Calculus 1 - Integration & Antiderivatives

Summary

Highlights

• Explore the wild world of integration, where adding and dividing become your calculus besties! 🧮
• Getting those antiderivatives involves remembering to always add the constant ‘C’. Unsolved wisdom! 🧙‍♂️
• Simplifying integrals is like a puzzle. Substitutions and rearrangements are your best moves! 🔄
• From trigonometry to exponentials, knowing integral forms is like having cheat codes for math missions! 🎮
• Discover the difference between definite and indefinite integrals. One finds the number treasure, the other maps an algebraic journey! 🗺️
• Remember the integral of 1/x is ln(x)! It’s the magician’s secret in calculus. 🔮

Key Takeaways

• Integration flips the power rule: add 1 to the exponent and divide! ✏️
• Always add a constant 'C' to indefinite integrals. It's the mystery spice! 🕵️‍♂️
• Use substitution to simplify complex integrals. Sub u, du what you gotta do! 🤔
• Definite integrals give you numbers. Indefinite integrals give you expressions. 🎲
• Know your trig integrals: sine, cosine, and friends. They’re vital! 📚
• Logarithms sneak in with 1/x: ln(x) is your go-to! 📈
• Exponential integrals love linear exponents. Keep it simple! 🌟

Overview

Chapters

• 00:00 - 00:30: Introduction to Integration and Antiderivatives The chapter titled 'Introduction to Integration and Antiderivatives' focuses on the concept of integration and finding antiderivatives of functions. It begins by revisiting the power rule from calculus, which states that the derivative of x raised to the power of n (where n is a constant) is equal to n times x raised to the power of n minus 1. The chapter then transitions to explaining how this rule is applied in the context of antiderivatives and integration, laying the groundwork for understanding these fundamental concepts in calculus.
• 00:30 - 01:30: Power Rule for Integration The chapter introduces the power rule for integration, highlighting the differences from differentiation. Instead of subtracting 1 from the exponent (as in differentiation), you add 1 to it and divide the result. Additionally, a constant of integration is included in the final result. The explanation leads into solving practical examples, like finding the derivative of X using the power rule of integration.
• 01:30 - 03:00: Examples Using Power Rule The chapter titled 'Examples Using Power Rule' explains how to use the power rule to find the derivative and antiderivative of functions. Using the power rule, if you have a function like X^3, the derivative is found by multiplying by the exponent (3) and reducing the exponent by one, resulting in 3x^2. To find the antiderivative of 3x^2, you add one to the exponent and divide by that new exponent.
• 03:00 - 04:30: Understanding Constants of Integration This chapter focuses on understanding the concept of constants of integration in calculus. It explains the process of integration and the importance of adding a constant to indefinite integrals. The transcript provides a basic example, starting with the integration of x squared, which results in x cubed. To complete the integral, a constant of integration is added. The chapter emphasizes that when taking the derivative of the integrated function plus the constant, the correct original function is obtained, exemplified by taking the derivative of x cubed plus a constant, which returns 3x squared plus the derivative of the constant, which is zero.
• 04:30 - 08:00: Finding Antiderivatives of Binomials and Trinomials In this chapter, the focus is on finding antiderivatives of binomials and trinomials. The discussion begins with the basic concept that the derivative of a constant is zero, thus when calculating indefinite integrals, a constant of integration must always be added since the original constant of a function is unknown. The chapter continues with an example on how to find the antiderivative of X to the fourth power using the power rule, which involves increasing the exponent by one and adjusting the coefficient accordingly.
• 08:00 - 15:00: Antiderivatives of Radical Functions The chapter on 'Antiderivatives of Radical Functions' involves discussions on calculating antiderivatives and emphasizes adding the constant 'C' as part of the result. It illustrates the process by showing an example of x raised to a power, specifically, converting the expression into X to the fifth power divided by 5 plus C. It further explains that this result can also be expressed in the form of 1/5 times X to the fifth plus C. The student is encouraged to practice with similar examples.
• 15:00 - 18:30: Integrating Trigonometric Functions The chapter 'Integrating Trigonometric Functions' delves into the process of finding antiderivatives for monomials, specifically focusing on trigonometric expressions. The discussion includes an example where students are encouraged to determine the antiderivative of expressions like X squared and X to the seventh. It is explained that the antiderivative involves increasing the exponent by one and then dividing by the new exponent, adding a constant C as the final term. This explanation is aimed at helping readers understand the step-by-step method to compute antiderivatives.
• 18:30 - 23:30: Indefinite vs. Definite Integrals The chapter titled 'Indefinite vs. Definite Integrals' introduces basic concepts of integration. It starts by explaining how to find the indefinite integral of simple functions with examples. Specifically, it covers the process of finding the antiderivative of x to the first power, which results in x squared divided by 2. It then moves on to more complex integrals, such as the indefinite integral of a fraction. The example of x over 4 with respect to x (indefinite integral) illustrates the technique of moving constants (like 1/4) outside the integral to simplify the integration process.
• 23:30 - 24:30: The Fundamental Theorem of Calculus The chapter focuses on the Fundamental Theorem of Calculus, illustrating its application through examples. It involves solving integrals by applying the power rule, including a step-by-step breakdown. For instance, integrating a function like ax DX through the power rule results in (1/8)x^2 + C and tackles further examples like integrating 8x^3.
• 24:30 - 29:00: Antiderivatives of Exponential Functions The chapter provides a step-by-step explanation on how to find the antiderivative of exponential functions with a focus on polynomials. The transcript illustrates the process of moving constants in front of the integral, adjusting the exponent, dividing by the new exponent, and simplifying the expression. It emphasizes the calculation involved in determining the antiderivative, using the example of x cubed, demonstrating that the antiderivative of x^3 is (1/4)x^4, simplified to 2x^4 when multiplied by a constant. The importance of adding the constant 'C' to the solution is also highlighted, showing the complete format of an antiderivative solution.
• 29:00 - 30:30: Using U-substitution for Antiderivatives The chapter discusses the process of finding antiderivatives when dealing with constants. To find the antiderivative of a constant, one must add a variable to it. The variable is determined by the differential present in the integral. For instance, the antiderivative of 4 with respect to x (dx) is 4x, and for 5 with respect to y (dy), it is 5y plus a constant C.
• 30:30 - 37:00: Finding Antiderivatives of Rational Functions This chapter discusses the technique of finding antiderivatives specifically for rational functions, using the power rule as a core strategy. The power rule is explained as it applies to constants, such as reinterpreting a constant like 4 as 4 multiplied by X to the 0 power. By adding 1 to the exponent (0 in this case) and then dividing by the new exponent, the antiderivative is determined. The explanation provides a foundational understanding necessary for dealing with constants in the context of antiderivatives.
• 37:00 - 38:20: Antiderivative of 1/x The chapter discusses the process of finding the antiderivative of the function 1/x. It includes an example problem involving the integration of a binomial expression 7x - 6 and emphasizes the importance of introducing a constant variable (represented as 'C') in the final answer to account for the indefinite integral nature.

Calculus 1 - Integration & Antiderivatives Transcription

• 00:00 - 00:30 now in this video we're going to go over integration how to find the antiderivative of certain functions and now if you recall according to the power rule the derivative of x raised to the nth power that is a variable raised to a constant is equal to n X raised to the N minus 1 now the power rule when dealing with antiderivatives or integration is like
• 00:30 - 01:00 this instead of subtracting the exponent by 1 you need to add 1 to the exponent and instead of multiplying you're going to divide the result of what you see here and plus you need to add the constant of integration and so that's basically the power rule for integration let's work on some examples so let's say if we wish to find the derivative of X
• 01:00 - 01:30 cube using the power rule we can see that our n value is stream so it's 3 X raised to the 3 minus 1 which is 2 so the derivative of X cube is 3x squared now let's say if we wish to find the antiderivative of 3 x squared well we need to do in this case is add 1 to the exponent and then divide by that new
• 01:30 - 02:00 exponent 2 plus 1 is 3 and 3 divided by 3 is 1 and so this gives us our original answer which is X cube but we need to add the constant of integration if we were to take the derivative of x cubed plus the constant of integration it will give us three x squared plus
• 02:00 - 02:30 zero the derivative of any constant is zero and so we don't know what this constant could be it could be five it could be negative eight it could be zero but when dealing with an indefinite integral you always need to add the constant of integration now let's find the antiderivative of X to the fourth go ahead and work on this example so using the power rule we need to add 1 to 4 and
• 02:30 - 03:00 divide by that result and let's add C so 4 plus 1 is 5 so it's going to be X to the fifth power divided by 5 plus C but you can also write your answer like this you could say it's 1/5 X to the fifth plus C go ahead and try these two examples find the antiderivative of x
• 03:00 - 03:30 squared and X to the seventh so feel free to take a minute and try those two so this one is going to be X at a third over 3 plus say you just add 1 to 2 and then 7 plus 1 is 8 so this is going to be X to the 8 over 8 plus C and so that's a simple way in which you could find the antiderivative of a monomial now what is the antiderivative of X
• 03:30 - 04:00 what's the answer so this is X to the first power so it's going to be x squared divided by 2 and that's it for that one now what about this what if we have a fraction what is the indefinite integral of x over 4 DX in this case what you could do is rewrite it let's move the 4 to the front and so this is the same as 1/4 antiderivative
• 04:00 - 04:30 of ax DX and then use the power rule so that's gonna be one-fourth times x squared over two plus C so four times two is eight the final answer is going to be one over eight x squared plus C now what about this one let's say if this is 8 X to the 3rd power what's the
• 04:30 - 05:00 answer for this one well let's move the 8 to the front and then let's find the antiderivative of X cube so 3 plus 1 is 4 and then let's divide by that result and now we could simplify it so 8 divided by 4 that's going to be 2 and the answer is going to be 2 X to the fourth power plus C
• 05:00 - 05:30 now what about a constant what is the antiderivative of four all you need to do in a situation like this is add a variable to it so since we have a DX we're going to add the variable X let's say if we have the integration of five dy the answer will be 5y plus C now
• 05:30 - 06:00 let's go back to the first example to show you a work using the power rule his we could do 4 is the same as 4 times X to the 0 power because anything raised to the 0 power is 1 so if we add 1 to 0 and then divide by 1 this will give us the same result for X and so that's what you could do when dealing with a constant so the antiderivative of
• 06:00 - 06:30 negative 7 let's say DZ this would be negative 7 times Z plus C so based on what you see here you need to introduce that variable and your answer now let's find the antiderivative of a binomial let's say 7x minus 6 go ahead and try that so what we need to do is integrate
• 06:30 - 07:00 each component of this expression separately so the antiderivative of 7x is going to be 7 X to the 2nd power over 2 and for a constant we simply need to add an X variable to it and then let's add the constant of integration and so this is the answer there's another one find the antiderivative of 6x squared
• 07:00 - 07:30 plus 4x minus 7 go ahead and find the antiderivative of that particular trinomial so let's add 1/2 to 2 plus 1 is 3 / the exponent and then let's add one to this exponent so one plus one is two and then divide by two and the antiderivative of negative seven is
• 07:30 - 08:00 negative 7x and now let's simplify so 6 divided by 3 is 2 and then 4 divided by 2 that's 2 and so this is going to be the answer for this problem so as you can see it's not really too difficult finding the antiderivative of a function as you work out a few problems it gets easier but now let's move on to the
• 08:00 - 08:30 radical functions what is the antiderivative of the square root of x go ahead and try that now the first thing you need to do is you need to rewrite the expression this is X to the first power and the index number is a two if it's not present so this is the same as X to the 1/2 as a rational exponent and now let's use the power rule so let's add one to the exponent
• 08:30 - 09:00 and then divide by that result so 1/2 plus 1 is the same as 1/2 plus 2 divided by 2 because 2 divided by 2 is still 1 and 1 plus 2 is 3 so 1/2 plus 1 is 3 over 2 so now we have X raised to the 3 over 2 / derivative now what I'm going
• 09:00 - 09:30 to do is multiply the top and bottom by 2/3 and so here the twos will cancel and a 3 so will cancel so 1/3 1/2 or 1/3 over 2 is the same as 2 over 3 all you need to do is flip the fraction so the final answer for this problem is 2 over 3x raised to the river 2 plus C now you
• 09:30 - 10:00 can leave your answer like this if you want to or you can convert it back into its radical form and so X to the three halves is the same as the square root of x cube now you really don't need to write the two you can put it there if you want to but it's the same so you can also leave your answer like this if you prefer let's work on another example find the antiderivative of the cube root of x to the fourth go
• 10:00 - 10:30 ahead and try that so step one is that we need to rewrite it this is X raised to the 4 over 3 the index number goes on the bottom and the exponent goes on the numerator of the fraction whenever you want to convert it from its radical form into a rational exponent so now let's
• 10:30 - 11:00 integrate the expression so this is going to be X raised to the 4 over 3 plus 1/4 over 3 plus 1 now we need to get common denominators so I'm going to replace 1 with 3 divided by 3 and so now we can add the numerators of those two fractions so 4 plus 3 is 7 and so what we have right now let's not forget plus C this is going to be X raised to the 7
• 11:00 - 11:30 over 3 divided by 7 over 3 plus C now X divided by 7 over 3 is the same as X times 3 over 7 if you wish to show your work what you could do is multiply like we did before by the reciprocal of 7 over 3 and so these will cancel and
• 11:30 - 12:00 the 7s will cancel and so the final answer is 3 over 7 X raised to the 7 over 3 plus C or we can write it like this 3 over 7 times the cube root of x to the 7 plus C and that's it now for practice let's try another one
• 12:00 - 12:30 let's try the fourth root of x to the seventh power to take a minute and work on that example so just like before we're going to rewrite it so this is X raised to the 7 over 4 and now let's add 1 so this is going to be 7 over 4 plus 1 divided by 7 over 4 plus 1 now 7 over 4 plus 1 we can replace the one with 4
• 12:30 - 13:00 over 4 so 7 plus 4 is 11 so we get X to the 11th over 4 / 11 over 4 plus C and so instead of dividing it by 11 over 4 we can multiply the X variable by the reciprocal which is 4 of 11 so it's 4 over 11 X to the 11th over 4 plus C and then we can rewrite it as 4 of 11 times
• 13:00 - 13:30 the fourth root of x to the 11 plus C and that's all you need to do for that example now let's talk about the integration of certain trigonometric expressions and so if you have a notepad you may want to write down some notes now recall that the derivative of sine X is cosine X so therefore the
• 13:30 - 14:00 antiderivative of cosine X DX has to be sine X plus C now the derivative of cosine X we know it's negative sine X plus C so the antiderivative of negative sine X DX has to be positive cosine X plus C which means that the
• 14:00 - 14:30 antiderivative of positive sine X DX must be negative cosine X if we reverse the sign and there's no point reverse in and the sign for the constant because Z could be 5 or negative 5 it doesn't matter now let's move on to the derivative of tangent X which is secant squared so the antiderivative of secant
• 14:30 - 15:00 squared it's tangent X plus C and so these are formulas that you just have to know if you don't know him make sure you write them down and as you do your homework you're going to become familiar with those expressions now the next one we know that the derivative of cotangent is negative cosecant squared so the antiderivative of negative cosecant squared is cotangent and the
• 15:00 - 15:30 antiderivative of positive cosecant squared is going to be negative cotangent X next up we have the derivative of secant X and so that's going to be secant X tangent X so the antiderivative of secant tangent as you know by now has to be well let's not
• 15:30 - 16:00 forget DX so this has to be secant X plus C and the last one the derivative of cosecant X is negative cosecant x cotangent X so the antiderivative of positive cosecant X cotangent X changing the sign has to be negative cosecant X plus C and so make sure you know those
• 16:00 - 16:30 common trigonometric expressions along with their derivatives and integrals so let's say if we wish to find the antiderivative of 4 sine X minus 5 cosine X plus 3 secant squared X now the antiderivative of sine is negative cosine and the antiderivative of cosine is positive
• 16:30 - 17:00 sine and the antiderivative of secant squared is tangent and so the answer is going to be negative for cosine X minus 5 sine X plus 3 tangent X plus C and that's it so as you can see it's important to know the integrals of common trig expressions so you can do
• 17:00 - 17:30 questions like these now let's talk about the difference between an indefinite integral and a definite integral an indefinite integral it looks like the expression on the left on the right I'm going to draw a definite integral a definite integral has the lower limit
• 17:30 - 18:00 of integration and the upper limit of integration and so it has those values the indefinite integral will give you a answer in terms of X so you get an algebraic expression where as a definite integral it will give you a number your final answer should be a specific number like 25 36 or something so let's go through both examples so to find the
• 18:00 - 18:30 indefinite integral of 6x squared we need to use the power rule and so the answer is going to be 2x cubed plus C so anytime you you're dealing with an indefinite integral you're going to get an expression in terms of some variable plus the constant of integration now when dealing with a definite integral you don't need to worry about the constant of integration it's going
• 18:30 - 19:00 to disappear now we still need to use the power rule on this expression so that's not going to change so we're going to get 2x cubed now we need to evaluate it from 1 to 2 so the first thing you need to do is replace X with 2 and then you need to subtract it by replacing X with 1 2 to the third power
• 19:00 - 19:30 is 8 and 1 to the third power is just 1 so it's going to be 16 minus 2 which is 14 and so that's how you can evaluate a definite integral using the limits of integration now let's say if we incorporated the constancy at this
• 19:30 - 20:00 point so once we integrated this expression and got to X cube what's going to happen if we introduce the constant of integration so when you plug in 2 you're going to get 2 times 2 to the third power plus C and then minus 2 times 1 to the third plus C so notice that C cancels C minus C will disappear
• 20:00 - 20:30 and so your answer will still be 14 and thus when dealing with definite integrals it's not necessary to add the constant of integration you could just ignore it go ahead and try this one let's say if we have 8x minus 3a DX and we wish to evaluate it from let's say 2 2 3 go ahead and find a value of that
• 20:30 - 21:00 definite integral so first let's integrate the expression this is going to be 8x squared over 2 minus 3x from 2 to 3 and then 8 divided by 2 is 4 so we can write that as 4x squared now the first thing we're going to do is plug in
• 21:00 - 21:30 the top number 3 so let's replace X with 3 and now let's replace X with 2 3 squared is 9 9 times 4 is 36 3 times 3 is 9 2 squared is 4 times 4 that's 16 3 times 2 is 6 36 minus 9 is 27 16 minus 6
• 21:30 - 22:00 is 10 and 27 minus 10 is 17 and so this is the answer now I'm going to take a minute to let you know that for those of you who want to support my patreon page I'm going to post the second part of this video so the video that you're currently watching is just the first half of the video but if you want to have access to the second part of the video feel free to check it out on my patreon page and you can also access
• 22:00 - 22:30 other video content that I have there as well that you might find interesting so feel free to take a look at that when you get a chance and let's get back to the lesson now just to give you some background info the process by which we evaluate the definite integral comes from the fundamental theorem of calculus one part of it says the integral from A to B of some function f of X DX where
• 22:30 - 23:00 this function is continuous on a closed interval from A to B is equal to the evaluation of the antiderivative which is capital F at B the minus the evaluation of the antiderivative F at a and so basically that's what we did we've integrated f of X to give us the antiderivative capital F and then we plugged in the top number first and then
• 23:00 - 23:30 the bottom number later and so the way we evaluate the definite integral it comes from the fundamental theorem of calculus now it's important to know that the integral of f of X DX that is the indefinite integral is equal to the antiderivative which is represented by capital F Plus see so it makes you understand this
• 23:30 - 24:00 capital F is the antiderivative of lowercase F and if you were to take the derivative of capital F it will give you a lowercase F so the derivative of capital f of X is equal to lowercase F of X now let's talk about exponential functions the derivative of an exponential function like e to the U is
• 24:00 - 24:30 e to the U times u prime now what about the antiderivative of e to the U now if u is a linear function let's say ax plus B you could use this formula it's e to u divided by u prime if it's a quadratic function a cubic function is not going to work so let me give you some examples let's say if we wish to find the
• 24:30 - 25:00 antiderivative of e 3x it's going to be e 3x divided by the derivative of X which is 1 plus C and so it's e to the X now let's say if we want to find the antiderivative of e to the 5x it's going to be well let's write the formula if you wish to follow it so remember it
• 25:00 - 25:30 only works if you as a linear function so you as 5x that means you prime is 5 so it's easy to you that's e to the 5x divided by 5 plus C and so that's it so let's say if we wish to find the integral of e to the negative 7x and also let's say e to the 3x minus 5 for the first one it's going to be e to the negative 7x divided by the derivative of
• 25:30 - 26:00 negative 7x which is negative 7 plus C and for the next one it's going to be e to the 3x minus 5 / the derivative of 3x minus 5 which is 3 and so that's a quick and simple way in which you can basically find the antiderivative of exponential functions if this is a linear function now let's work on a similar problem but using a
• 26:00 - 26:30 different technique to get the answer so for this one we know the antiderivative of e to the 8x it's e to the 8x over 8 plus C but let's use another method to get that answer and the method that we're going to use is something called u substitution we're going to make the u variable equal to 8x now we're going to find the differential of U which is D U and the derivative of 8x is 8 times the
• 26:30 - 27:00 differential of X or DX so D U is 8 DX next we're going to solve for DX by dividing both sides by 8 so D u over 8 is equal to DX now let's replace a tax with the u variable and let's replace DX with D u divided by 8 now it's important to understand that the antiderivative of
• 27:00 - 27:30 e to the X is e to the X plus C and this is something that you need to commit to memory so make sure you know that expression so right now we have e to you and then DX is d u over 8 and we're gonna move the 8 to the front so it's 1 over 8 integral of e to you bu and the antiderivative of each of you is just easy to you just as the antiderivative of e to the X is e to the X so this is 1
• 27:30 - 28:00 over 8 e 2 u plus C now we need to replace the U variable with 8x so the final answer is 1 over 8 e to the 8x plus C and so that's how you could show you work using the technique known as use substitution now let's try another example what is the aunt derivative of 4x e raised to the x
• 28:00 - 28:30 squared DX now for this we need to use u substitution again but this time we're going to make u equal to x squared D u that's going to be the derivative of x squared which is 2x times DX and now let's get DX by itself let's divide both sides by 2x and so this disappears so DX is d u divided by 2x just like we did
• 28:30 - 29:00 before let's replace x squared with the u variable unless replace DX with D u over 2 X so this is going to be integral 4x e 3u times D u over 2x now we can cancel these two 4x divided by 2x is 2
• 29:00 - 29:30 and let's move the 2 to the front so it's 2 integral e to the UD u which is equal to 2 e to the U plus C now the last thing we need to do is replace U with x squared so the final answer is going to be 2 e to the x squared plus C and so u substitution is a common technique that you need to be familiar with we're dealing with integration now
• 29:30 - 30:00 let's talk about how to find the antiderivative of rational functions so what is the antiderivative of 1 over x squared how can we find the answer here now what we need to do is we need to rewrite the expression and the way we can do this is by moving the X variable to the top and as you do that the sign of the exponent changes from positive 2 to negative 2 and so at this point we
• 30:00 - 30:30 could use the power rule so what we can do is add 1 to the exponent and then divide by that result and let's not forget to add the constant of integration so negative one I mean negative two plus one is negative one and at this point after you integrate it you can move the X variable back to the bottom where the exponent will change from negative one to positive one and so our answer is negative one over X or X to the first
• 30:30 - 31:00 power and so this is the antiderivative of 1 over x squared it's negative 1 over X now let's try another example let's find the antiderivative of 1 over X cubed so go ahead and work on that problem so feel free to take a minute and try step one is to rewrite the problem by moving the X variable to the top step two is to integrate let's add 1
• 31:00 - 31:30 to the exponent and then divide by that result so negative 3 plus 1 is negative 2 so we have X to the negative 2 power divided by negative 2 so after you integrate it and then rewrite it and simplify so let's move the X variable to the bottom and so this becomes we can leave a 1 on top this negative side and
• 31:30 - 32:00 we could transfer it to the top the 2 has to stay on the bottom but now we have an x squared on the bottom as well so the answer is a negative 1 over 2 x squared plus C now what about this one let's integrate 8 over X to the 4 DX so go ahead and try it now because we have a constant on the top I recommend moving it to the front at the same time let's
• 32:00 - 32:30 move the X variable to the top so this is going to be 8 times the integral of x to the negative 4 DX and now let's find an antiderivative so let's use the power rule for integration negative 4 plus 1 is negative 3 now what we need to do is take the X variable and
• 32:30 - 33:00 move it to the bottom and we could take the negative side and move it to the top so we're gonna have negative 8/3 X cubed plus C and so that's it for this example now let's integrate this expression 1 over 4x minus 3 squared DX go ahead and try that now for this type of problem I
• 33:00 - 33:30 recommend you using u substitution let's make u equal to 4x minus 3 so D U is going to be the derivative of 4x minus 3 which is just 4 and then multiply that by DX now let's solve for DX so DX is going to be d u divided by 4 if you divide both sides by 4 here so
• 33:30 - 34:00 we're gonna be placed in this expression with the u variable and we're gonna replace the DX with D U divided by 4 so this is going to be the integral of 1 over u squared times D u over 4 now let's take the 4 and move it to the front and let's take the U variable move it to the top so this becomes 1 over 4 integration U to the negative 2 D u now let's use the power rule so let's add 1 to the
• 34:00 - 34:30 exponent so negative 2 plus 1 that's going to be negative 1 and then we need to divide by negative 2 plus 1 or negative 1 now let's move the u variable to the bottom and let's move the negative sign to the top so this is going to be negative run over 4 times u plus C now the last thing we need to do is replace the u variable with 4x minus 3 so the final answer is going to be a
• 34:30 - 35:00 negative 1 over 4 times four X minus three plus C and so that's what we need to do in this problem here's another similar example find the antiderivative of 7 over 5 X minus 3 to the fourth power go ahead and try that problem so first let's use u-substitution let's make you equal to
• 35:00 - 35:30 5x minus 3 so d-u is going to be the derivative of 5x minus 3 which is 5 times DX now what I like to do is I always like to solve for DX it just it helps me to avoid mistakes and so D you over 5 is DX now let's replace 5x minus 3 with the you variable unless replace DX with D u over 5 so this is going to
• 35:30 - 36:00 be the integral of 7 over U to the 4 times D u over 5 now let's take the 7 and the 5 and move it to the front and let's take the you variable and move it to the top so this is going to be 7 over 5 integral U to the negative 4 D u so now let's use the power rule negative 4
• 36:00 - 36:30 plus 1 is negative 3 and Nitze divided by negative 4 plus 1 now let's move the you variable to the bottom and let's move the negative sign to the top so this is going to be negative 7 on the bottom we have 5 times 3 which is 15 and then you to the 3rd power plus C now the last step is to replace the U variable
• 36:30 - 37:00 with 5x minus 3 let's see if I can fit it here so it's going to be negative 7 over 15 times 5x minus 3 to the third power plus the constant of integration and so this is the final answer that's all we need to do in that example now what is the antiderivative of 1 over X go ahead and
• 37:00 - 37:30 try that problem now if we try to approach this problem like the other problems instinctively we may decide to move the X variable to the top and this will give us X to the negative 1 if we use the power rule at this point it's going to be negative 1 plus 1 divided by negative 1 plus 1 plus C negative 1 plus 1 is 0 X to the 0 is 1 but 1 divided by
• 37:30 - 38:00 0 is undefined and so we're going to have some issues here so there must be something else that we could do you simply need some known that the antiderivative of 1 over X is Ln X because if you recall the derivative of Ln U with respect to X it's going to be u prime over u and so the derivative of
• 38:00 - 38:30 Ln X is going to be if u equals x u prime is the derivative of X which is 1 so it's 1 over X so if the derivative of Ln X is 1 of X the antiderivative of 1 over X is Ln X so that's one of those formulas that you need to commit to memory try this problem find the antiderivative of 7 over X DX now this
• 38:30 - 39:00 is very similar to the previous problem we just have a number on top let's move that constant to the front so this becomes 7 integral 1 over X DX now we know the antiderivative of 1 over X you just need to know that it's Ln X and so this is our answer 7 Ln X plus C now what about this one and one over X plus five what is the antiderivative of that this is going to
• 39:00 - 39:30 be Ln X plus five plus C now sometimes some textbooks they may have an absolute value with the natural log symbol because X plus five can sometimes be negative for certain X values and so you might see an absolute value expression here now we can get the same answer using use substitution so if we were to
• 39:30 - 40:00 make u equal to X plus 5 D U is going to be the derivative of x plus five which is 1 times DX so let's replace X plus 5 with U and D ax with two you and so this gives us Ln U plus C and then we need to replace U with X plus five and so we get Ln X plus five plus C and so you could
• 40:00 - 40:30 show your work used in both techniques