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| Integral
calculus |
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The fundamental theorem of
calculus |
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The fundamental theorem of differential calculus |
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Evaluating definite integrals
using indefinite integrals |
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The fundamental theorem of integral calculus |
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Evaluating the definite integral examples |
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The fundamental theorem of calculus |
| The
theorem that states the relationship between integration and
differentiation, that is, between areas and tangent
lines, is called the fundamental theorem
of calculus |
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The fundamental theorem of differential calculus |
| If
f (x) is continuous on
closed interval [a,
b] and F
(x)
is defined to be |
 |
| then,
F
is differentiable on (a,
b)
such that F' (x)
= f (x) for all x
in (a,
b).
This means that |
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| Evaluating definite integrals
using indefinite integrals |
| To
evaluate the definite integral |
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we should find one primitive
function F(x)
or antiderivative of |
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the function f(x),
and since the indefinite integral |
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is a primitive
function of f (x)
then, as two |
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| primitives of the same function can differ only by a
constant, we can write |
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| To
find the value of the constant C
that belongs to the lower limit a,
substitute x = a
to both sides of the |
| above equality, and
since |
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then C
= -
F(a),
so that |
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| Therefore,
the
definite and indefinite integrals are related by the fundamental theorem
of calculus. |
| This
result shows that integration is inverse of differentiation. |
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| The fundamental theorem of integral calculus |
| If
f(x)
is integrable on the interval [a,
b]
and F(x)
is an antiderivative of f
on (a,
b),
then |
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| The
right side of the above equation we usually write |
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| so
that, |
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| Thus,
to evaluate the definite integral we need to find an atiderivative F
of f,
then evaluate F(x)
at x = b
and |
| at
x
= a, and calculate the
difference F(b)
-
F(a). |
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| Evaluating
the definite integral examples |
| Example:
Find
the area under the line f(x)
= x
+ 1
over the interval [1,
5]. |
| Solution:
To evaluate the
definite integral we need to find the atiderivative
F
of f
(x)
= x
+ 1, evaluate F(x) |
| at x
= 5
and at x
= 1,
and calculate the difference |
| F(5)
-
F(1). |
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 |
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| Cavalieri
- Gregory formula for quadrature of the parabola |
| Example:
Let define the
surface area enclosed by the arc of the parabola f
(x)
= Ax2
+ Bx
+ C
and x-axis
over the interval [a,
b]. |
| Solution: |
 |
| where
the expression in the square brackets |
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| Thus,
the quadrature of the quadratic function leads to calculation of the
three ordinates or values of the function
f (a), |
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f ((a+ b)/2)
and f
(b). |
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| Example:
Find
the area enclosed by the sine function f
(x)
= sin x
and the x-axis
over the interval [p/2,
3p/2].
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Solution: |
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| Example:
Evaluate |
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| Solution: |
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| Example:
Find
the area enclosed by the graph of the cubic
f(x) = (-1/3)x3
+ 4x
and the positive part of the x-axis. |
| Solution: The
limits of integrations are defined by the roots of the cubic that we can
find by solving the |
| equation
f(x)
= 0, |
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| The
limits of integration, x1
= 0 and x3
= 2Ö3,
so |
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| Example:
Evaluate |
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| Solution: |
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