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Integrals of Trigonometric Functions
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Integrals of Hyperbolic Functions
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Integrals of Exponential and Logarithmic Functions
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Integrals of Simple Functions
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Integral (Indefinite)
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Derivatives Basic
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Differentiation Rules
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Integral (Definite)
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Integral (Indefinite)
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Integrals of Simple Functions
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Math Formulas
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Calculus
> Integral (Indefinite)
Integral (Indefinite)
Given a function,
f(x)
, an anti-derivative of
f(x)
is any function
F(x)
such that
F´(x) = f(x)
If
F(x)
is any anti-derivative (or primitive function) of
f(x)
then the most general anti-derivative of
f(x)
is called an indefinite integral and denoted,
∫f(x)dx = F(x) + c
, where c is a constant
In this definition the ∫ is called the integral symbol, f(x) is called the integrand,
x
is called the integration variable and the “
c
” is called the constant of integration and can take any real value. By assigning different value to c, we obtain different values of the integral.
Properties of Indefinite Integrals
1.
∫ a dx = ax + c
2.
∫a f(x) dx = a ∫ f(x) dx
3.
∫ - f(x) dx = - ∫ f(x) dx
4.
∫ [f(x) ± F(x)] dx = ∫f(x) dx ± ∫F(x) dx
5. The Substitution Rule:
If
u = g(x)
is a differentiable function whose range is an interval I and
f
is continuous on I, then
∫f(g(x))g´(x)dx = ∫f(u)du
Standard Substitution
1. For terms of the form x
2
+ a
2
or square root of x
2
+ a
2
, put x = a tan θ or a cot θ.
2. For terms of the form x
2
- a
2
or square root of x
2
- a
2
, put x = a sec θ or a cosec θ.
3. For terms of the form a
2
- x
2
or square root of a
2
- x
2
, put x = a sin θ or a cos θ.
4. If both
,
are present, then put x = a cos θ.
5. For the type
, put x = a cos
2
q + b sin
2
q.
6. If n is not equal to minus one, the integral of u
n
du is obtained by adding one to the exponent and divided by the new exponent. This is called the General Power Formula.
where C is the constant of Integration.
7. Integration by Parts: If u and v be two functions of x, then integral of the product of these two functions is given by:
This method is mainly used when the integrand is the product of two functions.
Note :
In applying the above rule care has to be taken in the selection of the first function(u) and the second function. If both of the functions are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually we use the following preference order for the first function.
(Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)
. In this order, the function on the left is always chosen as the first function. This rule is called as
ILATE.
Not listed in the properties were integrals of products and quotients. The reason for this is simple. Just like with derivatives each of the following will NOT work.
∫ [f(x) × F(x)] dx ≠ ∫f(x) dx ×∫F(x) dx
and
∫ [f(x) / F(x)] dx ≠ ∫ [f(x) dx] / ∫ [F(x) dx]
Example 1
: Evaluate the integral
Solution
:
Be careful to not think of the third term as
x
to a power for the purposes of integration. Using that rule on the third term will NOT work. The third term is simply a logarithm. Also, don’t get excited about the 15. The 15 is just a constant and so it can be factored out of the integral. In other words, here is what we did to integrate the third term.
Example 2
: Calculate the indefinite integral
Solution:
Example 3:
∫ (1-2x
2
)
3
dx
Solution:
∫ (1-2x
2
)
3
dx
If we let
u=1-2x
2
and
n=3
, then
du= - 4xdx
. But there is no
x
in the given integrand. It is easy to insert
-4
in the integrand and offset this by placing
-1/4
before the integral sign but nothing can be done about the missing factor
x
. We therefore expand (1-2x
2
)
3
and integrate term by term.
∫ (1-2x
2
)
3
dx
=
∫ [1
3
- 3(1
2
)(2x
2
) + 3(1)(2x
2
)
2
- (2x
2
)
3
] dx
=
∫(1 - 6x
2
+ 12x
4
– 8x
6
)dx
=∫(1 - 6x
2
+ 12x
4
- 8x
6
)dx
=x – 6x
3
/3 + 12x
5
/5 – 8x
7
/7 + C
=x – 2x
3
+ 12x
5
/5 – 8x
7
/7 + C
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