1: Find Inside Function 2: Set U substitution and find DU. Make sure it equals to the original. If it does not make sure you compensate by dividing or multiplying. 3: Transform Integral 4: Compute
Making a shapeless potato into cube shape with out cutting it by mathematics. Maths is ultimate tool for solving every day problems, it is really magic.
I just wanted to say that you are amazing and I'm really glad that you do what you do :) I might not need this right now, but I know exactly where to look if and when I eventually learn this :D
Hello, I'm new to the channel so I'm not sure if a video on this topic has been covered yet. It seems like ultimately the goal for math teachers is to over simplify the subject, and more or less structure the axioms like easy to remember quotes like PEMDAS or FOIL. I know those are simple examples but I can't count the number of times a lecturer or a tutor completely confused me by using an odd combination of work methods to make sense of the problem and get the answer. Obviously there's an element that's obvious to math teachers for why they use certain work methods. So the question is, for learning math at any level, do these combinations of methods for solving equations just need to be brute-force memorized, or are their simpler logic points to latch on to which make the methods which are needed more obvious?
My advice is try to understand why the simple (shortcut) techniques work and if there's a better, faster way to do it for YOU, by listening to multiple different people explaining the idea of concern, because everyone is different. That's why maths is beautiful in the first place and why it's the language of the universe: everything has to, and does make sense, so make sure if you use a technique, you understand why it works; that's why mathematicians don't take conjectures to be true unless there exists a proof and as a result, conjectures become theorems. For example, FOIL: if you have (a+b)(c+d), and set c+d = e for a second, we see (a+b)(c+d) = (a+b)e = ae + be = a(c+d) + b(c+d) = ac + ad + bc + bd. So, we can see that to figure out the product, it takes multiple steps which wouldn't be convenient if it's an intermidiate step in a longer, more complicated problem. To counteract this, teachers would teach the shortcut directly (since it, generally, takes time to explain why and might be too challenging for some students or require an advanced course) by telling you to multiply the First, Outer, Inner, and Last terms of the two sums together individually and take the sum of all the products. To summarize this, they use FOIL, although I would say distributing a and b individually to c+d is a better and faster way that also works for sums with more than 2 terms each and that have a different number of terms like (a+b+c)(d+e+f+g) = ad + ae + af + ag + bd + be + bf + bg + cd + ce + cf + cg. (I challenge you to prove this only using substitution and a(b+c) = ab + ac. The answer is at the end of the reply.) This can be justified by (a+b+c)(d+e+f+g) = a(d+e+f+g) + b(d+e+f+g) + c(d+e+f+g) which can be easily seen by setting d+e+f+g = h, for example. Oh, and the distributive property can be seen by deriving its reverse (factoring): ab + ac = {a + a + ... + a} + {a + a + ... + a} (b a's) (c a's) = {a + a + ... + a} (b+c a's) = a(b+c), or we can say: if I have b apples and c apples, I have b+c apples. Now you can work backwards to distribute. There are math websites where you can ask questions about any mathematical idea there is, like mathstackexchange; if you can't understand something, post a question and people will answer if it's a good one. (a+b+c)(d+e+f+g) = (a+b+c)h (h=d+e+f+g) = (a+i)h (i=b+c) = ah + ih = ah + (b+c)h = ah + bh + ch = a(d+e+f+g) + b(d+e+f+g) + c(d+e+f+g) = a(j+k) + b(j+k) + c(j+k) (j=d+e, k=f+g) = aj + ak + bj + bk + cj + ck = a(d+e) + a(f+g) + b(d+e) + b(f+g) + c(d+e) + c(f+g) = ad + ae + af + ag + bd + be + bf + bg + cd + ce + cf + cg
Integration by substitution is literally the chain rule worked in reverse! However, this is easier said than actually do because it requires lots of practice, i.e., working lots of problems. This is because the integrand will usually be in reduced form or sometimes algebraically re-arranged in a way that obscures the function F(x) you’re trying to find. Although there are some mnemonics that can sometimes help you take the first step, relying on them is by no means guarantee for success nor is it a valid substitute for doing lots of integrals- through lots of dedicated practice in finding anti-derivatives, patterns will emerge over time so that you can develop your own intuition for how to attack an integral rather than relying on a ‘recipe’ for finding anti-derivatives. If you rely on a mnemonic, you will only be able to solve the easy problems but will struggle with any problem that you likely encounter on most calculus 2 exams. Bottom line, if you decide to rely on ‘shortcut’ of a mnemonic, and don’t apply yourself to finding lots of anti-derivatives, you should expect to earn only a passing grade (C) in Calc 2, and possibly a B. If you want to crush Calc 2, put in the effort to build your anti-differentiation skills. Not only will you be rewarded with a better Calc 2 grade, you’ll also be well prepared for the Calculus ‘feature attraction’ called Differential Equations where the theme of the entire course is dedicated to finding anti-derivatives!!
Oh this is right on time, I would be glad if you could help. I look for a resolution to this weird question bugging my mind: According to Leibnitz notation, the first derivative of y = f(x) is y' = dy/dx. And the second derivative is y ' ' = d^2 y / dx^2 (d-squared-y over d-x-squared)... Now I developed a confusion: what does that really mean ? is it (d^2 y) / (dx)^2 , OR , (d^2 y) / (d (x^2)) ?
We can see that the second derivative is: d/dx (dy/dx) = (ddy)/(dxdx) = d^2y/(dx)^2 (if we treat the d/dx as a fraction of variables, not an operator) This also makes more sense than d^2y/d(x^2) because d/d(x^2) means we are taking the derivative with respect to x^2. Although the notation should be d/dx dy/dx, it's cut to just d^2y/(dx)^2 and then to d^2y/dx^2. Neither necessarily "make sense," but d^2y/(dx)^2 is not ambiguous as in neither of the y or x is raised to the second power. However the notation is similar to that in metric areas and volumes like cm^3 which should really be (cm)^3.
@@stardust-r8z Well thank you! Now it makes sense to treat dx^2 as (dx)^2 = (dx) * (dx)... ( and the numerator as d*d y), hence the following has usual meaning : (d^2 y/ dx^2) * dx = d^2 y / dx = d* ( dy / dx) = d( y' ) = y' ' * dx ...
lol. I learned cal 1 with legendary Krista King and then tried out cal 2 with you, unfortunately there were several critical pieces missing such as u sub and differential equations in switching the route, I had to find the pieces in the jungles myself.
One of the super and stunning lectures ever i saw sir,keep on uploading videos so that we could understand any chapter with ease❤❤.
1: Find Inside Function
2: Set U substitution
and find DU. Make sure it equals to the original. If it does not make sure you compensate by dividing or multiplying.
3: Transform Integral
4: Compute
Making a shapeless potato into cube shape with out cutting it by mathematics. Maths is ultimate tool for solving every day problems, it is really magic.
I just wanted to say that you are amazing and I'm really glad that you do what you do :)
I might not need this right now, but I know exactly where to look if and when I eventually learn this :D
One of the coolest methods ever!!
Thank you for this great video. I like how you don’t skip any steps.
Hello, I'm new to the channel so I'm not sure if a video on this topic has been covered yet. It seems like ultimately the goal for math teachers is to over simplify the subject, and more or less structure the axioms like easy to remember quotes like PEMDAS or FOIL. I know those are simple examples but I can't count the number of times a lecturer or a tutor completely confused me by using an odd combination of work methods to make sense of the problem and get the answer. Obviously there's an element that's obvious to math teachers for why they use certain work methods. So the question is, for learning math at any level, do these combinations of methods for solving equations just need to be brute-force memorized, or are their simpler logic points to latch on to which make the methods which are needed more obvious?
My advice is try to understand why the simple (shortcut) techniques work and if there's a better, faster way to do it for YOU, by listening to multiple different people explaining the idea of concern, because everyone is different.
That's why maths is beautiful in the first place and why it's the language of the universe: everything has to, and does make sense, so make sure if you use a technique, you understand why it works; that's why mathematicians don't take conjectures to be true unless there exists a proof and as a result, conjectures become theorems.
For example, FOIL:
if you have (a+b)(c+d), and set c+d = e for a second, we see
(a+b)(c+d)
= (a+b)e
= ae + be
= a(c+d) + b(c+d)
= ac + ad + bc + bd.
So, we can see that to figure out the product, it takes multiple steps which wouldn't be convenient if it's an intermidiate step in a longer, more complicated problem.
To counteract this, teachers would teach the shortcut directly (since it, generally, takes time to explain why and might be too challenging for some students or require an advanced course) by telling you to multiply the First, Outer, Inner, and Last terms of the two sums together individually and take the sum of all the products. To summarize this, they use FOIL, although I would say distributing a and b individually to c+d is a better and faster way that also works for sums with more than 2 terms each and that have a different number of terms like (a+b+c)(d+e+f+g) = ad + ae + af + ag + bd + be + bf + bg + cd + ce + cf + cg. (I challenge you to prove this only using substitution and a(b+c) = ab + ac. The answer is at the end of the reply.)
This can be justified by (a+b+c)(d+e+f+g) = a(d+e+f+g) + b(d+e+f+g) + c(d+e+f+g) which can be easily seen by setting d+e+f+g = h, for example.
Oh, and the distributive property can be seen by deriving its reverse (factoring):
ab + ac
= {a + a + ... + a} + {a + a + ... + a}
(b a's) (c a's)
= {a + a + ... + a}
(b+c a's)
= a(b+c), or we can say: if I have b apples and c apples, I have b+c apples. Now you can work backwards to distribute.
There are math websites where you can ask questions about any mathematical idea there is, like mathstackexchange; if you can't understand something, post a question and people will answer if it's a good one.
(a+b+c)(d+e+f+g)
= (a+b+c)h (h=d+e+f+g)
= (a+i)h (i=b+c)
= ah + ih
= ah + (b+c)h
= ah + bh + ch
= a(d+e+f+g) + b(d+e+f+g) + c(d+e+f+g)
= a(j+k) + b(j+k) + c(j+k)
(j=d+e, k=f+g)
= aj + ak + bj + bk + cj + ck
= a(d+e) + a(f+g) + b(d+e) + b(f+g) + c(d+e) + c(f+g)
= ad + ae + af + ag + bd + be + bf + bg + cd + ce + cf + cg
Integration by substitution is literally the chain rule worked in reverse! However, this is easier said than actually do because it requires lots of practice, i.e., working lots of problems. This is because the integrand will usually be in reduced form or sometimes algebraically re-arranged in a way that obscures the function F(x) you’re trying to find. Although there are some mnemonics that can sometimes help you take the first step, relying on them is by no means guarantee for success nor is it a valid substitute for doing lots of integrals- through lots of dedicated practice in finding anti-derivatives, patterns will emerge over time so that you can develop your own intuition for how to attack an integral rather than relying on a ‘recipe’ for finding anti-derivatives. If you rely on a mnemonic, you will only be able to solve the easy problems but will struggle with any problem that you likely encounter on most calculus 2 exams.
Bottom line, if you decide to rely on ‘shortcut’ of a mnemonic, and don’t apply yourself to finding lots of anti-derivatives, you should expect to earn only a passing grade (C) in Calc 2, and possibly a B. If you want to crush Calc 2, put in the effort to build your anti-differentiation skills. Not only will you be rewarded with a better Calc 2 grade, you’ll also be well prepared for the Calculus ‘feature attraction’ called Differential Equations where the theme of the entire course is dedicated to finding anti-derivatives!!
Can you PLEASE make a video if possible on the book "DICTIONARY OF MATHEMATICS TERMS" by Douglass Downing Ph.D
Oh this is right on time, I would be glad if you could help. I look for a resolution to this weird question bugging my mind: According to Leibnitz notation, the first derivative of y = f(x) is y' = dy/dx. And the second derivative is y ' ' = d^2 y / dx^2 (d-squared-y over d-x-squared)... Now I developed a confusion: what does that really mean ? is it (d^2 y) / (dx)^2 , OR , (d^2 y) / (d (x^2)) ?
We can see that the second derivative is:
d/dx (dy/dx)
= (ddy)/(dxdx)
= d^2y/(dx)^2
(if we treat the d/dx as a fraction of variables, not an operator)
This also makes more sense than d^2y/d(x^2) because d/d(x^2) means we are taking the derivative with respect to x^2.
Although the notation should be d/dx dy/dx, it's cut to just d^2y/(dx)^2 and then to d^2y/dx^2. Neither necessarily "make sense," but d^2y/(dx)^2 is not ambiguous as in neither of the y or x is raised to the second power. However the notation is similar to that in metric areas and volumes like cm^3 which should really be (cm)^3.
@@stardust-r8z Well thank you! Now it makes sense to treat dx^2 as (dx)^2 = (dx) * (dx)... ( and the numerator as d*d y), hence the following has usual meaning : (d^2 y/ dx^2) * dx = d^2 y / dx = d* ( dy / dx) = d( y' ) = y' ' * dx ...
@@4thesakeofitname No worries! Yeah, I can imagine this notation helps a lot in differential equations.
Sir, plz make a video on cylinderical coordinate and spherical coordinate system ...
Calculus 3: Lecture 11.7 Cylindrical and Spherical Coordinates
th-cam.com/video/QbY7DYwmu3U/w-d-xo.html
Thanks a lot
lol. I learned cal 1 with legendary Krista King and then tried out cal 2 with you, unfortunately there were several critical pieces missing such as u sub and differential equations in switching the route, I had to find the pieces in the jungles myself.
Woah, it says you have 15k likes with 3k views on my end. Insane
👍
❤️
I am from Bangladesh