Examples)

THE FUNDAMental THEOREM OF CALCULUSILLUSTRATIONS OF THE FUNDAMENTAL THEOREM

In this section, we will illustrate both parts of the Fundamental Theorem by applying them in solving various

integrals. Examples 1 through 12 will deal with Part 1 of the Theorem, while examples 13 to 19 will deal with Part 2.

To be able to evaluate an integral using FTC 2, you need to know how to compute an antiderivative for any given

function. The table below lists some general functions and their corresponding antiderivative formulas:

Function General Antiderivative

f(x) + g(x) F(x) + G(x)

cf(x) cF(x)

cos x sin x

sec 2 x tan x

sec x tan x sec x

xn (n ≠ -1)

Study the following examples carefully.

Example 1Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionAlways remember this: to differentiate an integral, the first step should be to ensure that the integrand is a

continuous function; because a discontinuous function IS NOT differentiable. Thus, since the integrand f(t) = 1 + 2t

is continuous, the FTC 1 gives:

Example 2Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionThe integrand f(t) = (2 + t4)5 is continuous. Therefore, using FTC 1 gives its derivative as:

xn+1

n + 1

g(x) = √1 + 2t dt∫ 0

x

g '(x) = √1 + 2x

g(x) = (2 + t4)5 dt∫ 1

x

g'(x) = (2 + x4)5

Example 3Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionThe integrand f(t) = t2 sin t is continuous; therefore,

Example 4Using Part 1 of the Fundamental Theorem, compute the derivative of

Solution The integrand f(x) = [1/(x + x2)] is a continuous function on the given interval. Therefore, FTC 1 gives the

derivative of g as

Example 5Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionHere, the integrand is F(t) = cos(t2). This function is continuous, we can therefore obtain a derivative. However,

there is one small problem: recall that the task is to differentiate the integral function F with respect to x which is

supposed to be the upper limit of the integral. But in this case, x has become the lower limit. Thus, to move any

further, what we need to do is “tweak” the integral so that x becomes the upper limit. To do this, we apply property 1

of definite integrals, so that

Thus, we'll be dealing with

Based on FTC 1, the derivative of F is

g(y) = (t2 sin t) dt∫ 2

y

g'(y) = y2 sin y

g(u) = [1/(x + x2)] dx∫ 3

u

g'(u) = 1/(u + u2)

F(x) = [cos(t2)] dt∫ x

2

F(x) = [cos(t2)] dt∫ x

2 = [cos(t2)] dt∫ 2

x–

F(x) = [cos(t2)] dt∫ 2

x–

F'(x) = – cos(x2)

Example 6Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionThis is pretty much like example 5 above; we have to apply property 1 of the definite integral before computing the

derivative of the integral. Using property 1 of integrals, F becomes

Thus, since F is continuous, its derivative is

Example 7Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionFrom the integral above, we find that the integrand is

F(t) = sin4 t

Observe the upper limit of the integral: 1/x. The upper limit is a function on its own. Thus, what we have here is a

function within a function, a situation which usually calls for the chain rule which, is widely used in differentiating

composite functions.

The first step to differentiating functions like this is to treat the upper limit as a function, and we do that by

representing it appropriately as such. Thus, let's assume

m = 1/x [or m(x) = 1/x)]This gives

So now, our task is to find

This is the point where the chain rule is put into use; we'll be differentiating two functions: the integral itself and the

function representing the upper limit. For the integral, its derivative is

F(x) = [tan θ] dθ∫ x

10

F(x) = [tan θ] dθ∫ 10

x–

F'(x) = – tan x

h(x) = [sin4t] dt∫ 2

1/x

h(x) = [sin4t] dt∫ 2

m

h'(x) = [sin4t] dt∫ 2

mEQUATION 1

d

dm

h'(x) = [sin4t] dt∫ 2

md

dx

and the derivative of the upper limit is

Using the chain rule, we put both equations 1 and 2 together so that

h'(x) = sin4 m × (-x-2)

Example 8Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionThe integrand f(r) = √1+r3 is continuous. Again observe the upper limit; it is a function on its own. We will therefore

have to make use of the chain rule here. Let's assume

b = x2 or b(x) = x2

Therefore,

h'(x) = √1 + b3 × 2x

h'(x) = 2x √1 + (x2)3

h'(x) = 2x √1 + x6

EQUATION 2m' (x)dm

dx= =

d

dx(1/x)

h'(x) = [sin4t] dt∫ 2

m dm

dx×

d

dm

h' (x) = 1x

sin4 × (-x-2)

1x2 –

1x

sin4h' (x) =

x2

– 1x

sin4

h' (x) =

h(x) = √1 + r3 dr∫ 2

x2

h'(x) = [√1 + r3] dr∫ 0

b d

dx

h'(x) = [√1 + r3] dr∫ 0

b d

db

db

dx×

Example 9Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionThe integrand f(t) = (cos t)/t is continuous on the given interval. The upper limit in this case is the root function

√ x. Therefore, let

m = √x or m(x) = √x

Thus,

Example 10Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionThe integrand here is f(t) = t + sin t which is continuous on the given interval. Let w represent the upper limit, i.e.

w = cos x or w(x) = cos x

Therefore,

y = dt∫ 3

√x cos tt

y' = ∫ 3

m d

dm

y ' = dt∫ 3

√x cos tt

ddx

dtcos tt

dm

dx×

×cos mm

1

2√xy' =

×cos √x√x

1

2√xy' =

cos √x

2xy' =

y = [t + sin t] dt∫ 1

cos x

y ' = [t + sin t] dt∫ 1

cos xddx

y ' = [t + sin t] dt∫ 1

wddw

dw

dx×

y' = (w + sin w) × - sin x

y' = - sin x (w + sin w)

y' = - sin x(cos x + sin [cos x])

y' = - sin x cos x – sin x sin(cos x)

Example 11Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionBy property 1 of definite integrals, we have

Let a(x) = 1 – 3x. Thus, we have

Hence, the derivative of y equals

Since a = 1 – 3x, then,

Example 12Using Part 1 of the Fundamental Theorem, compute the derivative of

SolutionUsing property 1 of definite integrals, we have

y = du∫ 1 -3x

1 u3

1 + u2

y = du∫ 1

1 -3x u3

1 + u2–

y = du∫ 1

a u3

1 + u2–

y ' = du∫ 1

a u3

1 + u2– d

dada

dx×

- 3×– a3

1 + a2y ' =

3a3

1 + a2y ' =

3(1 – 3x)3

1 + (1 – 3x)2y ' =

y = [sin 3t] dt∫ 1/x2

0

Let

m(x) = 1/x2 = x-2

Thus, we have

Hence, the derivative y' equals

y' = - sin3 m × -2x-3

In the next set of examples, we will illustrate Part 2 of the Fundamental theorem by using it to evaluate different

kinds of definite integrals

Example 13Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionTo evaluate an integrals using FTC 2, the first step to be taken, like in using FTC 1, is to ensure that the integrand is

continuous on that particular interval. In this case, the integrand is the function y = f(x) = x5 and it's continuous on

the interval [1, 3]. Next, we have to figure out an antiderivative for the integrand.

Using the general antiderivative formula

(Where c is an arbitrary constant)

We find that the derivative of f in this case is

F(x) = x6/6

Recall the FTC 2 formula:

y = [sin 3t] dt∫ 0 1/x2

–

y = [sin 3t] dt∫ 0m

–

y ' = [sin 3t] dt∫ 0m

– ddm

dm

dx×

y' = 1x2–sin3 × (-2x-3)

x3

y' =

1x2

2 sin3

x5 dx∫ -1

3

x n+1

n + 1F(x) = + C

f(x) dx∫ a

b= F(b) – F(a)

This means the antiderivative will be evaluated at the two values a and b. In this case, a = -1 and b = 3. Therefore,

= [(3)6/6] - [(-1)6/6]

= (729/6) - (1/6)

= 728/6

= 364/3 Hence

Example 14Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionThe integrand f(x) = 4x + 3 is continuous on the interval [2, 8]. Using the general antiderivative formula, we find

that the antiderivative F of f is

F(x) = 2x2 + 3x

Therefore, Part 2 of the fundamental theorem gives

(where a = 2 and b = 8)

Example 15

Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionHere, we have the integrand f(y) = 1 + 3y - y2 , whose antiderivative is given by

f(x) dx∫ a

b= F(b) – F(a)

x5 dx∫ -1

3 = 364/3 ≈ 121.3

(4x + 3) dx∫ 2

8

(4x + 3) dx∫ 2

8= F(b) – F(a)

(4x + 3) dx∫ 2

8 = [2(8)2 + 3(8)] - [2(2)2 + 3(2)]

= (128 + 24) - (8 + 6)

= 152 - 14

= 138

(1 + 3y – y2) dy∫ 0

4

F(y) = 3y2

2y + –

y3

3

Therefore, Part 2 of the fundamental theorem gives

(where a = 0 and b = 4)

= 4 + 24 – (64/3) - 0

= 20/3Therefore,

Example 16Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionFor ease of simplification, this integral can be better expressed as

Using the general antiderivative formula, the antiderivative of the integrand becomes

F(t) = -t-3 = -1/t3

Using FTC 2, we have

(Where a = -1 and b = 1)

Hence, the value of the integral is -2. Or is it!!!!

So far, it seems we've performed the calculation correctly and obtained the right answer.

(1 + 3y – y2) dy∫ 0

4= F(b) – F(a)

(1 + 3y – y2) dy∫ 0

4 = –3(4)2

2(4) + –

(4)3

33(0)2

2(0) + –

(0)3

3

(1 + 3y – y2) dy∫ 0

4= 20/3 ≈ 6.667

(3/t4) dt∫ -1

1

= F(b) – F(a) (3t–4) dt∫ -1

1

(3t–4) dt∫ -1

1= [-1/(1)3] – [-1/(-1)3]= (-1/1) - (-1/-1)

= -1 - 1

= - 2

(3t–4) dt∫ -1

1

Illustration 1: Graph of y = 3/t4. Observe the discontinuity in the interval [-1, 1]

There is however, one small problem: THIS ANSWER IS WRONG. The reason for this is because the integrand f(t)

= 3t-4 is NOT CONTINOUS ON THE GIVEN INTERVAL [-1, 1]. The graph above clearly shows the discontinuity.

This is why it is specifically mentioned in the Fundamental Theorem; to be able to evaluate or differentiate an

integral, the function (integrand) concerned MUST BE CONTINUOUS ON THE INTERVAL [a, b]. Therefore the

correct answer should be

Example 17Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionThe integrand f(θ) = cosθ is continuous on the interval [π, 2π]. The antiderivative formula gives F as

F(θ) = sin θ Using FTC 2, we have

DOES NOT EXIST (3t–4) dt∫ -1

1

(cos θ) dθ∫ π

2π

(cos θ) dθ∫ π

2π= F(b) – F(a)

(where a = π and b = 2π). So,

If you use a calculator or computer you might get something like 5.91991876344e-13 which is indeed very close to

zero.

Example 18Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionWe are about to evaluate the integral of a piecewise defined function. One good way to start is by graphing the

function f so as to get a visual of the area we are looking for. That way, we'll know how to approach the problem. The

graph is drawn above. Notice how both functions seem to form one continuous line. This is because both functions

have the same value when x = 1. Thus, both functions join at that point (indicated by the black dot).

Illustration 2: Graphs of y = x4 (red line) and y = x5 (blue line). Notice that both functions have the same value when

x = 1. They therefore seem to form one continuous line.

The graph below shows the areas bounded by each graph. The yellow area represents the area under the function y

= x5 on the interval [1, 2], while the small, grey area represents the area under the function y = x4 on the interval [0,

1]:

(cos θ) dθ∫ π

2π= sin 2π - sin π

= 0

x4 if 0 ≤ x < 1 = f(x) dx∫ 0

2

x5 if 1 ≤ x ≤ 2

From the graph, it is evident that we will have to find two areas:

which happen to be adjacent to each other. Thus, the sum of the areas will represent the value of the integral

This is simply another way of illustrating property 7 of the definite integral. Using FTC 2,

(where a = 0, b = 1 and the antiderivative F(x) = 0.2x5). Thus,

AND

(where a = 1, b = 2 and the antiderivative F(x) = x6/6). Thus,

Hence,

AND (x4) dx∫ 0

1 (x5) dx∫ 1

2

f(x) dx∫ 0

2

(x4) dx∫ 0

1= F(b) – F(a)

(x4) dx∫ 0

1= F(1) – F(0) = 0.2(1)5 - 0.2(0)5

= 0.2

(x5) dx∫ 1

2= F(b) – F(a)

(x5) dx∫ 1

2 = [(2)6/6] - [(1)6/6]

= 64/6 - 1/6 = 63/6 = 10.5

f(x) dx∫ 0

2= (x4) dx∫ 0

1 (x5) dx∫ 1

2+

= 0.2 + 10.5 = 10.7

Example 19Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.

SolutionLike example 18 above, we start by graphing f:

A graph of the piecewise function y = f(x). The orange area represents the area under the function y = sin x on the

interval [0, π], while the green area represents the area under the function y = x on the interval [-π, 0]

From the graph, we'll be dealing with two areas:

Also, from the graph, we find that

Let's deal with each integral one at a time. Thus, we have

(where a = 0, b = π and the antiderivative F(x) = -cos x). Therefore,

x if -π ≤ x ≤ 0= f(x) dx∫ -π

π

sin x if 0 ≤ x ≤ π

AND (x) dx∫ -π

0 (sin x) dx∫ 0

π

+ f(x) dx∫ -π

π (sin x) dx∫ 0

π (x) dx∫ -π

0= –

f(x) dx∫ -π

π (sin x) dx∫ 0

π (x) dx∫ -π

0= –

(sin x) dx∫ 0

π= F(b) – F(a)

AND

(where a = -π, b = 0 and F(x) = x2/2). Therefore,

Hence,

So,

Unlike example 18, this area is a result of the difference between adjacent areas.

Example 20Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact

area.

y = 3√x , 0 ≤ x ≤ 27

SolutionUsing the definite integral notation, we can express the required area as

But first, let's estimate the area using a graph (which will involve the Riemann method. This graph shows the

required area represented by the integral above, divided into 9 equal sub areas.

From the graph, we find that a = 0, b = 27 and n = 9. Therefore,

∆x = (b – a)/n∆x = (27 – 0)/9∆x = 3

(sin x) dx∫ 0

π= F(π) – F(0)

= -cos(π) - [-cos (0)]

= 1 - [-1] = 2

(x) dx∫ -π

0= F(b) – F(a)

(x) dx∫ -π

0= F(0) – F(– π)

= [(02)/2] - [(-π)2/2]

= 0 - π2/2 = - π2/2

f(x) dx∫ -π

π= 2 - (- π2/2)

= 2 + (π2/2)

= (4 + π2 )/2

f(x) dx∫ -π

π= (4 + π2 )

2

(3√x) dx∫ 0

27

This means

x1 = a + ∆x = 0 + 3 = 3x2 = a + 2∆x = 0 + 2(3) = 6x0 = a + 3∆x = 0 + 3(3) = 9x4 = a + 4∆x = 0 + 4(3) = 12x5 = a + 5∆x = 0 + 5(3) = 15x6 = a + 6∆x = 0 + 6(3) = 18x7 = a + 7∆x = 0 + 7(3) = 21x8 = a + 8∆x = 0 + 8(3) = 24x9 = a + 9∆x = 0 + 9(3) = 27

Hence, an estimate for the area is given by the Riemann sum

which equals

R9 = f(xi)∆x + f(x2)∆x + f(x3)∆x + f(x4)∆x + .....+ f(x8)∆x + f(x9)∆x

R9 = 3f(xi) + 3f(x2) + 3f(x3) + 3f(x4) + ......+ 3f(x8) + 3f(x9)

R9 = 3[(3√3) + (3√6) + (3√9) + (3√12) + (3√15) + (3√18) + (3√21) + (3√24) + (3√27)]

R9 = 3[1.4422+ 1.8171+ 2.0801 + 2.2894 + 2.4662 + 2.6207 + 2.7589 + 2.8845 + 3.0000]R9 = 4.3266 + 5.4513 + 6.2403 + 6.8682 + 7.3986 + 7.8621 + 8.2767 + 8.8635 + 9.0000R9 = 64.0773

From the diagram, observe that right endpoints were used in the approximating rectangles. This means that the

above estimate is an overestimate; the actual area is less than 64.0773.

Using the midpoint rule, we obtain a better estimate: 60.9836.

R9 = ∑n=9

i=1f(xi)∆ x

Now, let's use Part 2 of the Fundamental Theorem to find the exact area. Recall the area we're dealing with:

where

f(x) = 3√x, a = 0, b = 27 and F(x) = ¾ (√x)4

Therefore,

So,

This example has clearly illustrated the power of the Fundamental Theorem; see how we were able to find the

required area quickly with just a few steps. Unlike the old method of using Riemann sums where only estimates are

obtained, the Fundamental Theorem gives PRECISE results.

Example 21Use a graph to give a rough estimate of the area of the region that lies beneath the given curve, Then find the exact

area.

y = sec2x, 0 ≤ x ≤ π/3

SOLUTIONThis time, let's use the midpoint rule to estimate the area. Here, we have a graph of y = sec2x plotted, and showing

the required area on the interval [0,π/3]:

(3√x) dx∫ 0

27

= F(b) – F(a) (3√x) dx∫ 0

27

= F(27) – F(0) = [¾ (3√27)4] – [¾ (3√0)4]

= [¾ (81)] – [0] = 60.75

= 60.75 (3√x) dx∫ 0

27

This next graph shows the same graph partitioned into 10 subintervals, where the curve touches the midpoint of

each approximating rectangle:

From the graph, we find that

A1 lies on the interval [0, π/30]A2 lies on the interval [π/30, π/15]

A3 lies on the interval [π/15, π/10]

A4 lies on the interval [π/10, 2π/15]

A5 lies on the interval [2π/15, π/6]

A6 lies on the interval [π/6, π/5]

A7 lies on the interval [π/5, 7π/30]

A8 lies on the interval [7π/30, 4π/15]

A9 lies on the interval [4π/15, 3π/10]

A10 lies on the interval [3π/10, π/3]

This means the midpoint of each interval equals

midpoint of A1 = ½[0 + π/30] = π/60

midpoint of A2 = ½[π/30 + π/15] = π/20

midpoint of A3 = ½[π/15 + π/10] = π/12

midpoint of A4 = ½[π/10 + 2π/15] = 7π/60

midpoint of A5 = ½[2π/15 + π/6] = 3π/20

midpoint of A6 = ½[π/6 + π/5] = 11π/60

midpoint of A7 = ½[π/5 + 7π/30] = 13π/60

midpoint of A8 = ½[7π/30 + 4π/15] = π/4

midpoint of A9 = ½[4π/15 + 3π/10] = 17π/60

midpoint of A10 = ½[3π/10 + π/3] = 19π/60

We find that a = 0, b = π/3, and we have used 10 subintervals; thus

∆x = (b – a)/n∆x = (π/3 – 0)/10∆x = π/30

We therefore estimate the area of A using the Riemann sum

Which equals

This gives

A = π/30[sec2(π/60)+ sec2(π/20) + sec2(π/12) + sec2(7π/60) + sec2(3π/20) + sec2(11π/60)

+ sec2(13π/60) + sec2(π/4) + sec2(17π/60) + sec2(19π/60)]

A = 0.10472[1.00275 + 1.02501 + 1.07179 + 1.14735 + 1.25962 + 1.42173 + 1.65575

+ 2 + 2.52497 + 3.37118]

A = 0.10472[16.48015] = 1.7258

Therefore, the estimated area under the curve y = sec2x on [0,π/3] using 10 approximating rectangles is 1.7258.

Now, let's do this the easy way:

Again , the function we're dealing with is y = sec2x on the interval [0,π/3]. So, an appropriate integral notation would

be

To evaluate the integral, we apply Part 2 of the Fundamental Theorem, and to do that, we need an antiderivative of

y = sec2x, which is F(x) = tan x .

Therefore,

Hence, the EXACT area is 1.7321. In other words,

=

= 1.732508 – 0 ≈ 1.7321

A = ∑n=10

i=1f(xi)∆ x

A = ∑n=10

i=1 (sec2 xi) ∆x ∑

n=10

i=1 (sec2 xi) (π/30)

∑n=10

i=1 (sec2 xi) = (π/30)

A = (sec2 x) dx∫ 0

π/3

(sec2 x) dx∫ 0

π/3= tan (π/3) – tan(0)

= 1.7321 (sec2 x) dx∫ 0

π/3

Example 22Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch.

SOLUTIONThe integrand y = x3 is continuous on the interval [-1,2]:

Using FTC 2, we have

Where

F(x) = x4/4, a = -1, and b = 2

Therefore,

We can also evaluate the integral by interpreting it as a difference of areas. This next graph shows the graph

partitioned into two regions: one part of the graph (labeled A2) lies below the x-axis, and the other part (labeled A1)

lies above the x-axis):

= F(b) – F(a)

= [16/4] – [¼ ] = 15/4

= 3.75

(x3) dx∫ –1

2

(x3) dx∫ –1

2

= [(2)4/4] – [(-1)4/4] (x3) dx∫ –1

2

The part of the graph below the x-axis is on the interval [-1, 0]. Thus, its area is

Note that the minus sign indicates the area is below the x-axis. On the other hand, the part of the graph above the x-

axis is on the interval [0, 2]. Therefore, its area is

From one of the definitions of a definite integral,

Which means,

Example 23Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch.

= –

= 4 – 0.25 = 3.75

∫ π/4

5π/2(sin x) dx

A2 = = [(0)4/4] – [(-1)4/4] = – 0.25 (x3) dx∫ –1

0

A1 = = [(2)4/4] – [(0)4/4] = 4 (x3) dx∫ 0

2

= A1 – A 2 f(x) dx∫ a

b

(x3) dx∫ –1

2 (x3) dx∫ 0

2 (x3) dx∫ –1

0

SOLUTIONLet's graph the integrand first, and understand what we're dealing with:

Using FTC 2,

Where

F(x) = – cos x, b = 5π/2, a = π/4

Which gives

If we go back to the graph above, we see that there are three regions: X, Y, and Z. If we were to interpret the given

integral as a difference of areas, then we'd have

where X is on the interval [π/4,π], Y is on the interval [2π,5π/2], and Z is on the interval [π,2π]. So,

= F(b) – F(a)

= – cos[5π/2] – [– cos(π/4)] Area =

= 0 – [– √2/2] = – √2/2 = 0.7071

= (X + Y) – Z

π/4∫ 5π/2(sin x) dx

π/4∫ 5π/2(sin x) dx

Area =π/4∫ 5π/2

(sin x) dx

= – cos [π] – [– cos(π/4)] = 1 – [– √2/2] = 1.7071

Area of X =π/4∫ π

(sin x) dx

And then

Now that we've computed all three areas, the total area becomes

This answer coincides with the one obtained when we first used FTC 2 to evaluate the integral.

Example 24Find the derivative of the function.

Here's a hint.

SOLUTIONThe integrand here is

Let's assume that

Using the provided hint,

= – 1 – 1

= – 2

= + –= [1.7071 + 1 ] – 2

= 2.7071 – 2

= 0.7071

= +

f(u) =u2 – 1 u2 + 1

= 0 – [– 1] = 1

= – cos [5π/2] – [– cos(2π)] Area of Y =2π∫ 5π/2

(sin x) dx

= – cos [2π] – [– cos(π)] Area of Z = π∫ 2π

(sin x) dx

π/4∫ 5π/2(sin x) dx π/4∫ π

(sin x) dx2π∫ 5π/2

(sin x) dx π∫ 2π

(sin x) dx

∫ 3x

2x f(u) du ∫ 0

2x f(u) du ∫ 3x

0 f(u) du

∫ 2x

3xdug(x) =

u2 – 1 u2 + 1

∫ 2x

3xdu

u2 – 1 u2 + 1

Let's solve the RHS of equation 1 one at a time. We start with

This kind of integral should look familiar; we evaluated similar integrals in examples 8, 9, 10, etc. So, first, we let

m(x) = 2x, which results in

At this point, we use FTC 1 alongside the chain rule, which gives:

So,

And now for the second part:

We let n(x) = 3x. Thus, we have

+

= – i

× 2=

ii

=∫ 2x

3xdu

u2 – 1 u2 + 1 ∫ 2x

0du

u2 – 1 u2 + 1 ∫ 0

3xdu

u2 – 1 u2 + 1

∫ 0

2xdu

u2 – 1 u2 + 1 + ∫ 0

3xdu

u2 – 1 u2 + 1

– du∫ 0

2x u2 – 1 u2 + 1

– ∫ 0

mdu

u2 – 1 u2 + 1

dmd – ∫ 0

mdu

u2 – 1 u2 + 1 dx

dm – m2 – 1

m2 + 1

= – 2 m2 – 1

m2 + 1 = – 2 (2x)2 – 1

(2x)2 + 1

= – 2 4x2 – 1

4x2 + 1

dxd

– du∫ 0

2x u2 – 1 u2 + 1

= – 2 4x2 – 1

4x 2 + 1

∫ 0

3xdu

u2 – 1 u2 + 1

∫ 0

ndu

u2 – 1 u2 + 1

Thus, we are to evaluate

Using FTC 1 and the chain rule, we have

This gives:

So,

Combining ii and iii gives

So, what have we learnt from this example?

In order to differentiate an integral whose endpoints are individual functions, the key is to split the main integral into

two simpler integrals, differentiate each and combine the results.

Let's try one more example like this one.

Example 25Find the derivative of the function.

=

iii

g'(x) =

+

y = ∫ √x

x3

(√t sin t) dt

dxd ∫ 0

ndu

u2 – 1 u2 + 1

∫ 0

ndu

u2 – 1 u2 + 1 dn

d

dxdn n2 – 1

n2 + 1 × 3

n2 – 1 n2 + 1 3 =

(3x)2 – 1 (3x)2 + 1 3

=9x2 – 1 9x2 + 1 3

∫ 2x

3xdu

u2 – 1 u2 + 1 dx

d

g'(x) = – 2 4x2 – 1

4x2 + 1

9x2 – 1 9x2 + 1 3

g'(x) =9x2 – 1 9x2 + 1 3 2 4x2 – 1

4x2 + 1 –

SOLUTIONThe first step to differentiating this integral is splitting it into two simpler integrals (on adjacent intervals). Like the

previous example, we have:

Let

So now, the next task is to find m'(x), n'(x), and then m'(x) + n'(x). Let's compute m'(x) first. Before we start, we

need to rewrite m:

Therefore,

We let r = √x, so that we have

Using FTC1 and the chain rule, we have

So,

Next, we solve n'(x):

∫ √x

x3(√t sin t) dt ∫√x

0(√t sin t) dt ∫ 0

x3(√t sin t) dt = +

m(x) = ∫√x

0(√t sin t) dt

n(x) = ∫ 0

x3(√t sin t) dt

∫√x

0 (√t sin t) dt m(x) = = ∫ 0

√x (√t sin t) dt –

dxdm'(x) = ∫ 0

√x (√t sin t) dt –

m'(x) = ∫ 0

r (√t sin t) dt – dx

d

m'(x) = ∫ 0

r (√t sin t) dt – dr

ddxdr

=

m'(x) =

√r sin r ×2√x

1

= √√x sin √x ×2√x

1

2√x

– 4√x sin √x=

2√x

– 4√x sin √x i

Which means

We let g =x3, so that

So now, we combine i and ii to give

Of course, the result can be simplified further using basic algebra. We end our study of the Fundamental Theorem

here. Next, we move on to the another concept of integral calculus: the indefinite integral. But before we begin, you

need to review ANTIDERIVATIVES.

calculus4engineeringstudents.com

n(x) = ∫ 0

x3(√t sin t) dt

dxdn'(x) = ∫ 0

x3(√t sin t) dt

dxdn'(x) = ∫ 0

g(√t sin t) dt

dgdn'(x) = ∫ 0

g(√t sin t) dt dx

dg

n'(x) = √g sin g × 3x2

n'(x) = 3x2 √x3 sin x3 ii

∫ √x

x3(√t sin t) dt dx

d= m'(x) + n'(x)

= 2√x

– 4√x sin √x+ 3x2 √x3 sin x3