evaluatingintegrals(substitution ruleii)

EVALUATING INTEGRALS - SUBStitutiONEVALUATING 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 1You may be given something like this:

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

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

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

Then,

Therefore, (iii) becomes

which equals

√4 + 3x dx 7∫ 0

] 7

0 √4 + 3x dx 7∫ 0 = √4 + 3x dx ∫ i

√4 + 3x dx ∫ ii

u ½ dx ∫ iii

dudx = 3 so that dx = du

3 iv

u ½ × ∫ du3 = 1

3 u ½ du ∫ = u3/2 3/2

13 + C

The final step is to evaluate

In other words, we are to compute

We therefore have

Therefore,

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

and then proceed to evaluate the indefinite integral:

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

Next,

2u3/2 3

13 + C = 2u3/2

9 + C = 2(4 + 3x)3/2 9

+ C

2(4 + 3x)3/2 9

+ C at the endpoints a = 0, b = 7.

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

0

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

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

0= 2(4 + 3(7))3/2

9 – 2(4 + 3(0))3/2 9

= 2(25)3/2 9 – 2(4)3/2

9= 26

√4 + 3x dx 7∫ 0 = 26

√4 + 3x dx 7∫ 0 = √4 + 3x dx ∫ i

√4 + 3x dx ∫ ii

u ½ dx ∫ iii

dudx = 3 so that dx = du

3 iv

Therefore, (iii) becomes

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

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

(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:

u ½ × ∫ du3 = 1

3 u ½ du ∫

u ½ du 25

∫ 4

13

√4 + 3x dx 7∫ 0 has been reduced to u ½ du

25

∫ 4

13

So now, we evaluate the integral

and evaluate with the Fundamental Theorem:

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

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.

u ½ du 25

∫ 4

13 = u3/2

3/2 ]25

4[1

3= 2u3/2

3 ]25

4[1

3= 2u3/2

9 ]25

4

2u3/2 9 ]25

4= 2(25)3/2

9 – 2(4)3/2 9

= 26

√4 + 3x dx 7∫ 0 = u ½ du

25

∫ 4

13 = 26

EXAMPLE 2Evaluate the definite integral, if it exists.

SolutionWe 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

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

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

At this point, we apply the Fundamental Theorem:

which results in

Therefore,

x2(1 + 2x3)5 dx 1

∫ 0

x2 u5 dx ∫ i

dudx = 6x2 and then x2 dx = du

6 ii

3∫ 1 u5 × du6

= 16

3∫ 1 u5 du

16

3∫ 1 u5 du = u6 6 ] 3

1[1

6= u6

36 ] 3

1

u6 36 ]3

1= (3)6

36– (1)6

36= 20.222 728

36≈

x2(1 + 2x3)5 dx 1∫ 0

≡ 16

3∫ 1 u5 du = 182 9

= 182 9


SolutionFirst 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,

Thus, (i) becomes

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

Using the Fundamental Theorem,

So,


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

and get this:

√π∫ 0 x cos(x2) dx

x cos u dx ∫ = cos u x dx ∫ i

dudx = 2x so that x dx = du

2

cos u × ½ du ∫ = cos u du ∫½ ii

π∫ 0 cos u du ½

π∫ 0 cos u du ½ = sin u ] π

0[12 ½ (sin π – sin 0)= = 0

√π∫ 0 x cos(x2) dx = 0

dx1x2 1 + 1

x√ 4

∫ 1

u = 1 + 1x u = 1 + x –1 OR

Next,

Put in (ii) to (i) get

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)–1 = 5/4.

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

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:

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

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

which equals

∫ u ½ 1x2 dx

∫

u ½

x2 dx= i

dudx = –1

x2which means

dxx2 = – du ii

u ½ × – du ∫ = – u ½ du ∫ iii

5/4∫ 2– u ½ du iv

f(x) dxb∫ a= – f(x) dxa∫ b

2∫ 5/4 u ½ du v

2∫ 5/4 u ½ du = u 3/2

3/2 ] 2

5/4

(2) 3/2 3/2 – (5/4) 3/2

3/2 = (√2)3 3/2

– (5)3/2

3/2 (4)3/2 =

= 4√2 3 –

(√5)3

3/2

(√4)3 = 4√2 3 –

5√5

3/28

Therefore,

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


SolutionFirst, we rewrite the integral:

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:

= 4√2 3

– 5√5 8

23

= 4√2 3

– 5√5 12

dx1x2 1 + 1

x√ 4

∫ 1 = 4√2 3

– 5√5 12

dx (2x – 3)2

2

∫ 0

dx 1 (2x – 3)2

2

∫ 0

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

reach the conclusion that


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

First, we rewrite the integral to ease simplification:

We put u = 1 + 2x to get

Then,

Thus, (ii) becomes

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

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

when x = 13, then u = 1 + 2(13) = 27.

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


Therefore,

dx (2x – 3)2

2

∫ 0 DOES NOT EXIST

13

∫ 0

dx

√(1 + 2x )2 3

13

∫ 0

dx

√(1 + 2x )2 3=

13∫ 0 (1 + 2x)–2/3 dx i

∫ u – 2/3 dx ii

dudx = and so dx = ½ du2

∫ u – 2/3 × ½ du = ∫ u – 2/3 du ½ iii

27∫ 1 u – 2/3 du ½

27∫ 1 u – 2/3 du ½ = u1/3 1/3 ]27

1[1

2= 3u 1/3 ]27

1[1

2

= ½ [3(27)1/3 – 3(1)1/3] = ½ [3(3) – 3(1)] = 3


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

We put u = x – 1 to get

Then,

This results in

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

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

when x = 1, then u = 1 – 1 = 0, and

when x = 2, then u = 2 – 1 = 1.

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


13

∫ 0 (1 + 2x)–2/3 dx = 3

2

∫ 1 x √ x – 1 dx

dudx = thus dx = du1

∫ x u ½ dx i

∫ x u ½ du ii

∫ (u + 1) u ½ du = ∫ (u3/2 + u ½) du iii

1

∫ 0 (u3/2 + u ½) du

1

∫ 0 (u3/2 + u ½) du u5/2 5/2 ]1

0+ u

3/2 3/2= = 2u5/2

5 ]1

0+ 2u3/2

3

= 2(1)5/2 5 + 2(1)3/2

3 2(0)5/2

5 + 2(0)3/2 3–

1615=


SolutionHere, the integrand is

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,

a

∫ 0 x √a2 – x2 dx

x √a2 – x2 f(x) =

∫ x u ½ dx i

= which means x dx = –½ dududx –2x

Therefore, (i) becomes

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.

Therefore, the new limits are a = 2, and b = 0. Then (ii) becomes

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

Finally, we evaluate using the Fundamental Theorem:

Therefore, for any value of a,

EXAMPLE 9Use 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

SolutionThe 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

∫ u ½ × –½ du = ∫ u ½ du –½ ii

u ½ du –½ 0∫ a2

iii

u ½ du ½ a2

∫ 0 iv

u ½ du ½ a2

∫ 0 u3/2 3/2 ]a2

0[1

2= = 2u3/2

3 ]a2

0[1

2= u3/2

3 ]a2

0

(a2)3/2 3=

(0)3/2 3

–= (√a2)3

3

a

∫ 0 x √a2 – x2 dx = (√a2)3 3

∆x = b – a n = π/5

Therefore, the intervals are

x0 = a = 0x1 = a + ∆x = 0 + (π/5) = π/5x2 = a + 2∆x = 0 + 2(π/5) = 2π/5x3 = a + 3∆x = 0 + 3(π/5) = 3π/5x4 = a + 4∆x = 0 + 4(π/5) = 4π/5x5 = a + 5∆x = 0 + 5(π/5) = π

Which means the midpoints are:

x1 = ½ (0 + π/5) = π/10x2 = ½ (π/5 + 2π/5) = 3π/10 x3 = ½ (2π/5 + 3π/5) = 5π/10 x4 = ½ (3π/5 + 4π/5) = 7π/10 x5 = ½ (4π/5 + π) = 9π/10

Thus, an estimate of the area under the figure is 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)] +

[2sin (9π/10) – sin 2(9π/10)]]A ≈ 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

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

We put u = 1 – cos x , which gives

Note that

which means

Therefore, (ii) becomes

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

when x = 0, then u = 1 – cos 0 = 0 , and

when x = π, then u = 1 – cos π = 2 .

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

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

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.

π∫ 0 (2sin x – sin 2x) dx

π∫ 0 (2sin x – sin 2x) dx = π∫ 0 (2sin x – 2sin x cos x) dx

= π∫ 0 2sin x(1 – cos x) dx i

∫ 2sin u dx ii

dudx = =– (– sin x) sin x

dusin xdx =

∫ 2sin u × dusin x = ∫ 2 du iii

2∫ 0 2 du = 2u ] 2 0 = 2(2) – 2(0) = 4

π∫ 0 (2sin x – sin 2x) dx = 4

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:

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:

THEOREM 2:

Compare Theorems 1 and 2 with the graphs below respectively:

A typical even function. Notice that f(-x) = f(x) for every value of x in the domain of f.For example, f(-2) = f(2), etc.

A typical odd function. Notice that f(-x) = -f(x) for every value of x in the domain of f.For example, f(-2) = -f(2), etc.

a∫ -a f (x) dx = 0 If f is ODD.

If f is EVEN. a∫ 0

a∫ -a f (x) dx = f (x) dx 2

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

EXAMPLE 1Evaluate the definite integral

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

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

a∫ 0 a∫ -a f (x) dx = f (x) dx 2 a∫ -a f (x) dx = 0

x2 sin x 1 + x6

π/2

∫ –π/2dx

x2 sin x 1 + x6f(x) =

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

x2 sin x 1 + x6

π/2

∫ –π/2dx = 0


SolutionThe 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


SolutionWe 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.

π/3∫ -π/3 sin 5 θ dθ

= 0 π/3∫ -π/3 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, thusthe value of the integralon that interval is ZERO.

a∫ -a x √x2 + a2 dx

EXAMPLE 4If f is continuous, and

find the value of

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

So then,

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

From the definition of an integral,

This means,

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, thusthe value of the integralon the interval [-a,a] is ZERO.

4∫ 0 f(x) dx = 10

2∫ 0 f(2x) dx

∫ f(u) dx i

dudx = and so dx = ½ du2

4∫ 0 f(u) du ½

4∫ 0 f(x) dx = 4∫ 0 f(u) du = 10

4∫ 0 f(u) du ½ = ½ (10) = 5

Therefore,

Hope you see how it works. Now here's your task:

Use the substitution rule to evaluate

EXAMPLE 5Suppose f is continuous on (which means ALL values of x). For the case where f(x) ≥ 0, prove that

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

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

First, we put u = –x. This gives

Then,

So, the integral becomes

Since u = –x. Then

when x = a, then u = –a , and

when x = b, then u = –b , and

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

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,

2∫ 0 f(2x) dx = 5

3∫ 0 xf(x2) dx if 9∫ 0 f(x) dx = 4 (assuming f is continuous)

b∫ a f(–x) dx -a∫ -b

f(x) dx = (A).

(B). b∫ a f(x + c) dx = b+c∫ a+c

f(x) dx

(A).

dudx = so that –du = dx–1

∫ f(u) dx i

∫ f(u) du – ii

-b∫ -a f(u) du –

iii

-b∫ -a f(u) du – –

= -a∫ -b

f(u) du

From the definition of an integral, we know that

So, we have

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:

Again, to prove the equation

we evaluate the left hand side. First, we put u = x + c. This changes the LHS to

Since u = x + c, then

Also, since u = 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

-a∫ -b f(u) du =

-a∫ -b f(x) dx

b∫ a f(–x) dx = -b∫ -a

f(u) du – = -a∫ -b

f(x) dx

The graphs here representy = f(x) [blue curve] and y = f(-x) [red curve]. This figure is graphical a illustration of what we justproved above.Essentially, what we simply proved was that the areas A and B are equal, i.e., A = B.

(B).

b∫ a f(x + c) dx = b+c∫ a+c f(x) dx

∫ f(u) dx i

dudx = which means du = dx1

From the definition of an integral,

And so, this proves that

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 6If a and b are positive numbers, show that

SolutionLike 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.

b+c∫ a+c f(u) du

b+c∫ a+c f(u) du = b+c∫ a+c

f(x) dx

b∫ a f(x + c) dx = b+c∫ a+c f(x) dx

b+c∫ a+c f(u) du =

1∫ 0 xa(1 – x)b dx = 1∫ 0 xb(1 – x)a dx

We put u = 1 – x to get

Next,

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

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

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

Therefore, it follows that

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

∫ xa ub dx i

dudx = which means –du = dx–1

0∫ 1 xa ub du –

= 1∫ 0 xa ub du ii

1∫ 0 (1 – u)a ub du

1∫ 0 (1 – x)a ub dx

1∫ 0 xa(1 – x)b dx = 1∫ 0 xb(1 – x)a dx

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

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

EXAMPLE 7Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production

of these calculators after t weeks is

(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.

SolutionLet'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

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

dxdt = 5000 1 – 100

(t + 10)2calculators/week

4∫ 25000 1 – 100

(t + 10)2 dt

∫ 5000(1 – 100u–2) dt = ∫ (5000 – 500000u–2) dt i

Note that

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


which equals

Since the curve is a slopefunction (a function thatmeasures the rate of calculatorproduction per week), then, fromthe 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 from2 to 4 instead of from 3 to 4 (whichis what you would expect)?Well, if you read the question carefully, you'll see that the rateof production is measured after t weeks. Thus, the beginning of the 3rdweek means the end of the 2nd week, which is why we compute the area [2,4] and NOT [3,4].

dudt = which means du = dt1

14∫ 12 (5000 – 500000u–2) du

14∫ 12 (5000 – 500000u–2) du = 5000u – 500000

–1 u ]14

12

= 500000u

5000u + ]14

12

500000(14)

5000(14) + – 500000(12)

5000(12) +

7400007

3050003

– ≈= 4047.619

≈ 4048 calculators

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 8Breathing 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.

SolutionFrom 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

we put u = 2πt/5, which gives

Next,

Therefore, becomes

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

This means that the function

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:

∫ ½ sin(2πt/5) dt = ½ ∫ sin(2πt/5) dt i

½ ∫ sin u dt ii

dudt = so that dt = 2π

55 du2π

½ ∫ sin u × 5 du2π

= ½ × 5/2π ∫ sin u du

= 5 4π ∫ sin u du iii

5 4π ∫ sin u du =

5 4π [– cos u + C] =

5 4π [C – cos u]

= 5 4π [C – cos (2πt/5)]

F(t) = 5 4π [C – cos (2πt/5)]

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,

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

Before we conclude this tutorial, try this exercise:

● If f is a continuous function such that

for all x, find an explicit formula for f(x).

(hint: Apply Part 1 of the Fundamental Theorem)

calculus4engineeringstudents.com

y = f(t) y = F(t)

F(0) = 5 4π [C – cos (2π × 0/5)] = 0

= 5 4π [C – cos 0] = 0

= 5 4π [C – 1] = 0

F(t) = 5 4π [1 – cos (2πt/5)]

x∫ 0f(t) dt = x sin x +

x∫0

dtf(t)

1 + t2

evaluatingintegrals(substitution ruleii)

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