Mathematics Dictionary
Dr. K. G. Shih
Integral : Anti-derivatives and substitution rule
Subjects
Symbol Defintion
Example : Exp(x) is e^x and is e to power x
Q01 |
- Anti-derivative
Q02 |
- Integration : Power rule of derivative ∫(x^n)dx
Q03 |
- Integration : Trigonometric functions by anti-differentiation
Q04 |
- Integration : Hyperbolic function by anti-differentiation
Q05 |
- Integration : Substitution rule
Q06 |
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Q07 |
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Q08 |
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Q09 |
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Q10 |
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Answers
Q01. Anti-derivative
Definition : Ant-derivative
d/dx(sin(x)) = cos(x) is called derivative of sin(x)
We can express it as d(sin(x)) = (cos(x))dx
Take integration on both sides
sin(x) = ∫ cos(x)dx which is called anti-derivative
Graphs in Analytic geometry
Anlytic geometry
What is anti-derivative ?
Curve of function
The derivative of y = F(x) which will change the graph of the function
The anti-derivative will re-store the graph of function
Slope and area
Derivative finds slope using limit (Section 1)
Integral finds area using summation (section 1)
Examples
The graph of y = sin(x) is a sine curve
The graph of d/dx(sin(x) becomes a cosine curve
The graph of ∫ cos(x)dx + C is sine curve again
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Q02. Integration : Power rule
Derivative : Power rule
d/dx(x^5) = 5*x^4
d/dx(x^(-5)) = - 5*x^(-6)
d/dx(x^(1/5)) = (1/5)*(x^(-4/5))
d/dx(x^(-1/5)) = (-1/5)*(x^(-6/5))
Anti-differentiation : Corresponding power rule
d(x^5) = (5*x^4)dx and hence ∫(5*x^4)dx = x^5 + C
d(x^(-5)) = (-5*x^(-6))dx and hence ∫(-5*x^(-6))dx = x^(-5) + C
d(x^(1/5)) = ((1/5)*(x^(-4/5)))dx and hence ∫((1/5)*(x^(-4/5)))dx = x^(1/5) + C
d(x^(-1/5)) = ((-1/5)*(x^(-6/5)))dx and hence ∫((-1/5)*(x^(-6/5)))dx = x^(-1/5) + C
Formula
∫(x^n)dx = (1/(n+1))*(x^(n+1)) + C
The value of n can be positive integer or negative integer
The value of n can be also real rational number
But n can not be equal to -1.
Note : ∫(x^(-1))dx = ln(x) + C
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Q03. Integration : Trigonometric functions
Derivatives and anti-differentiations
Since d/dx(sin(x)) = +cos(x).......... Hence ∫(cos(x))dx = +sin(x)
Since d/dx(cos(x)) = -sin(x).......... Hence ∫(cos(x))dx = -cos(x)
Since d/dx(tan(x)) = +sec(x)^2........ Hence ∫(sec(x)^2)dx = tan(x)
Since d/dx(csc(x)) = -csc(x)*cot(x)... Hence ∫(csc(x)*cot(x))dx = -csc(x)
Since d/dx(sec(x)) = +sec(x)*tan(x)... Hence ∫(sec(x)*tan(x))dx = +sec(x)
Since d/dx(cot(x)) = -csc(x)^2........ Hence ∫(csc(x)^2)dx = -cot(x)
Derivatives and anti-differentiations
Since d/dx(arcsin(x)) = +1/Sqr(1-x^2)...... ∫(1/Sqr(1-x^2))dx = +arcsin(x)
Since d/dx(arccos(x)) = -1/Sqr(1-x^2)...... ∫(1/Sqr(1-x^2))dx = -arccos(x)
Since d/dx(arctan(x)) = +1/(1+x^2)......... ∫(1/(1+x^2)dx = arctan(x)
Since d/dx(arccsc(x)) = -1/(x*Sqr(x^2-1)).. ∫(1/(x*Sqr(x^2-1)))dx = -arccsc(x)
Since d/dx(arcsec(x)) = +1/(x*Sqr(x^2-1)).. ∫(1/(x*Sqr(x^2-1)))dx = +arcsec(x)
Since d/dx(arccot(x)) = -1/(1+x^2)......... ∫(1/(1+x^2)dx = -arccot(x)
Howe to prove ?
section 10 : Integral of trigonometric functions
Section 12 : Integral of inverse trigonometric functions
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Q04. Integration : Hyperbolic functions
Derivatives and anti-differentiations
Since d/dx(sinh(x)) = +cosh(x)............. ∫(cosh(x))dx = +sinh(x)
Since d/dx(cosh(x)) = +sinh(x)............. ∫(cosh(x))dx = +cosh(x)
Since d/dx(tanh(x)) = +sech(x)^2........... ∫(sech(x)^2)dx = tanh(x)
Since d/dx(csch(x)) = -csch(x)*coth(x)..... ∫(csch(x)*coth(x))dx = -csch(x)
Since d/dx(sech(x)) = -sech(x)*tanh(x)..... ∫(sech(x)*tanh(x))dx = -sech(x)
Since d/dx(coth(x)) = -csch(x)^2........... ∫(csch(x)^2)dx = -coth(x)
Derivatives and anti-differentiations
Since d/dx(arcsinh(x)) = +1/Sqr(x^2+1)...... ∫(1/Sqr(x^2+1))dx = +arcsinh(x)
Since d/dx(arccosh(x)) = +1/Sqr(x^2-1)...... ∫(1/Sqr(x^2-1))dx = +arccosh(x)
Since d/dx(arctanh(x)) = +1/(1-x^2)......... ∫(1/(1-x^2)dx = arctanh(x)
Since d/dx(arccsch(x)) = -1/(x*Sqr(x^2+1)).. ∫(1/(x*Sqr(x^2+1)))dx = -arccsch(x)
Since d/dx(arcsech(x)) = -1/(x*Sqr(x^2-1)).. ∫(1/(x*Sqr(x^2-1)))dx = -arcsech(x)
Since d/dx(arccoth(x)) = +1/(1-x^2)......... ∫(1/(1-x^2)dx = +arccoth(x)
Howe to prove ?
section 10 : Integral of trigonometric functions
Section 12 : Integral of inverse trigonometric functions
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Q05. Integration : Substitution rule
Why ?
No derivative of function is available for finding anti-derivative
For example : Integral of ∫x*Sqr(1 + x^2)dx has no corresponding derivative
What ?
Substitution rule in the integration is corresponding to chain rule in derivative
We use new variable in the function
How ?
Let function be F(x)
Let u = G(x) and du = (G'(x))dx
∫F(G(x))*(G'(x))dx = ∫F(u)du
Example : evaluate ∫(2*x)*Sqr(1 + x^2)dx
Let 1 + x^2 = u and du = (2*x)dx
Hence ∫(2*x)*Sqr(1 + x^2)dx
= ∫Sqr(u)du
= (1/(1+1/2)*(u^(1+1/2)
= (2/3)*(u^(3/2))
= 2*(1 + x^2)^(3/2) + C
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Q06. New
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Q07. Answer
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Q08. Answer
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Q09. Answer
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Q10. Answer
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