@Ghaiyst Hey! For 1/(x^4+x^2), you factor out x^2 and you get 1/(x^2(x^2+1)). Since you're dealing with quadratic factors, the partial fraction decomp will be A/x + (Bx+C)/(x^2)+(Dx+E)/(x^2+1). When you multiply out, you'll find that A=0, B=0, C=1, D=0 and E=-1, which means your integral is now 1/(x^2)-1/(x^2+1). Since the integral of 1/(x^2+1) is tan^(-1)x, you get -1/x-tan^(-1)x, which you can simplify to -(xtan^(-1)x+1)/x. Hope that helps!! :D
I can't even begin to tell you how much this helped me. your videos are clear and concise and I like that you don't waste time explaining the basics. fantastic job, keep up the good work!
You could do it either way. If you leave it as xdx, then you'd substitute du/2 for xdx. Otherwise, if you solve it the way I did, you just make the substitution for dx, and then cancel out the x's from inside your integral. :)
try solving your equation for u, for x. you should get x=(16-u)/3. then if you plug that in for x, and plug u in for 16-3x, and plug du/(-3) in for dx, you should end up with just u's, and be able to simplify enough that you can integrate. sometimes if your u-substitution doesn't work, consider solving your equation for u, for x. sometimes this can be the trick that unlocks the whole thing. :)
you have to change the limits of integration if you don't plan to back-substitute. but if you plan to plug back in for x at the end before you evaluate, then you don't have to both changing limits of integration. :)
@barbarossaaaa Yes! Or.... you can always leave the limits of integration in terms of x, complete the trig. sub problem, then back-substitute x's for theta's before evaluating at 1 and 4. Both ways will work. :)
Really nice refresher video for those who studied this long time ago. I searched the web and every single tut is about explaining every single steps, where I need just one single explanation "How was the limit of integration?" Nice video!
yep! that's totally okay. just evaluate the definite integral like you would any other, plugging in the top number first (4), and then subtracting whatever you get when you plug in the lower number (16). :)
you make me feel smart by making me understand Cal nice and smoothly. I love all of your math resources, I admire your work. Keep up with the good work :)
You never add C for definite integrals, only for indefinite integrals. Remember that definite integrals are the ones where you're integrating on an interval, in this case, from 0 to 4. Hope that helps! :D
hey! I'm a junior in high school and taking calc and I have to thank you because your videos have helped me a lot and my grades look so beautiful compared to the first unit we had in calc and I had no study skills *cries tears of joy* you are the queen of calc
does this same process in which you change the limits of integration in terms of u work for trig substitution as well? lets say the trig formula called for using x = asinΘ and the upper and lower limits are 1 and 4 on the original problem. do i just plug in x= 4 and x=1 into x = asinΘ and solve for x and those are my new limits of integration?
thank you very much , i was confused why should we change the intervals of the integration .. but since we don't have to .. i guess im doing well :) .. appreciate your effort best of wishes
Also by interval is from (0,4) so when I plug in 0 and 4 into 16-3x, it is (16,4). Is that possible since the higher number will be on the bottom of the line??? I don't know if that made sense .
No, because this is a definite integral. We don't put "+C" on definite integrals, because the "+C's" are just going to cancel each other out, and add to zero (0). If this was an indefinite integral, however, then we do need the "+C."
A wonderful video . Great and thorough explanation. U really provided a great insight into these integrals miss... Im gonna watch more and more of your videos now in order to make myself more good in Calculus.
You are performing invalid mathematics when you divide by 2x, substitute into the integrand, and cancel the x factors. Why? Because x can be zero, and you cannot divide by zero. The proper method of substitution would divide both side of du=2xdx by 2 and substitute 1/2 du for xdx in the integrand.
Heyy :D, i am understanding U-substitution but the questions that come up in my paper tend to be a harder version of u substuition. For example, the x's dont cancel out....am i meant to rearrange u to get x and then sub it in? Also, your videos are life-saving! Thank you!
Yes, oftentimes if the u's don't cancel out, you need to solve the equation for u for x, and then plug in for x. If that doesn't work, you may have to use integration by parts or partial fractions, or trig sub, in addition to u-substitution. Or, it's possible that you didn't choose the right value for u, so you might try a different one. Hope that helps!! :D
Hello Ma'am, can you please help me with this question(please take this question into consideration as i do not have any non virtual teacher to ask my doubts as I am learning these things on my own) Evaluate ∫x√(a^2 - x^2)dx the lower bound is 0 and the upper bound is "a". Please do help me. I will be grateful.
Hi Samarth, You'll want to use trig substitution to solve the integral. Specifically, you need to use a sin substitution. You can follow the steps in this video: ruclips.net/video/Hkf0WPjmtQ8/видео.html :)
Slightly flawed, if the integral is with respect to u, then the limits will be. So the integral on the left should have x=0 and x=4 as the limits, not 0 and 4.
This stuff can definitely be very tricky. Please show me how to intergrate 1/(x^4 + x^2) and 1/(x^2 +1) using partial fractions. I can't really seem to do it, because you cant factor x^2 + 1, but I did first substitute and did some stuff and almost came to an answer, but it was all over the place by then and I just got mind twisted! ugh! lol
@Ghaiyst Hey! For 1/(x^4+x^2), you factor out x^2 and you get 1/(x^2(x^2+1)). Since you're dealing with quadratic factors, the partial fraction decomp will be A/x + (Bx+C)/(x^2)+(Dx+E)/(x^2+1). When you multiply out, you'll find that A=0, B=0, C=1, D=0 and E=-1, which means your integral is now 1/(x^2)-1/(x^2+1). Since the integral of 1/(x^2+1) is tan^(-1)x, you get -1/x-tan^(-1)x, which you can simplify to -(xtan^(-1)x+1)/x. Hope that helps!! :D
I can't even begin to tell you how much this helped me. your videos are clear and concise and I like that you don't waste time explaining the basics. fantastic job, keep up the good work!
You could do it either way. If you leave it as xdx, then you'd substitute du/2 for xdx. Otherwise, if you solve it the way I did, you just make the substitution for dx, and then cancel out the x's from inside your integral. :)
try solving your equation for u, for x. you should get x=(16-u)/3. then if you plug that in for x, and plug u in for 16-3x, and plug du/(-3) in for dx, you should end up with just u's, and be able to simplify enough that you can integrate. sometimes if your u-substitution doesn't work, consider solving your equation for u, for x. sometimes this can be the trick that unlocks the whole thing. :)
you have to change the limits of integration if you don't plan to back-substitute. but if you plan to plug back in for x at the end before you evaluate, then you don't have to both changing limits of integration. :)
@barbarossaaaa Yes! Or.... you can always leave the limits of integration in terms of x, complete the trig. sub problem, then back-substitute x's for theta's before evaluating at 1 and 4. Both ways will work. :)
Really nice refresher video for those who studied this long time ago. I searched the web and every single tut is about explaining every single steps, where I need just one single explanation "How was the limit of integration?"
Nice video!
yep! that's totally okay. just evaluate the definite integral like you would any other, plugging in the top number first (4), and then subtracting whatever you get when you plug in the lower number (16). :)
@Gracesaves130 You're welcome!! Glad it helps. :)
you make me feel smart by making me understand Cal nice and smoothly. I love all of your math resources, I admire your work. Keep up with the good work :)
Thanks! :D
@AWohlfort I'm so glad!! Thanks for letting me know. :)
I'm so glad you found it helpful. :)
You never add C for definite integrals, only for indefinite integrals. Remember that definite integrals are the ones where you're integrating on an interval, in this case, from 0 to 4. Hope that helps! :D
hey! I'm a junior in high school and taking calc and I have to thank you because your videos have helped me a lot and my grades look so beautiful compared to the first unit we had in calc and I had no study skills *cries tears of joy* you are the queen of calc
i'm so glad the videos have helped, and that your grades have gotten so much better! that's awesome, keep up the good work!
does this same process in which you change the limits of integration in terms of u work for trig substitution as well? lets say the trig formula called for using x = asinΘ and the upper and lower limits are 1 and 4 on the original problem. do i just plug in x= 4 and x=1 into x = asinΘ and solve for x and those are my new limits of integration?
Glad you liked it! :)
best u-substitution playlist on youtube!
Great format!
Thorough explanations!
Easy to follow!
Fine work. Thank you so much.
+hettygreene You're welcome, I'm so glad you like the videos!
thank you very much , i was confused why should we change the intervals of the integration .. but since we don't have to .. i guess im doing well :) .. appreciate your effort
best of wishes
Awesome explanation! It was very helpful with understanding my lecture from today's class. Thank you! :D
I wasn't sure when I should change the limits and when not to, so thank you for explaining! All of your videos are amazing!!
Aw thanks! I'm glad it could help! :)
I have been struggling soooo much with substitution for definite integrals until now... Thank you SO much!
you're welcome!! i'm so glad it clicked! :D
You are so excellent! I am so happy I found your channel!
I'm honored! Thank you so much! :D
you're welcome! glad it helped! :)
You're welcome!
Also by interval is from (0,4) so when I plug in 0 and 4 into 16-3x, it is (16,4). Is that possible since the higher number will be on the bottom of the line??? I don't know if that made sense .
Glad I can help! :)
x(sqrt(16-3x)) I am confused about getting dx value for this one. I don't know how it can get canceled out with the x. The dx value I get is du/-3.
Awesome! Thanks for letting me know. :)
i was wondering why did u take dx = du/2x. shouldn't it be xdx = du/2. just want to know
try du/dx = 2x, du = 2xdx, du/2 = xdx du/dx is not a fraction, but can be treated as one.
you are my professor. thanks a lot
When integrating and finding the anti-derivative, aren't you supposed to add a constant "C"? Shouldn't it be 1/2(2/3u^3/2) + C?
No, because this is a definite integral. We don't put "+C" on definite integrals, because the "+C's" are just going to cancel each other out, and add to zero (0). If this was an indefinite integral, however, then we do need the "+C."
This is insightful. Thank you!
You're welcome, glad you enjoyed it! :)
I was going to comment the exact same thing !! thank you so much!!!
you are just amazing !! i have been fallow your channel since i started university and it help me a lot thank you so much :) :)
You're welcome, I'm so glad they've helped!!
A wonderful video . Great and thorough explanation. U really provided a great insight into these integrals miss... Im gonna watch more and more of your videos now in order to make myself more good in Calculus.
+Hamza Khan Thank you so much, I hope you enjoy the rest of the videos!
This video is great! Thank you so much for the help!
+Yoma Ogbevire You're welcome, I'm so glad you liked it!
@lighthaq knowledge is certainly best shared :)
thanks for your videos,they've been very useful for me!.
Best regards from Mexico.
You're welcome, I'm so glad they've helped!!
Thanks for sharing the knowledge =)
Thank you so much for this video!!!
+Hibs A You're welcome!
Very nice explanation
Thank you! :)
Awww, thank you so much!! :D
This is powerful ✊
No words to say BUT Thank You so much sweet heart 👍👍👍👍👍👍👍👍👍👍👍👍
BIG thumbs up! thank you for great explanation of the substitution rule sensei! you're like a calculus ninja.
+mike genova Thank you so much, I'm glad you liked it!
This rocks! Thank you!
Art Owner You're welcome, glad you liked it!
very^1000 helpful!! thank you!
Wow very helpful!!
You are performing invalid mathematics when you divide by 2x, substitute into the integrand, and cancel the x factors. Why? Because x can be zero, and you cannot divide by zero. The proper method of substitution would divide both side of du=2xdx by 2 and substitute 1/2 du for xdx in the integrand.
thank you!!!
Yo, thanks. I appreciate the help.
Heyy :D, i am understanding U-substitution but the questions that come up in my paper tend to be a harder version of u substuition. For example, the x's dont cancel out....am i meant to rearrange u to get x and then sub it in? Also, your videos are life-saving! Thank you!
Yes, oftentimes if the u's don't cancel out, you need to solve the equation for u for x, and then plug in for x. If that doesn't work, you may have to use integration by parts or partial fractions, or trig sub, in addition to u-substitution. Or, it's possible that you didn't choose the right value for u, so you might try a different one. Hope that helps!! :D
@justxforxcommentsx Thank you so much, I'm so glad it helped!! :D
THANKS A LOT . You are really great. you help me a lot with my study. i love you
you're welcome!! :D
Hello Ma'am, can you please help me with this question(please take this question into consideration as i do not have any non virtual teacher to ask my doubts as I am learning these things on my own)
Evaluate ∫x√(a^2 - x^2)dx the lower bound is 0 and the upper bound is "a".
Please do help me. I will be grateful.
Hi Samarth, You'll want to use trig substitution to solve the integral. Specifically, you need to use a sin substitution. You can follow the steps in this video: ruclips.net/video/Hkf0WPjmtQ8/видео.html :)
Cannot thank you enough for your help. I try to spread a word about your wonderful channel whenever I can. Thank you.
thanx. got it right
Awesome! :D
Slightly flawed, if the integral is with respect to u, then the limits will be.
So the integral on the left should have x=0 and x=4 as the limits, not 0 and 4.
Thank you Krista! I would have gotten you a cup coffee for saving life. :)
I'll still take the cup of coffee :)
coffee cup isn't enough 😂
:D
This stuff can definitely be very tricky. Please show me how to intergrate 1/(x^4 + x^2) and 1/(x^2 +1) using partial fractions. I can't really seem to do it, because you cant factor x^2 + 1, but I did first substitute and did some stuff and almost came to an answer, but it was all over the place by then and I just got mind twisted! ugh! lol
@TheIntegralCALC thanks. I think I overthought this one.