THE FUNDAMental THEOREM OF CALCULUS ILLUSTRATIONS 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
x (n ≠ -1)
xn+1 n+1
n
Study the following examples carefully.
Example 1 Using Part 1 of the Fundamental Theorem, compute the derivative of
g(x)
=
∫
x
√ 1 + 2t
0
dt
Solution
Always 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:
g '(x)
=
√ 1 + 2x
Example 2
Using Part 1 of the Fundamental Theorem, compute the derivative of
g(x)
=
∫ (2 + t ) x
1
4
5
dt
Solution The integrand f(t) = (2 + t4)5 is continuous. Therefore, using FTC 1 gives its derivative as:
g'(x)
=
(2 + x4)5
Example 3
Using Part 1 of the Fundamental Theorem, compute the derivative of
g(y)
∫ (t y
=
2
2
sin t
)
dt
Solution The integrand f(t) = t2 sin t is continuous; therefore,
g'(y)
=
y2 sin y
Example 4 Using Part 1 of the Fundamental Theorem, compute the derivative of
g(u)
∫
=
[1/(x + x )]
u
dx
2
3
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
g'(u)
=
1/(u + u2)
Example 5 Using Part 1 of the Fundamental Theorem, compute the derivative of
F(x)
∫ [cos(t )] 2
=
2
x
dt
Solution
Here, 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
F(x)
∫ [cos(t )] 2
=
2
x
dt
Thus, we'll be dealing with
F(x)
=
–
∫ [cos(t )] x
2
Based on FTC 1, the derivative of F is
F'(x)
=
– cos(x2)
2
dt
=
–
∫ [cos(t )] x
2
2
dt
Example 6
Using Part 1 of the Fundamental Theorem, compute the derivative of
F(x)
∫
=
10 x
[tan θ]
dθ
Solution This 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
–
F(x) =
∫
x 10
[tan θ]
dθ
Thus, since F is continuous, its derivative is
F'(x)
=
– tan x
Example 7
Using Part 1 of the Fundamental Theorem, compute the derivative of
h(x)
∫
=
1/x 2
[sin t]
dt
4
Solution From 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
[or
= 1/x
m(x) = 1/x)
]
This gives
h(x)
=
∫
m 2
[sin t] 4
dt
So now, our task is to find
h'(x)
=
d dx
∫
m 2
[sin t] 4
dt
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
h'(x)
=
d dm
∫
m 2
[sin t] 4
dt
EQUATION 1
and the derivative of the upper limit is
m' (x)
dm
=
d
=
dx
dx
(1/x)
EQUATION 2
Using the chain rule, we put both equations 1 and 2 together so that
h'(x)
d
=
h'(x)
dm
=
h' (x)
–
=
h' (x)
×
dt
dm dx
(-x-2) × (-x-2)
1 x2
–
4
2
1 x
sin4
[sin t]
m
×
sin4 m
=
h' (x)
∫
1 x
sin4 1 x
sin4
= x2
Example 8
Using Part 1 of the Fundamental Theorem, compute the derivative of
h(x)
∫ x √1 + r 2
=
3
2
dr
Solution The integrand f(r) = √1+r
3
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
b(x) = x2
or
Therefore,
h'(x)
=
h'(x)
=
h'(x)
=
d
∫
dx
d
∫
db
√1 + b 3
b 0
b 0
× 2x
h'(x)
=
2x √1 + (x2)3
h'(x)
=
2x
√1 + x
6
[√1 + r ] dr 3
[√1 + r ] dr 3
×
db dx
Example 9
Using Part 1 of the Fundamental Theorem, compute the derivative of
y
∫
=
√x
cos t t
3
dt
Solution The 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
m(x) = √x
or
Thus,
y'
=
y'
=
y'
=
d dx
∫
√x
cos t t
3
d
∫
dm cos m m
y'
=
y'
=
m 3
cos t t
dm
×
dt
dx
1
×
cos √x √x
dt
2√x 1
×
2√x
cos √x 2x
Example 10 Using Part 1 of the Fundamental Theorem, compute the derivative of
y
∫
=
[t
cos x 1
+ sin t
] dt Solution
The 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'
=
d dx
∫
y'
=
d dw
∫ [t + sin t] dt
cos x 1
w
1
[t
+ sin t
] dt ×
dw dx
y'
=
(w + sin w)
y'
=
- sin x (w + sin w)
y'
=
- sin x(cos x + sin [cos x])
y'
=
- sin x cos x – sin x sin(cos x)
×
- sin x
Example 11 Using Part 1 of the Fundamental Theorem, compute the derivative of
y
∫
=
u3
1
du
1 + u2
1 -3x
Solution
By property 1 of definite integrals, we have
y
–
=
∫
u3
1 -3x
du
1 + u2
1
Let a(x) = 1 – 3x. Thus, we have
y
–
=
∫
u3
a 1
1 + u2
du
Hence, the derivative of y equals
y'
=
y'
=
y'
d da
–
– a3 1 + a2
∫
a 1
u3 1+u
2
×
du
da dx
× -3
3a3
=
1 + a2
Since a = 1 – 3x, then,
y'
3(1 – 3x)3
=
1 + (1 – 3x)2
Example 12
Using Part 1 of the Fundamental Theorem, compute the derivative of
y
=
∫
0
1/x2
[sin t] dt 3
Solution Using property 1 of definite integrals, we have
y
–
=
∫
2
1/x
[sin t] dt 3
0
Let
m(x) = 1/x2 =
x-2
Thus, we have
y
–
=
∫
m
[sin t] dt 3
0
Hence, the derivative y' equals
y'
y' y'
y'
d dm
=
–
– sin3
1 x2
2 sin3
1 x2
=
[sin t] dt
0
dm
×
3
dx
× -2x-3
= - sin3 m =
∫
m
× (-2x-3)
x3 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 13
Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
3 -1
x5 dx
Solution To 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) = x 5 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
F(x)
x n+1 n+1
=
We find that the derivative of f in this case is
F(x) =
x6/6
Recall the FTC 2 formula:
∫
b a
f(x) dx
=
F(b)
–
F(a)
+
C
(Where c is an arbitrary constant)
This means the antiderivative will be evaluated at the two values a and b. In this case, a = -1 and b = 3. Therefore,
∫
b
f(x) dx
a
=
F(b)
–
F(a)
= [(3)6/6] - [(-1)6/6] =
(729/6) - (1/6)
=
728/6
=
364/3
Hence
∫
3 -1
x5 dx
=
364/3
≈
121.3
Example 14 Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
8
(4x
2
+ 3) dx
Solution The 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
∫
8 2
(4x
+ 3) dx
=
F(b)
–
F(a)
(where a = 2 and b = 8)
∫
8 2
(4x
+ 3) dx
[2(8)
=
2
+ 3(8)] -
[2(2)
=
(128 + 24) - (8 + 6)
=
152 - 14
=
138
2
+ 3(2)]
Example 15 Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
4 0
(1 +
3y – y2) dy
Solution
Here, we have the integrand f(y) = 1 + 3y - y2 , whose antiderivative is given by
F(y)
=
y
+
3y2 2
–
y3 3
Therefore, Part 2 of the fundamental theorem gives
∫
4 0
(1 +
3y – y2) dy
=
F(b)
–
F(a)
(where a = 0 and b = 4)
∫
4
(1 +
0
3y – y2) dy
=
Therefore,
∫
4
(1 +
0
3y – y2) dy
3(4)2 2
+
(4)
=
4 + 24 – (64/3) -
=
20/3
=
≈
20/3
–
(4)3 3
–
(0)
+
3(0)2 2
0
6.667
Example 16
Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
1
(3/t ) 4
-1
dt
Solution For ease of simplification, this integral can be better expressed as
∫
1
(3t ) –4
-1
dt
Using the general antiderivative formula, the antiderivative of the integrand becomes
F(t)
= -t-3
= -1/t3
Using FTC 2, we have
∫
1 -1
(3t ) –4
dt
=
F(b)
–
F(a)
(Where a = -1 and b = 1)
∫
1 -1
(3t ) –4
dt
=
[-1/(1) ] 3
=
(-1/1)
=
-1 - 1
=
-2
-
–
[-1/(-1) ] 3
(-1/-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.
–
(0)3 3
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
∫
1
(3t ) –4
-1
DOES NOT EXIST
dt
Example 17 Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
2π π
(cos θ) dθ
Solution The 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
∫
2π π
(cos θ) dθ
=
F(b)
–
F(a)
(where a = π and b = 2π). So,
∫
2π π
(cos θ) dθ
=
sin 2π - sin π
=
0
If you use a calculator or computer you might get something like 5.91991876344e-13 which is indeed very close to zero.
Example 18 Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
2 0
f(x) dx
=
x4
if 0 ≤ x < 1
x5
if 1 ≤ x ≤ 2
Solution We 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]:
From the graph, it is evident that we will have to find two areas:
∫
1 0
(x4) dx
∫
AND
2 1
(x5) dx
which happen to be adjacent to each other. Thus, the sum of the areas will represent the value of the integral
∫
2 0
f(x) dx
This is simply another way of illustrating property 7 of the definite integral. Using FTC 2,
∫
1 0
=
(x4) dx
F(b)
–
F(a)
(where a = 0, b = 1 and the antiderivative F(x) = 0.2x5). Thus,
∫
1 0
(x4) dx
=
F(1)
=
0.2
–
=
F(0)
0.2(1)5 - 0.2(0)5
AND
∫
2 1
(x5) dx
=
F(b)
–
F(a)
(where a = 1, b = 2 and the antiderivative F(x) = x6/6). Thus,
∫
2 1
(x5) dx
=
[(2)6/6] - [(1)6/6]
=
64/6 - 1/6
=
63/6
=
Hence,
∫
2 0
f(x) dx
= =
∫
1 0
(x4) dx
0.2 + 10.5
+ =
∫
2 1
10.7
(x5) dx
10.5
Example 19
Evaluate the integral using Part 2 of the Fundamental Theorem, or explain why it doesn't exist.
∫
π -π
if -π ≤ x ≤ 0
x
=
f(x) dx
sin x
if 0 ≤ x ≤ π
Solution
Like 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:
∫
0 -π
(x) dx
∫
AND
π 0
(sin x) dx
Also, from the graph, we find that
∫ ∫
π -π π -π
f(x) dx
f(x) dx
=
=
∫
π
∫
π
0
0
(sin x) dx
(sin x) dx
–
+
–
∫
Let's deal with each integral one at a time. Thus, we have
∫
π 0
(sin x) dx
=
F(b)
–
F(a)
(where a = 0, b = π and the antiderivative F(x) = -cos x). Therefore,
0 -π
∫
0 -π
(x) dx
(x) dx
∫
π 0
(sin x) dx
=
F(π)
=
–
F(0)
-cos(π) - [-cos (0)]
=
1 - [-1]
=
2
AND
∫
0 -π
(x) dx
=
F(b)
–
F(a)
(where a = -π, b = 0 and F(x) = x2/2). Therefore,
∫
0 -π
(x) dx
=
F(0)
F(– π)
–
=
[(0 )/2]
=
0 - π2/2
-
2
[(-π) /2] 2
=
- π2/2
Hence,
∫
π -π
f(x) dx
2 - (- π2/2)
= =
2 + (π2/2)
=
(4 + π2 )/2
So,
∫
π -π
f(x) dx
(4 + π2 )
=
2
Unlike example 18, this area is a result of the difference between adjacent areas.
Example 20
Use 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
Solution
Using the definite integral notation, we can express the required area as
∫
27 0
( √x) 3
dx
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 ∆x ∆x
= (b – a)/n = (27 – 0)/9 = 3
This means
x1 x2 x0 x4 x5 x6 x7 x8 x9
= = = = = = = = =
a a a a a a a a a
+ + + + + + + + +
∆x 2∆x 3∆x 4∆x 5∆x 6∆x 7∆x 8∆x 9∆x
= = = = = = = = =
0 0 0 0 0 0 0 0 0
+ + + + + + + + +
3 2(3) 3(3) 4(3) 5(3) 6(3) 7(3) 8(3) 9(3)
= = = = = = = = =
3 6 9 12 15 18 21 24 27
Hence, an estimate for the area is given by the Riemann sum
n=9
R9
=
∑ f(x )∆ x i
i=1
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 + 9.0000
R9
=
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.
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,
27 0
( √x) 3
a = 0,
dx
b = 27
and
=
F(b)
F(a)
=
F(27)
F(x) = ¾ (√x)4
Therefore,
∫
27 0
( √x) 3
dx
∫
27 0
( √x) 3
dx
=
–
F(0)
[¾ (81)]
=
So,
–
–
[¾ ( √27) ]
=
[0]
3
=
4
–
[¾ ( √0) ] 3
4
60.75
60.75
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 21
Use 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
SOLUTION
This 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]:
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 ∆x ∆x
= (b – a)/n = (π/3 – 0)/10 = π/30
We therefore estimate the area of A using the Riemann sum
n=10
A
∑ f(x )∆ x
=
i
i=1
Which equals
n=10
A
=
∑
i=1
n=10
(sec 2 xi) ∆x
∑
=
i=1
(sec 2 xi) (π/30)
n=10
=
(π/30)
∑ (sec
2
i=1
xi)
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 = sec 2x on the interval [0,π/3]. So, an appropriate integral notation would be
A
∫
=
π/3 0
(sec x) 2
dx
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,
∫
π/3 0
(sec x) 2
=
dx
tan (π/3) –
=
tan(0)
1.732508 – 0
≈
Hence, the EXACT area is 1.7321. In other words,
∫
π/3 0
(sec x) 2
dx
=
1.7321
1.7321
Example 22
Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch.
∫
2 –1
(x ) 3
dx
SOLUTION The integrand y = x is continuous on the interval [-1,2]: 3
Using FTC 2, we have
∫
2 –1
(x ) 3
dx
=
F(b) – F(a)
Where
F(x) = x4/4,
a = -1,
and
b=2
Therefore,
∫
2 –1
(x ) 3
dx
=
[(2) /4]
=
[16/4]
=
3.75
4
–
[(-1) /4]
– [¼ ]
4
=
15/4
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):
The part of the graph below the x-axis is on the interval [-1, 0]. Thus, its area is
A2
∫
=
0 –1
(x ) 3
[(0) /4]
=
dx
4
[(-1) /4]
–
4
=
– 0.25
Note that the minus sign indicates the area is below the x-axis. On the other hand, the part of the graph above the xaxis is on the interval [0, 2]. Therefore, its area is
A1
∫
=
2 0
(x ) 3
=
dx
[(2) /4]
–
4
[(0) /4] 4
=
From one of the definitions of a definite integral,
∫
b a
=
f(x) dx
A1 – A 2
Which means,
∫
2
(x ) –1 3
dx
=
∫
=
4
2 0
dx
–
∫
0.25
=
3.75
(x ) 3
–
0 –1
(x ) 3
dx
Example 23 Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch.
∫
5π/2 π/4
(sin x)
dx
4
SOLUTION
Let's graph the integrand first, and understand what we're dealing with:
Using FTC 2,
∫
5π/2 π/4
(sin x)
dx
=
F(b) – F(a)
Where
F(x)
a = π/4
b = 5π/2,
= – cos x,
Which gives
Area
=
∫
5π/2 π/4
(sin x)
dx
[
]
–
[– √2/2]
– cos 5π /2
= =
0
=
– √2/2
[– cos(π/4)]
–
=
0.7071
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
Area
=
∫
5π/2 π/4
(sin x)
=
dx
(X + Y) – Z
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,
Area of X
=
∫
π π/4
(sin x)
dx
– cos [π] –
=
= =
1
–
[– √2/2]
1.7071
[– cos(π/4)]
∫
=
Area of Y
5π/2 2π
(sin x)
dx
=
– cos [5π/2] –
= =
0
[– cos(2π)]
[– 1 ]
–
1
And then
∫
=
Area of Z
2π π
(sin x)
– cos [2π] –
=
dx
= =
[– cos(π)]
–1 – 1 –2
Now that we've computed all three areas, the total area becomes
∫
5π/2 π/4
(sin x)
∫
=
dx
= = =
π π/4
(sin x)
[1.7071 + 1 ] 2.7071
–
∫
+
dx
–
5π/2 2π
(sin x)
dx
–
∫
2
2
0.7071
This answer coincides with the one obtained when we first used FTC 2 to evaluate the integral.
Example 24
Find the derivative of the function.
∫
3x 2x
u2 – 1
du
u2 + 1
Here's a hint.
∫
3x 2x
f(u) du
=
∫
0
f(u) du
2x
+
∫
f(u) du
SOLUTION
The integrand here is
f(u)
u2 – 1
=
u2 + 1
Let's assume that
g(x)
3x 0
=
Using the provided hint,
∫
3x 2x
u2 – 1 u2 + 1
du
2π π
(sin x)
dx
∫
u2 – 1
3x
u2 + 1
2x
=
∫
=–
∫
du
0
u2 – 1
2x
u2 + 1
2x
u2 – 1
0
u2 + 1
du
du
+
∫
+
∫
3x
u2 – 1
0
u2 + 1
3x
u2 – 1
0
u2 + 1
du
i
du
Let's solve the RHS of equation 1 one at a time. We start with
∫
–
2x
u2 – 1
0
u2 + 1
du
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
∫
–
m
u2 – 1
0
u2 + 1
du
At this point, we use FTC 1 alongside the chain rule, which gives:
d
=
–
2
=
–
2
∫
–
dm
m
u2 – 1
0
u2 + 1
m2 – 1
=
m2 + 1
du
–
4x2 – 1
dm
dx
=
(2x)2 – 1
2
(2x)2 + 1
ii
4x2 + 1
So,
d
dx
–
∫
2x
u2 – 1
0
u
du
+ 1
2
=
And now for the second part:
∫
3x
u2 – 1
0
u2 + 1
du
We let n(x) = 3x. Thus, we have
∫
n
u2 – 1
0
u2 + 1
du
–
2
4x2 – 1 4x 2 + 1
–
m2 – 1 m2 + 1
×
2
Thus, we are to evaluate
∫
d
dx
n
u2 – 1
0
u2 + 1
du
Using FTC 1 and the chain rule, we have
d
∫
dn
n
u2 – 1
0
u2 + 1
dn
du
=
dx
n2 – 1 n2 + 1
×
3
This gives:
n2 – 1
3
=
3
3
=
n2 + 1
(3x)2 – 1
9x2 – 1
(3x)2 + 1
iii
9x2 + 1
So,
g'(x)
∫
d
=
dx
u2 – 1
3x
u2 + 1
2x
du
Combining ii and iii gives
g'(x)
=
g'(x)
=
–
4x2 – 1
2
+
4x2 + 1
9x2 – 1
3
9x
2
3
–
+ 1
9x2 – 1 9x2 + 1
4x2 – 1
2
4x2 + 1
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 25 Find the derivative of the function.
y
=
∫
x3 √x
(√t
sin t
)
dt
SOLUTION
The first step to differentiating this integral is splitting it into two simpler integrals (on adjacent intervals). Like the previous example, we have:
∫
x3 √x
(√t sin t) dt
∫
=
0 √x
(√t sin t) dt
+
∫
x3 0
(√t sin t) dt
Let
m(x) =
n(x) =
∫ ∫
0 √x
x3 0
(√t sin t) dt
(√t sin t) dt
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:
m(x) =
∫
0
(√t sin t) dt
√x
=
–∫ 0
√x
Therefore,
m'(x)
=
–∫ 0
√x
d dx
(√t sin t) dt
We let r = √x, so that we have
m'(x)
=
–∫ 0 r
d dx
(√t sin t) dt
Using FTC1 and the chain rule, we have
m'(x)
=
–∫ 0 r
d dr
=
√r sin r
=
√√x sin √x
=
–
4
(√t sin t) dt
×
×
dr dx
1 2√x 1 2√x
√x sin √x 2√x
So,
m'(x)
=
Next, we solve n'(x):
–
4
√x sin √x 2√x
i
(√t sin t) dt
n(x) =
∫
x3 0
(√t sin t) dt
Which means
n'(x)
=
d dx
∫
x3 0
(√t sin t) dt
We let g =x3, so that
n'(x)
=
d dx
∫
g
n'(x)
=
d dg
∫
g
n'(x)
=
√g sin g × 3x2
n'(x)
=
3x2 √x3 sin x3
0
0
(√t sin t) dt (√t sin t) dt
dg dx
ii
So now, we combine i and ii to give
d dx
∫
x3 √x
(√t sin t) dt
= =
m'(x) + n'(x) –
4
√x sin √x 2√x
+
3x2 √x3 sin x3
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.
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