Sometimes it is difficult to integrate a function using the simple formulas of integrals. We have to introduce a second variable to make the integration easier.
Example 1
Calculate ∫(x+1)⁵dx
If it was the integral of x⁵ it would be easy to calculate. We can introduce another variable to make the integral easy to calculate. If we make x + 1 equal to u we will just have to calculate the integral of u. So let's write u = x+1
Then our integral becomes ∫u⁵dx. We calculate the integral as if it was the integral of x⁵.
∫u⁵dx = u⁵⁺¹/5+1 + C = 1/6 u⁶ + C
Let's substitute u we get ∫u⁵dx = 1/6 (x+1)⁶ + C
Example 2
Calculate ∫√4x+3dx
Let's write the expression as a power:
∫√4x+3dx = ∫(4x+3)¹/² dx
Let's write u = 4x+3 du = 4dx then dx = du/4. Let's substitute 4x+3 and dx
∫√4x+3dx = ∫u¹/².du/4 = 1/4∫u¹/²du = 1/4(u¹/²+¹/¹/²+¹) + C = 1/4(u³/²/3/2) + C = (1/4)(2/3)u³/² + C = 1/6u³/² + C = 1/6√(4x+3)³ + C
If you are interested in learning more about integrals visit Center for Integral Development
Friday, May 4, 2018
Friday, April 27, 2018
Using derivative rules to find the integrals of some basic functions
Integral of the power function
Let f be f(x) = xⁿ its antiderivative is F(x) = 1/n+1x^n+1+C (Read F(x) equals 1 over n+1 times x exponent n+1 plus the constant of integration C) . You can rewrite the expression by utilizing a horizontal bar for the fraction 1 over n+1. You can also rewrite the expression x^n+1 by putting n+1 besides x and above. The rewriting will make the expression clearer.
The antiderivative of f(x) being the indefinite integral, we can write ∫f(x)dx = 1/n+1.x^n+1+C (Read sum f(x) dx equals 1 over n+1 times x exponent n+1 plus the constant of integration C).
In order to prove this, we have to demonstrate that the derivative of 1/n+1.x^n+1+C is xⁿ.
Let's calculate the derivative of F(x) = 1/n+1.x^n+1+C
F'(x) = n+1/n+1x^n+1-1 (Read n+1 over n+1 times x exponent n+1-1. By simplifying and reducing F'(x) = xⁿ
Integral of the inverse function
The integral of the inverse function f(x) = 1/x is ∫1/xdx = lnx +C.
Proof
The derivative of F(x) = lnx is F'(x) = 1/x
Integral of the exponential function
The integral of the exponential function f(x) = e^x is ∫e^xdx = e^x + C. The term e^x reads e exponential x.
Proof
The derivative of F(x) = e^x is F'(x) = e^x.
Rules of integrable functions
If f and g are integrable functions and C is a constant, then
∫[f(x)+g(x)] = ∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)] = ∫f(x)dx-∫g(x)dx
∫kf(x)dx = k∫f(x)dx
Example
Solution
Using the basic rules of integration we have:
If you are interested in learning more about integrals visit Center for Integral Development
Let f be f(x) = xⁿ its antiderivative is F(x) = 1/n+1x^n+1+C (Read F(x) equals 1 over n+1 times x exponent n+1 plus the constant of integration C) . You can rewrite the expression by utilizing a horizontal bar for the fraction 1 over n+1. You can also rewrite the expression x^n+1 by putting n+1 besides x and above. The rewriting will make the expression clearer.
The antiderivative of f(x) being the indefinite integral, we can write ∫f(x)dx = 1/n+1.x^n+1+C (Read sum f(x) dx equals 1 over n+1 times x exponent n+1 plus the constant of integration C).
In order to prove this, we have to demonstrate that the derivative of 1/n+1.x^n+1+C is xⁿ.
Let's calculate the derivative of F(x) = 1/n+1.x^n+1+C
F'(x) = n+1/n+1x^n+1-1 (Read n+1 over n+1 times x exponent n+1-1. By simplifying and reducing F'(x) = xⁿ
Integral of the inverse function
The integral of the inverse function f(x) = 1/x is ∫1/xdx = lnx +C.
Proof
The derivative of F(x) = lnx is F'(x) = 1/x
Integral of the exponential function
The integral of the exponential function f(x) = e^x is ∫e^xdx = e^x + C. The term e^x reads e exponential x.
Proof
The derivative of F(x) = e^x is F'(x) = e^x.
Rules of integrable functions
If f and g are integrable functions and C is a constant, then
∫[f(x)+g(x)] = ∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)] = ∫f(x)dx-∫g(x)dx
∫kf(x)dx = k∫f(x)dx
Example
Solution
Using the basic rules of integration we have:
If you are interested in learning more about integrals visit Center for Integral Development
Friday, March 2, 2018
The formula of the antiderivative of a function simply explained
Let's consider the function f(x) = x² +3 its derivative is f '(x) = 2x. The function f '(x) = 2x is the derivative of the function f(x) = x² + 3. The function f(x) = x² +3 is the antiderivative of the function
f '(x) = 2x. The antiderivative is the function from which the derivative was calculated.
If we consider the function f(x) = 2x it can be considered as the derivative of another function F. This function which is F(x) = x² + 3 is the antiderivative of the function f. Actually, there are several functions whose derivative is f(x) = 2x. Some examples of these functions are F(x) = x²+1, F(x) = x²-3, F(x) = x² + 6. There is a bunch of functions whose derivative is f '(x) = 2x. We can't list them all. For this reason we represent all them by the function F(x) = x²+C.
Notation and notion of indefinite integral
The process of finding the antiderivative of a function is called antidifferentiation or integration. If F is the antiderivative of a function f we write ∫f(x)dx = F(x) +C. The left side of this equation is called integral of f(x). The function f(x) is called the integrand. The letter C is the constant of integration. The symbol dx means that the function is integrated with respect to x.
Formula to calculate the antiderivative of the function power
Let's retake the function f(x) = 2x. Its antiderivative is F(x) = x² + C or F(x) = 2/2x² + C. Let's consider another function g(x) = 3x. Its antiderivative can be G(x) = 3/2x²+C. For the function h(x) = x² its antiderivative is H(x) = 1/3x³+C. For the function i(x) =5 x³ its antiderivative is I(x) = 5/4x⁴ + C
By observing each of these functions and their antiderivative, two points draw our attention.
1) The exponent of x in the antiderivative function is one degree more than the exponent of x in the given function.
2) The coefficient of x in the antiderivative function is obtained by dividing the coefficient of x in the given function by the exponent of x in the antiderivative function.
Let's come back to the examples above. In the function antiderivative F(x) the exponent of x is 2 and the exponent of x in the function f is 1. The coefficient of x 2/2 in the function F(x) is equal to the coefficient 2 of x in the function f(x) divided by the exponent 2 of x in the antiderivative function.
In the function antiderivative G(x) the exponent of x is 2 and the exponent of x in the function g(x) is 1. The coefficient 3/2 of x in the function G(x) is equal to the coefficient 3 of x in the function f(x) divided by the exponent 2 of x in the function G(x). The same is true for the functions H(x) and I(x).
From these observations, we can deduct a rule. In order to find the antiderivative of f((x) = xⁿ add 1 to the exponent and divide the coefficient by the new exponent,
The formula that generalizes this procedure is as follows:
If f(x) = kxⁿ its derivative is given by F(x) = k/n+1xⁿ⁺¹ + C.
That's all for today. To learn more about Calculus visit Center for Integral Development
f '(x) = 2x. The antiderivative is the function from which the derivative was calculated.
If we consider the function f(x) = 2x it can be considered as the derivative of another function F. This function which is F(x) = x² + 3 is the antiderivative of the function f. Actually, there are several functions whose derivative is f(x) = 2x. Some examples of these functions are F(x) = x²+1, F(x) = x²-3, F(x) = x² + 6. There is a bunch of functions whose derivative is f '(x) = 2x. We can't list them all. For this reason we represent all them by the function F(x) = x²+C.
Notation and notion of indefinite integral
The process of finding the antiderivative of a function is called antidifferentiation or integration. If F is the antiderivative of a function f we write ∫f(x)dx = F(x) +C. The left side of this equation is called integral of f(x). The function f(x) is called the integrand. The letter C is the constant of integration. The symbol dx means that the function is integrated with respect to x.
Formula to calculate the antiderivative of the function power
Let's retake the function f(x) = 2x. Its antiderivative is F(x) = x² + C or F(x) = 2/2x² + C. Let's consider another function g(x) = 3x. Its antiderivative can be G(x) = 3/2x²+C. For the function h(x) = x² its antiderivative is H(x) = 1/3x³+C. For the function i(x) =5 x³ its antiderivative is I(x) = 5/4x⁴ + C
By observing each of these functions and their antiderivative, two points draw our attention.
1) The exponent of x in the antiderivative function is one degree more than the exponent of x in the given function.
2) The coefficient of x in the antiderivative function is obtained by dividing the coefficient of x in the given function by the exponent of x in the antiderivative function.
Let's come back to the examples above. In the function antiderivative F(x) the exponent of x is 2 and the exponent of x in the function f is 1. The coefficient of x 2/2 in the function F(x) is equal to the coefficient 2 of x in the function f(x) divided by the exponent 2 of x in the antiderivative function.
In the function antiderivative G(x) the exponent of x is 2 and the exponent of x in the function g(x) is 1. The coefficient 3/2 of x in the function G(x) is equal to the coefficient 3 of x in the function f(x) divided by the exponent 2 of x in the function G(x). The same is true for the functions H(x) and I(x).
From these observations, we can deduct a rule. In order to find the antiderivative of f((x) = xⁿ add 1 to the exponent and divide the coefficient by the new exponent,
The formula that generalizes this procedure is as follows:
If f(x) = kxⁿ its derivative is given by F(x) = k/n+1xⁿ⁺¹ + C.
That's all for today. To learn more about Calculus visit Center for Integral Development
Saturday, February 10, 2018
Formal definition of limits
The formal definition of limits is mostly skipped in teaching about limits. The intuitive notion is the most taught. The formal definition of limits is interesting and is derived from the intuitive notion.
The intuitive definition says that the limit of a function f is L when x approaches a real number "a" if f becomes closer and closer to "L". We write lim f(x) = L when x➡ a .
When x comes closer and closer to a number "a" both to the left and right of that number, x gets values in the neighborhood of a. The variable x in approaching to "a" comes to a certain distance to "a" both from the left and the right. As this distance is very small we call it δ . The variable "x' takes values in the interval ⦐a-ẟ a+ẟ[. The set of values "x" are translated into the equation ᥣx-aᥣ<ε.
The independent variable f(x) comes to a certain distance of "L" both from the right and the left. As this distance is very small we call it δ. The independent variable f(x) takes values in the interval ]f(x)-ε f(x)+ε[.
When x approaches "a" f(x) approaches "L". In other words you give me an ϵ such that ❙f(x)-a❙<ε I will find you a δ that satisfies the equation ❙x-a❙<δ.
The formal definition of limits is stated:
Method
You give me the ε from the equation ❘f(x)-L❘<ε. I'll transform this equation in ❘x-xindice0❘<δ in order to find δ.
However when the inequality ❘f(x)-L❘<ϵ is transformed in a second degree inequality the process becomes more complicated. Follow the process of solving the example below to solve similar examples..
Interested in taking online Calculus and tutoring in the same subject visit Center for Integral Development
The intuitive definition says that the limit of a function f is L when x approaches a real number "a" if f becomes closer and closer to "L". We write lim f(x) = L when x➡ a .
When x comes closer and closer to a number "a" both to the left and right of that number, x gets values in the neighborhood of a. The variable x in approaching to "a" comes to a certain distance to "a" both from the left and the right. As this distance is very small we call it δ . The variable "x' takes values in the interval ⦐a-ẟ a+ẟ[. The set of values "x" are translated into the equation ᥣx-aᥣ<ε.
The independent variable f(x) comes to a certain distance of "L" both from the right and the left. As this distance is very small we call it δ. The independent variable f(x) takes values in the interval ]f(x)-ε f(x)+ε[.
When x approaches "a" f(x) approaches "L". In other words you give me an ϵ such that ❙f(x)-a❙<ε I will find you a δ that satisfies the equation ❙x-a❙<δ.
The formal definition of limits is stated:
You give me the ε from the equation ❘f(x)-L❘<ε. I'll transform this equation in ❘x-xindice0❘<δ in order to find δ.
However when the inequality ❘f(x)-L❘<ϵ is transformed in a second degree inequality the process becomes more complicated. Follow the process of solving the example below to solve similar examples..
Interested in taking online Calculus and tutoring in the same subject visit Center for Integral Development
Friday, December 1, 2017
Some considerations about the study of Calculus
Calculus has been invented a few hundred years ago by Sir Isaac Newton and Leibniz Gottfried to study the motion of planets and moons. After the work of these pioneers, several mathematicians have widened the field of Calculus by developing concepts and methods. The applications of Calculus have been extended to the study of phenomena in the physical, biological and social sciences.
Calculus is based on a few simple ideas and these have allowed the development of applications in different fields. The study of Calculus is based on a multi-representational approach to the concepts, methods, and applications represented numerically. analytically and graphically. The interesting element in the study of Calculus is that its core ideas are closed related. For example, the study of limits, derivatives, and integrals form a whole.
Calculus is the study of change and this is best modeled by the study of the behavior of functions. Functions have been studied in Pre-calculus, Different combinations of functions such as addition, multiplication, division and composition have been studied. Other properties of functions have been explored. The study of Calculus is more concerned about the behavior of functions closed to certain points. For example, the study of the different values of a function as the dependent variable comes closer and closer to a certain value leads to the notion of limits. The study of the slope of a tangent line to the graph of a function leads to the notion of derivative. The study of the area between curves leads to the study of integrals.
Calculus is based on a few simple ideas and these have allowed the development of applications in different fields. The study of Calculus is based on a multi-representational approach to the concepts, methods, and applications represented numerically. analytically and graphically. The interesting element in the study of Calculus is that its core ideas are closed related. For example, the study of limits, derivatives, and integrals form a whole.
Calculus is the study of change and this is best modeled by the study of the behavior of functions. Functions have been studied in Pre-calculus, Different combinations of functions such as addition, multiplication, division and composition have been studied. Other properties of functions have been explored. The study of Calculus is more concerned about the behavior of functions closed to certain points. For example, the study of the different values of a function as the dependent variable comes closer and closer to a certain value leads to the notion of limits. The study of the slope of a tangent line to the graph of a function leads to the notion of derivative. The study of the area between curves leads to the study of integrals.
The study of limits helps in the understanding of the derivatives and integrals. The limit of the slope of a secant line to a curve allows to find the slope of a tangent line to this curve. The slope of the tangent line is the derivative of the function. The limit of the sum of the rectangles of the area between two curves leads to a better approximation of the area between these curves. This leads to the notion of integrals.
Interested in learning more about Calculus visit this site Mathematical Education Center
Interested in learning more about Calculus visit this site Mathematical Education Center
Thursday, August 3, 2017
Note to the blog readers
Hello Readers,
I have taken a break since the middle of June in publishing posts about Calculus. These posts are additions to the 2 AP Calculus courses that you can visit at Mathematical Education Center. I'll be back soon for great content. You can browse the blog to see previous posts in Calculus, math learning, study skills, etc. If you find the content of this blog useful share it to others. If you want to learn more about Calculus you can subscribe to the courses I mentioned earlier. If you are interested in tutoring in Math, French, ESL and Spanish face-to-face and online visit New Direction Services at www.ndes.biz
I have taken a break since the middle of June in publishing posts about Calculus. These posts are additions to the 2 AP Calculus courses that you can visit at Mathematical Education Center. I'll be back soon for great content. You can browse the blog to see previous posts in Calculus, math learning, study skills, etc. If you find the content of this blog useful share it to others. If you want to learn more about Calculus you can subscribe to the courses I mentioned earlier. If you are interested in tutoring in Math, French, ESL and Spanish face-to-face and online visit New Direction Services at www.ndes.biz
Friday, June 16, 2017
Implicit differentiation
Implicit differentiation involves differentiating implicit functions. An implicit function is an implicit relation between variables. Differentiating an implicit function leads to differentiate the independent variable with respect to the dependent variable. It's basically finding the derivative using the notation dy/dx.
Two methods can be used:
1) You explicit the function
Example 1
Find the derivative of 3xy = 2
Let's explicit the function: y = 2/3x
Let's calculate the derivative: dy/dx = d/dx(2/3x)
= -2(3x)'/(3x)²
= -6./9x²
= -2/3x²
2) If expliciting is not possible, you make transformations in order to find the derivative.
The rules and formulas used to calculate the derivative of different forms of functions apply in the calculations of the derivative of an implicit function.
Since an implicit function is a relationship between the independent and the dependent variable the the application of the derivative rules might seem odd. Let's familiarize ourselves with the derivatives of some expressions where the derivative rules are applied.
Example 2. Let's y be a function of x find the derivative of y³ with respect to x.
Let's u = y³. we have two functions: u and y. U is a function of y and y is a function of x. U is a composite function. The chain rule has to be applied in order to find the derivative. The formula to apply here is du/dx = du/dy.dy/dx
du/dx = d(y³)/dy.dy/dx
= 3y²dy/dx.
Example 3 Find the derivative of u = 2x²y
du/dx = d(2x²y)
Let's apply the constant rule
du/dx = 2d(x²y)
Let's apply the product rule:
du/dx = 2[d/dx(x²)y+x² dy/dx]
= :2(2xy+x²dy/dx)
du/dx = 4xy+2x²dy/dx
Example 4 Find the derivative of 3y³+x²y = x-3
Let's differentiate both sides:
d/dx(3y³+x²y) = d/dx(x-3)
3y²dy/dx+2xy+x²dy/dx = 1
3y²dy/dx+x²dy/dx = 1-2xy
(3y²+x²)dy/dx = 1-2xy
dy/dx = 1-2xy/3y²+x²
Practice. Find the derivatives of the implicit functions:
1) x²+y² = 15
2) 3y²-siny = x²
3) x²+2xy-y = 2
Interested in learning more about derivatives and Calculus visit Mathematical Education Center
Two methods can be used:
1) You explicit the function
Example 1
Find the derivative of 3xy = 2
Let's explicit the function: y = 2/3x
Let's calculate the derivative: dy/dx = d/dx(2/3x)
= -2(3x)'/(3x)²
= -6./9x²
= -2/3x²
2) If expliciting is not possible, you make transformations in order to find the derivative.
The rules and formulas used to calculate the derivative of different forms of functions apply in the calculations of the derivative of an implicit function.
Since an implicit function is a relationship between the independent and the dependent variable the the application of the derivative rules might seem odd. Let's familiarize ourselves with the derivatives of some expressions where the derivative rules are applied.
Example 2. Let's y be a function of x find the derivative of y³ with respect to x.
Let's u = y³. we have two functions: u and y. U is a function of y and y is a function of x. U is a composite function. The chain rule has to be applied in order to find the derivative. The formula to apply here is du/dx = du/dy.dy/dx
du/dx = d(y³)/dy.dy/dx
= 3y²dy/dx.
Example 3 Find the derivative of u = 2x²y
du/dx = d(2x²y)
Let's apply the constant rule
du/dx = 2d(x²y)
Let's apply the product rule:
du/dx = 2[d/dx(x²)y+x² dy/dx]
= :2(2xy+x²dy/dx)
du/dx = 4xy+2x²dy/dx
Example 4 Find the derivative of 3y³+x²y = x-3
Let's differentiate both sides:
d/dx(3y³+x²y) = d/dx(x-3)
3y²dy/dx+2xy+x²dy/dx = 1
3y²dy/dx+x²dy/dx = 1-2xy
(3y²+x²)dy/dx = 1-2xy
dy/dx = 1-2xy/3y²+x²
Practice. Find the derivatives of the implicit functions:
1) x²+y² = 15
2) 3y²-siny = x²
3) x²+2xy-y = 2
Interested in learning more about derivatives and Calculus visit Mathematical Education Center
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