluate indefinite integrals. Generally speaking, to evaluate an indefinite integral ∫ f(x) dx, we follow these guidelines: 

Introduce a new variable u into the integral, and let this variable represent ➔ a complicated part of the integral, or ➔ an expression whose derivative also occurs in the integral.



Make the appropriate substitution.



Differentiate u with respect to x, and make the appropriate substitution.



At this point, the integral should have been reduced to a form that can easily be evaluated using the antiderivative formula. Evaluate the indefinite integral.



After evaluating the integral and obtaining a final result, be sure to return to the original variable x.

The method described above also works for definite integrals; the only difference being that the integral will have to be evaluated at the endpoints a and b (i.e, the limits of the integral). In other words, you apply Part 2 of the Fundamental Theorem. Here's an illustration:

EXAMPLE 1

You may be given something like this:

∫

7 0

√4 + 3x dx

So, where do we start? First, from the Fundamental Theorem, we know that

∫

7 0

√4 + 3x dx

∫ √4 + 3x

=

dx

]

7 0

i

Therefore, the first major step should be to evaluate the indefinite integral:

∫ √4 + 3x

dx

ii

At this point, we follow the routine: We let u = 4 + 3x , so that (ii) becomes

∫

Then,

du dx = 3

u ½ dx

so that

iii

dx =

du 3

iv

Therefore, (iii) becomes

∫ which equals

u½ ×

du 3

=

1 3

∫

u ½ du

=

1 u3/2 3 3/2

+ C

1 2u3/2 3 3

2u3/2 + C 9

=

+ C

2(4 + 3x)3/2 9

=

+ C

The final step is to evaluate

2(4 + 3x)3/2 9

+ C

at the endpoints a = 0, b = 7.

In other words, we are to compute

2(4 + 3x)3/2 9

]

The arbitrary constant C isn't of utmost importance here, so we ignore it.

7 0

We therefore have

2(4 + 3x)3/2 9

]

7 0

=

2(4 + 3(7))3/2 9

=

2(25)3/2 9

–

2(4 + 3(0))3/2 9

–

2(4)3/2 9

=

26

Therefore,

∫

7 0

√4 + 3x dx

26

=

So, what did we do to solve this definite integral? First, we had to evaluate its “indefinite integral form” to produce its antiderivative, and then evaluated the antiderivative at the endpoints. Thus, we have applied both the Fundamental Theorem and the Substitution Rule.

There is, however, another way. It's slightly similar to the method used above, but one thing is done differently: the limits of integration are changed. Remember, when we talk of the “limits of integration”, we're referring to the endpoints a and b of the definite integral.

Here's how we change the limits of integration: (We'll use the example above) The first few steps are basically the same: First, we recognize that

∫

7 0

√4 + 3x dx

=

∫ √4 + 3x

dx

i

and then proceed to evaluate the indefinite integral:

∫ √4 + 3x

dx

ii

using the substitution u = 4 + 3x , we get

∫

Next,

du dx = 3

u ½ dx

so that

iii

dx =

du 3

iv

Therefore, (iii) becomes

∫

u½ ×

du 3

=

1 3

∫

u ½ du

This is point where the calculation takes another turn!!

Remember we used the substitution u = 4 + 3x for the integral above. It is this substitution that is used to find the new limits of integration. ➔ From (i), we see that the limits of integration are x = a = 0 (the lower limit), and x = b = 7 (the upper limit). ➔ We then say that, given the substitution u = 4 + 3x, 

when x = 0, then u = 4 + 3(0) = 4, and



when x = 7, then u = 4 + 3(7) = 25.

Hence, the new limits of integration are x = a = 4 (the new lower limit), and x = b = 25 (the new upper limit). ➔ Therefore, the new integral is

1 3

∫

25 4

u ½ du

which can be evaluated easily. So, using the substitution rule,

∫

7 0

√4 + 3x dx

has been reduced to

1 3

∫

25 4

u ½ du

(but they are equal. The definite integral has simply been reduced to a standard form, but with different limits of integration. That's all.). Geometrically speaking, you'll also see that both integrals are always equal:

So now, we evaluate the integral

1 3

∫

25 4

u ½ du

1 3

=

[ 3/2 ] u3/2

25

1 3

=

4

[ ] 2u3/2 3

25

=

4

2u3/2 9

]

25 4

and evaluate with the Fundamental Theorem:

]

2u3/2 9

25

=

4

2(25)3/2 9

–

2(4)3/2 9

=

26

Therefore, using the substitution rule and the Fundamental Theorem, we find that

∫

7 0

√4 + 3x dx

=

1 3

∫

25 4

u ½ du

=

26

This time, there's no need to return to the original variable x because we're evaluating a definite integral. In summary, there are two possible ways to evaluate a definite integral: ✔

Evaluate the “indefinite integral form” and then apply the Fundamental Theorem, or

✔

Change the limits of integration,thereby reducing the relatively complicated definite integral to its simplest possible form. Then evaluate using the Fundamental Theorem.

Whatever method you choose is completely acceptable, as long as you know what you are doing!!

Let's study more examples.

EXAMPLE 2

Evaluate the definite integral, if it exists.

∫

1 0

x2(1 + 2x3)5 dx Solution

We want to evaluate the integral, but first we need to find out if this integral exists on the specified interval. On the basis of graphical evidence, we find that this integral indeed does exist on the interval [0,1]. This means it can be evaluated.

The nature of the integrand can also determine whether an integral exists on an interval. In this case, we see that the integrand is definitely a polynomial. And

from laws of continuity, polynomials are continuous at EVERY number in

their domains. This automatically means that the integral above does exist.

Note that polynomials, trigonometric functions, root functions and rational functions are always continuous at every number in their domains.

So, now that we know that the integral exists, let's evaluate it: We let u = 1 + 2x 3 , which gives

∫ x2 u5

dx

i

Observe that I've discarded of the upper and lower limits. We are going to replace them anyway. Next,

du 2 dx = 6x

and then

du 6

x2 dx =

ii

Next, we change the limits of integration: 

when x = 0, then u = 1 + 2(0)

3

= 1, and



when x = 1, then u = 1 + 2(1)

3

= 3.

Therefore, the new limits are a = 1, and b = 3. And the integral becomes

∫

u5 × du

3 1

1 6

=

6

∫

3 1

u5 du

At this point, we apply the Fundamental Theorem:

1 6

∫

3

u5 du

1

=

1 6

[6 ] u6

3

u6

=

36

1

]

3 1

which results in

u6 36

]

3

1

=

(3)6 36

(1)6 36

–

=

728 36

=

182 9

=

182 9

Therefore,

∫

1 0

x2(1 + 2x3)5 dx

≡

1 6

∫

3 1

u5 du

≈

20.222

EXAMPLE 3

Evaluate the definite integral, if it exists.

∫

√π 0

x cos(x2) dx Solution

First we need to test for continuity. If we graph the integrand f(x)

=

x cos(x2), we see that the integrand is

continuous on the interval [0,√π]. So, we can go ahead to evaluate it.

We put u = x 2, which results in

∫

Next,

du dx = 2x

x cos u dx so that

∫ cos u

=

x dx

i

du 2

x dx =

Thus, (i) becomes

∫ cos u

× ½ du

=

½ ∫ cos u du

ii

Next, we change the limits of integration. Since u = x 2, 

when x = 0, then u = (0)2 = 0, and



when x = √π, then u = (√π)2 = π.

Therefore, the new limits are a = 0, and b = π. So, the integral becomes

½ ∫ 0 cos u du π

Using the Fundamental Theorem,

½ ∫ 0 cos u du

=

1 2

∫

=

0

π

[ ] sin u

π

=

0

½ (sin π – sin 0)

=

So, √π 0

x cos(x2) dx

EXAMPLE 4 Evaluate the definite integral, if it exists.

∫

4 1

1 x2

√

1+ 1 x

dx

Solution

If we graph the integrand, we find that it is continuous on the interval [1,4]. Thus, the integral exists.We put

u = and get this:

1+ 1 x

OR

u = 1 + x –1

0

∫

1 u ½ dx x2

∫

=

u½

dx

x2

i

Next,

du dx =

–1 x

dx x2

which means

2

=

ii

– du

Put in (ii) to (i) get

∫

u ½ × – du

=

∫

–

u ½ du

iii

We then change the limits of integration. Since u = 1 + x–1 , then 

when x = 1, then u = 1 + (1)–1

=

2, and



when x = 4, then u = 1 + (4)

=

5/4.

–1

Therefore, the new limits are a = 2, and b = 5/4. So, (iii) becomes

–

∫

5/4 2

u ½ du

iv

Take a good look at the integral above. Do you notice anything unusual? Well, there is. Observe the limits: the upper limit is less than the lower limit, which should not be so. To correct this anomaly, we apply one of the properties of the definite integrals which we had studied earlier:

∫

b a

f (x) dx

∫

–

=

a b

f (x) dx

So, to correct (iv), we multiply the integral by a minus sign. This will “swap” the limits, resulting in

∫

2 5/4

u ½ du

v

Next, we evaluate (v) using the Fundamental Theorem:

∫

2 5/4

u ½ du

u 3/2 3/2

=

]

2

5/4

which equals

(2) 3/2 (5/4) 3/2 = 3/2 – 3/2

=

4√2 3

=

(5)3/2 (4)3/2

(√2)3 – 3/2

3/2

(√5)3

–

(√4)3 3/2

=

4√2 3

–

5√5 8 3/2

4√2 3

=

2 3

5√5 8

–

4√2 3

=

–

5√5 12

Therefore,

∫

4 1

1 x2

√

1+ 1 x

dx

=

4√2 3

–

5√5 12

The result is the precise value of the integral. If we want an approximation, we'd get something like 0.95392.

EXAMPLE 5

Evaluate the definite integral, if it exists.

∫

2 0

dx (2x – 3)2

Solution First, we rewrite the integral:

∫

2 0

1 (2x – 3)2

dx

There are two ways to test whether this integral exists. One way is to use the principle of continuity. If this integral were to exist, then it means that the integrand should be continuous on the interval [0,2].

We find that the integrand is a rational function, a kind of function which is not defined when the denominator is zero. For this particular integrand, we find that the denominator is zero when x = 1.5 (and this value falls in the range [0,2]). This is an indication that the integrand is NOT CONTINUOUS on [0,2] (or in other words, the integrand is DISCONTINUOUS at x = 1.5), and so the integral DOES NOT EXIST. We can also make use of graphical evidence; if we graph the integrand, we get this:

Notice the discontinuity in the graph. Again, this shows that the integrand is not continuous on [0,2]. Therefore, we reach the conclusion that

∫

2 0

dx (2x – 3)2

DOES NOT EXIST

EXAMPLE 6

Evaluate the definite integral, if it exists.

∫

13

dx 3

√(1 + 2x )2

0

Solution

We graph this function and find that it is continuous on [0,13]. Therefore, the integral exists.

First, we rewrite the integral to ease simplification:

∫

13

dx

√(1 + 2x ) 3

0

∫

=

2

13

(1 + 2x)–2/3 dx

0

i

We put u = 1 + 2x to get

∫u

– 2/3

dx

ii

Then,

du dx =

2

and so

dx = ½ du

Thus, (ii) becomes

∫u

– 2/3

× ½ du

∫u

½

=

– 2/3

du

iii

Next, we change the limits of integration. Since u = 1 + 2x, 

when x = 0, then u = 1 + 2(0)



when x = 13, then u = 1 + 2(13)

=

1, and =

27.

Therefore, the new limits are a = 1, and b = 27. And so, (iii) becomes

½ Using the Fundamental Theorem,

½

∫

27 1

u

– 2/3

du

= =

Therefore,

∫

27 1

1 2

u – 2/3 du

[ 1/3 ] u1/3

27

=

1

½ [3(27)1/3 – 3(1)1/3]

1 2

[3u ] 1/3

27

1

=

½ [3(3) – 3(1)]

= 3

∫

13

= 3

(1 + 2x)–2/3 dx

0

EXAMPLE 7 Evaluate the definite integral, if it exists.

∫

2

x √ x – 1 dx

1

Solution This integrand exists because the integrand is continuous on [1,2].

We put u = x –

1 to get

∫ xu

dx

½

i

Then,

du dx =

thus

1

dx = du

This results in

∫ xu

du

½

ii

We need to eliminate x in (i) above, and we do that by making another substitution: Since u = x – 1, then

x = u + 1. Therefore, (ii) becomes

∫ (u + 1) u

du

½

∫

=

(u3/2 + u ½)

du

iii

Next, we change the limits of integration. Since u = x – 1, 

when x = 1, then u = 1 – 1



when x = 2, then u =

2– 1

= =

0, and 1.

Therefore, the new limits are a = 0, and b = 1. This means that (iii) becomes

∫

1 0

(u3/2 + u ½)

du

Using the Fundamental Theorem,

∫

1 0

(u

3/2

+ u ) ½

du

= = =

u5/2 5/2

u3/2

+ 3/2

2(1)5/2 + 5 16 15

]

1

=

0

2(1)3/2 3

–

2u5/2 2u3/2 + 3 5 2(0)5/2 + 5

]

2(0)3/2 3

1

0

Evaluate the definite integral, if it exists.

∫

a 0

EXAMPLE 8

x √a2 – x2 dx

Solution Here, the integrand is

x √a2 – x2

f(x) =

Note that a is a number. Just to have an idea of what we'll be dealing with, we graph f with several values of a:

From the figure above, we find that the integral exists on [0,a]. we put

u = a2 – x2 to get

Next,

∫ xu du dx = –2x

½

dx which means

i

x dx = –½ du

Therefore, (i) becomes

∫u

½

× –½ du

–½ ∫ u ½ du

=

ii

The next step is to change the limits of integration. Since u = a2 – x2 , 

when x = 0, then u = a2 – (0)2 =

a2 , and



when x = a, then u = a2 – (a)2

0.

=

2

Therefore, the new limits are a = , and b = 0. Then (ii) becomes

–½

∫

0

a2

iii

u ½ du

You'll notice that, given any value of a, the upper limit of (iii) will always be less than the lower limit. So, the solution is to swap the limits (which is done by multiplying the integral by a minus sign). So, (iii) becomes

½

∫

a2 0

iv

u ½ du

Finally, we evaluate using the Fundamental Theorem:

½

∫

a2 0

u

½

1 2

=

du

[ 3/2 ] u3/2

=

0

–

(a2)3/2

=

a2

3

(0)3/2 3

1 2

[ ] 2u3/2 3

=

a2 0

=

u3/2 3

]

a2 0

(√a2)3 3

Therefore, for any value of a,

∫

a 0

x √a2 – x2 dx

(√a2)3

=

3

EXAMPLE 9 Use a graph to give a rough estimate of the area of the region that lies under the given curve. Then find the exact area.

y = 2sin x – sin 2x, 0 ≤ x ≤ π Solution The method used to estimate the area under a curve depends on the shape of the required area. If the area takes on a distinct shape (like a triangle, trapezium, semicircle, square, e.t.c), then the area is found by using the appropriate area formula. However, if the shape is irregular, we make use of approximating rectangles and the Riemann Sum. The figure below shows the graph of y

=

2sin x – sin 2x, with five approximating rectangles. We will use the

Midpoint Rule to estimate the area under the curve. Notice that the curve touches the midpoints of the approximating rectangles.

We find that a = 0, b = π, and n = 5. This means that

∆x =

b–a n

= π/5

Therefore, the intervals are

x0 x1 x2 x3 x4 x5

a a a a a a

= = = = = =

= 0 + ∆x + 2∆x + 3∆x + 4∆x + 5∆x

= = = = =

0 0 0 0 0

+ + + + +

(π/5) 2(π/5) 3(π/5) 4(π/5) 5(π/5)

= = = = =

π/5 2π/5 3π/5 4π/5 π

Which means the midpoints are:

x1 = x2 = x3 = x4 = x5 = Thus, an estimate of the area

½ (0 + π/5) ½ (π/5 + 2π/5) ½ (2π/5 + 3π/5) ½ (3π/5 + 4π/5) ½ (4π/5 + π) under the figure is

= π/10 = 3π/10 = 5π/10 = 7π/10 = 9π/10 given by the Riemann Sum

A

=

f (x1) ∆x + f (x2) ∆x + ............. + f (x5) ∆x

A

=

π/5 [2sin (π/10) – sin 2(π/10)] + [2sin (3π/10) – sin 2(3π/10)] +

[

[2sin (5π/10) – sin 2(5π/10)] + [2sin (7π/10) – sin 2(7π/10)] +

]

A

≈

[2sin (9π/10) – sin 2(9π/10)] π/5[0.0302 + 0.6670 + 2.0000 + 2.5691 + 1.2058]

A

≈

π/5[6.4721]

≈

4.0665

Therefore, the area under the curve y = 2sin x – sin 2x on the interval [0,π] using five approximating rectangles, is roughly 4.0665. Using more approximating rectangles, we obtain a better estimate. For instance we find that, using ten approximating rectangles the value of the area is about 4.0161. If we use twenty rectangles, we'd get something like 4.00411.

Enough with the guessing and approximations !! Let's use calculus to find the precise value of the area.

From the question, it is clear that the area we're looking for can be represented by the integral

∫

π 0

(2sin x – sin 2x) dx

This integral needs to be expressed in a form that can be easily evaluated using the substitution rule. From the trigonometric double angle formulas, we know that sin 2x

=

2sin x cos x

Therefore, the integral can be alternatively expressed as

∫

π 0

∫ ∫

=

(2sin x – sin 2x) dx

=

π 0 π 0

(2sin x – 2sin x cos x) dx 2sin x(1 – cos x) dx

i

We put u = 1 – cos x , which gives

∫ 2sin u dx

ii

Note that

du dx

= – (– sin x)

=

sin x

which means

dx

=

du sin x

Therefore, (ii) becomes

∫

du sin x

2sin u ×

=

∫

2 du

iii

Next, we change the limits of integration. Since u = 1 – cos x , then 

when x = 0, then u = 1 – cos 0



when x = π, then u = 1 – cos π

= =

0 , and 2.

Hence, the new limits are now a = 0, b = 2. This changes (iii) to

∫

2 0

2 du

=

2u

]

2 0

=

2(2) – 2(0)

=

4

Therefore, the exact area is 4. In other words,

∫

π 0

(2sin x – sin 2x) dx

=

4

Next, we study how the Substitution Rule is used to evaluate integrals whose integrands are symmetric. We also attempt to use the Rule to prove some identities, and finally apply the Rule to real situations.

A function, irrespective of its type, sometimes possesses a property called SYMMETRY. By definition, if there is a function f such that ➔ f(–x) = f(x) for every number in its domain, then f is called an EVEN FUNCTION. ➔ f(–x) = – f(x) for every number in its domain, then f is called an ODD FUNCTION. Geometrically, an even function is symmetric about the y-axis, and an odd function is symmetric about the x-axis.

Based on the definitions above, we find that if f is even, and have plotted it for f(x) ≥ 0, the rest of the graph is obtained by simply reflecting about the y-axis. On the other hand, if f is odd, and have plotted it for f(x) ≥ 0, the remaining part of the graph is obtained by reflecting about the x-axis (180o rotation about the origin).

Below are some examples:

A typical e ve n function.

Notice that f(-x) = f(x) for e ve ry value of x in the domain of f.

For e xample , f(-2) = f(2), etc.

A typical odd function.

Notice that f(-x) = -f(x) for e ve ry value of x in the domain of f.

For e xample , f(-2) = -f(2), etc.

There are of course, some exceptions. Some functions are NEITHER odd nor even. Based on the geometric properties of odd and even functions described above, two important Theorems that make it easier for us to evaluate the integrals of such functions have been developed. Suppose f is continuous on [-a, a], therefore, THEOREM 1:

∫

a -a

f (x) dx = 0

If f is ODD.

THEOREM 2:

∫

a -a

f (x) dx

=

2

∫

a 0

f (x) dx

If f is EVEN.

Compare Theorems 1 and 2 with the graphs below respectively:

∫

a -a

f (x) dx

=

2

∫

a 0

∫

f (x) dx

a -a

f (x) dx = 0

You can see the Theorems make perfect sense. So now, let's illustrate with a few examples:

EXAMPLE 1 Evaluate the definite integral

∫

π/2

–π/2

x2 sin x 1 + x6

dx Solution

First we need to determine whether the integrand is odd or even. We find that the integrand is

f(x) =

x2 sin x 1 + x6

We compute f(–x) and find that it is odd. The graph of f below also verifies that it is indeed an odd function:

Observe from the graph that the areas on either side of the y-axis are equal, and therefore cancel each other out, making the resultant area ZERO.

Based on Theorem 1 and the graph, we say that

∫

π/2

–π/2

x2 sin x 1 + x6

dx

= 0

EXAMPLE 2 Evaluate the definite integral

∫

π/3

sin 5 θ dθ

-π/3

Solution

The integrand here is f(θ) = sin 5 θ.

We compute compute f(–θ) and find that it is odd. The graph (on the following page) also says the same.

We see that the areas on either side of the y-axis cancel out because they're equal. Also, based on Theorem 1, since f is odd, it means

∫

π/3 -π/3

= 0

sin 5 θ dθ

Again, we see that the areas on either side of the y-axis are equal, and therefore cancel each other out, thus the value of the integral on that interval is ZERO.

EXAMPLE 3 Evaluate the definite integral

∫

a -a

x √x2 + a2

dx

Solution

We plot the integrand for several values of a, and we end up with basically the same kind of graph, only that the curve becomes almost straight as a increases. The figure above is a graph of f with a = 2.

From Theorem 1 and the graph above, we say that the value of the integral is zero.

Like examples 1 and 2, we see that the areas on either side of the y-axis are equal, and therefore cancel each other out, thus the value of the integral on the interval [-a,a] is ZERO.

EXAMPLE 4 If f is continuous, and find the value of

∫

2

∫

4 0

f(x) dx = 10

f(2x) dx

0

Solution To evaluate the integral, we apply the substitution rule. We let u = 2x to give

∫

So then,

du dx

=

2

f(u) dx

i

dx = ½ du

and so

Also, since u = 2x, then 

when x = 0, then u = 2(0)

=

0, and



when x = 2, then u = 2(2)

=

4.

Thus, the new limits are a = 0, and b = 4. Therefore, we rewrite (i) to get

½ From the definition of an integral,

∫

4 0

f(u) du

=

∫

4 0

∫

f(u) du 4 0

f(x) dx

This means,

½

∫

4 0

f(u) du

=

½ (10)

=

5

=

10

Therefore,

∫

2 0

f(2x) dx

=

5

Hope you see how it works. Now here's your task: Use the substitution rule to evaluate

∫

3 0

xf(x2 ) dx

∫

if

9 0

(assuming f is continuous)

f(x) dx = 4

EXAMPLE 5

Suppose f is continuous on  (which means ALL values of x). For the case where f(x) ≥ 0, prove that

∫

(A).

∫

(B).

b a b

a

f(–x) dx

=

∫

f(x + c) dx

=

∫

-a -b b+c a+c

f(x) dx f(x) dx

Draw a diagram to interpret each equation as an equality of areas.

Solution

Evaluating the left hand side of both equations above is the key to proving them.

(A).

First, we put u =

–x. This gives

∫

i

f(u) dx

Then,

du dx

=

–1

–du =

so that

dx

So, the integral becomes

∫ f(u)

– Since u =

–x. Then

du



when x = a, then u = –a , and



when x = b, then u = –b , and

ii

Thus, the new limits are a = –a, and b = –b. Thus, (ii) becomes

∫

-b -a

f(u) du

iii

–

Note that we are evaluating this integral with the assumption that b is greater that a. If that's the case, then from (iii) above, it is clear that the upper limit of the integral will always be less than its lower limit, regardless of the values of the limits. To solve this problem, we swap the limits; a step which is implemented by multiplying (iii) by a minus sign. Therefore,

∫

-b -a

f(u) du

=

∫

-a -b

f(u) du

– –

From the definition of an integral, we know that

∫

-a -b

f(u) du

∫

=

-a -b

f(x) dx

So, we have

a

f(–x) dx

=

–

∫

b

∫

-b -a

f(u) du

∫

=

-a -b

f(x) dx

Geometrically speaking, we see that these integrals are equal. In the figure below, we graph both on the same graph, on either side of the y-axis:

The graphs here represent y = f(x) [blue curve] and y = f(-x) [red curve]. This figure is graphical a illustration of what we just proved above.

Essentially, what we simply proved was that the areas A and B are equal, i.e., A = B.

(B).

Again, to prove the equation

∫

b a

f(x + c) dx

=

∫

b+c a+c

we evaluate the left hand side. First, we put u =

∫ Since u =

du dx

f(x) dx

x + c. This changes the LHS to

i

f(u) dx

x + c, then

=

Also, since u =

1

which means

du =

dx

x + c, then



when x = a, then u =

a + c , and



when x = b, then u =

b + c.

Thus, the new limits are a = a + c, and b = b + c. Thus, (i) becomes

∫

b+c a+c

f(u) du

From the definition of an integral,

∫

b+c a+c

∫

=

f(u) du

b+c

f(x) dx

a+c

And so, this proves that

∫

b

f(x + c) dx

a

=

∫

b+c a+c

f(u) du

=

∫

b+c a+c

f(x) dx

We also find that these two integrals are geometrically equal:

(Note that c is a constant). We have proved that, even though both functions differ by a constant c, the area under the curve y = f(x+c) on the interval [a,b] is equal to the area under y = f(x) on [a+c, b+c]. In other words, A = B. Now that we can prove the equality of areas of two different functions, here's a task for you: ➔ Show that the area under the graph of y = sin √x from 0 to 4 is the same as the area under the graph of y = 2x sin x from 0 to 2.

EXAMPLE 6

If a and b are positive numbers, show that

∫

1 0

xa(1 – x)b dx

=

∫

1 0

xb(1 – x)a dx

Solution Like example 5, we evaluate the left hand side of the equation. If the evaluation is correct, we end up with the expression in the right hand side.

We put u = 1 – x to get

∫

i

xa u b dx

Next,

du dx

–1

=

–du =

which means

dx

Also, since u = 1 – x, then 

when x = 0, then u =

1–0 = 1,



when x = 1, then u =

1 – 1 = 0.

Hence, the new limits are a = 1, b = 0. Thus, (i) becomes

∫

0 1

xa u b du

∫

=

1 0

xa u b du

ii

–

Since u = 1 – x, then x = 1 – u and so (ii) becomes

∫

1 0

(1 – u)a u b du

which, by the definition of an integral, is equivalent to

∫

1 0

(1 – x)a u b dx

Therefore, it follows that

∫

1 0

xa(1 – x)b dx

=

∫

1 0

xb(1 – x)a dx

The two graphs above are visual proof that the two integrals are equal. Notice that in both integrals, the powers a and b are swapped, but since they are both continuous on [0,1], and a and b are positive numbers, their areas are equal. Here, I put a = 1 and b = 2 to produce the graphs above. You can draw other graphs by changing the values

of a and b (just make sure they are positive numbers).

If we put a = 3 and b = 5, we get this:

EXAMPLE 7

Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production of these calculators after t weeks is

dx dt

=

100 (t + 10)2

5000 1 –

calculators/week

(Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers' unfamiliarity with new techniques). Find the number of calculators produced from the beginning of the third week to the end of the fourth week.

Solution

Let's use a graph to visualize the situation. If we draw a graph of the function dx/dt, we have the graph on the next page.

From the question, we are looking for a “total change” in the number of calculators manufactured within the given time interval [2,4]. Since the integral of a rate of change equals total change, it is clear that the solution to this problem can be found by evaluating the integral

∫

4 2

100 (t + 10)2

5000 1 –

dt

To evaluate the integral, we make use of the substitution rule. So, we put u = t + 10 which reduces the integral to

∫ 5000(1 – 100u

–2

) dt

=

∫ (5000 – 500000u

–2

) dt

i

Since the curve is a slope function (a function that measures the rate of calculator production per week), then, from the Total Change Theorem, the area under the curve from 2 to 4 represents the number of calculators manufactured from the beginning of the 3rd week to the end of the 4th week.

You might be tempted to ask: Why are we computing the area from 2 to 4 instead of from 3 to 4 (which is what you would expect)?

Well, if you read the question carefully, you'll see that the rate of production is measured after t weeks. Thus, the beginning of the 3rd week means the end of the 2nd week, which is why we compute the area [2,4] and NOT [3,4].

Note that

du dt

=

1

du =

which means

dt

Also, since u = t + 10, then 

when x = 2, then u = 2 + 10 = 12 , and



when x = 4, then u = 4 + 10 = 14.

Hence, a = 12, b = 14. Therefore, (i) becomes

∫

14 12

(5000 – 500000u –2 ) du

Using the Fundamental Theorem,

∫

14 12

(5000 – 500000u –2 ) du

=

5000u – 500000

=

5000u

+

–1 u 500000 u

which equals

5000(14) = ≈

740000 7

500000 (14)

–

305000 3

≈

+ –

4048 calculators

5000(12)

+

4047.619

500000 (12)

] ]

14

12 14

12

Thus, the number of calculators produced from the beginning of the third week (i.e., the end of the second week) to the end of the fourth week is approximately 4048.

EXAMPLE 8

Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 seconds. The maximum rate of air flow into the lungs is about 0.5 Liters/second. This explains, in part, why the function f(t) = ½ sin(2πt/5) has often been used to model the rate of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time t.

Solution From the question, we see that the function f(t) = ½ sin(2πt/5) measures the rate of air flow in liters/second.

Since we are looking for the volume of inhaled air in the lungs, it means we find the antiderivative F of f. In other words, we evaluate

∫ ½ sin(2πt/5)

dt

=

½

∫

sin(2πt/5) dt

i

we put u = 2πt/5, which gives

½

∫ sin u

dt

ii

Next,

du dt

=

2π 5

dt =

so that

5 du 2π

Therefore, becomes

½

∫ sin u

×

5 du 2π

= =

½ × 5/2π 5 4π

∫ sin u du

∫ sin u du

iii

we then evaluate (iii) using the general antiderivative formula:

5 4π

∫ sin u du

=

5 [– cos u + C] 4π

=

5 [C – cos (2πt/5)] 4π

=

5 4π

[C

– cos u ]

This means that the function

F(t)

=

5 [C – cos (2πt/5)] 4π

represents the volume of inhaled air in the lungs at time t. However, the function is “incomplete”, as we need to find the value of C. To do that, we investigate the function y = f(t) using its graph:

y = f(t)

y = F(t)

Using the geometry of antiderivatives, we use the the graph of y = f(t) to graph y = F(t). Since f(0) = 0, then it means F(0) = 0. Therefore,

F(0)

= = =

5 4π

[C

– cos (2π × 0/5)]

5 [C – cos 0] 4π 5 [C – 1] = 4π

=

=

0

0

0

Thus, C – 1 = 0, and so C = 1. Thus, the precise function is

F(t)

=

5 4π

[1

– cos (2πt/5)]

Before we conclude this tutorial, try this exercise:

●

If f is a continuous function such that

∫

x 0

f(t) dt

=

x sin x

+

∫

x 0

for all x, find an explicit formula for f(x). (hint: Apply Part 1 of the Fundamental Theorem)

calculus4engineeringstudents.com

f(t) 1 + t2

dt