Ok once we found the anti derivative we can plug in 1 for upper boundary minus zero plugged in for lower boundary. f(b)-f(a):{ 1^(3)*e^1}-{3*1^2*e^1}+{6*1e^1}-{6*e^1}+6=-2e+6 result
you technically here apply the formula of IBP. and steps are: 1- decide which function is u ( which you need to make du by differentiating u) and 2- pick up dv ( which you will make up v by integrating dv) and plug into formula.
Step 1: apply change of variable. Let -5x=t After that you will integrate and obtain : -1/125 * int ( e^t * t^2) dt Step2: Now apply integration by parts formula Let u= t^2 ( then du=2t dt) and dv= e^t ( then v is anti derivative of dv which is e^t ) Step3: now plug in everything into: uv- int( v du) formula integrate and simply.
Without u sub directly apply integration by parts formula. Let u=x^2 then du: 2x dx and let dv= e^-5x then v is the anti derivative : -1/5 * e^-5x the best use tabular method. You can check integration by part video I made that also shows practical method. If you check other IBP videos I made you will be fluent in these kind of integrals.
Apply substitution first let t=-x Then dt=-dx Rewrite the integral in terms of t Int(e^t*t^3 dt Now start integration by parts :(uv-int ( v du) Let u=t^3 and dv=e^t Then du: 3t^2 and v=e^t
Gracias por el bonito comentario. ¿Necesitas algún otro ejemplo paso a paso? se trata solo de la fórmula y el enchufe en sus valores u y dv primero, luego derivan u para obtener du e integran v para obtener v y enchufar en la fórmula: uv-integral of v du
muchas gracias, you had helped me a lot with my homework!
The tabular method help me a lot to understand how many times I should integrate
Thanks!
Cool.
Even faster and neater.
I didn't know this method until now. It's so useful, thank you so much!!!!
Tabular method omg it's wonderful
I need to know how to do this: -5x^3e^x dx
Anna please integral la limit 0 to 1 apply panna answer Enna varum . answer mattum chat pannuga please please Anna .
Ok once we found the anti derivative we can plug in 1 for upper boundary minus zero plugged in for lower boundary. f(b)-f(a):{ 1^(3)*e^1}-{3*1^2*e^1}+{6*1e^1}-{6*e^1}+6=-2e+6 result
Thank you so much anna 🥰
i can't really tell what's an integral and what might be an open parentheses, seeing as I'm comparing the work I did to your process
you technically here apply the formula of IBP. and steps are: 1- decide which function is u ( which you need to make du by differentiating u) and 2- pick up dv ( which you will make up v by integrating dv) and plug into formula.
try to use big bracket that wraps outside carefully and inside seperately.
Tabular method needs to be nerfed
Thank you so much par sayar
thanks bro,,,,
Can you help me how to integrate e^-5x x^2 dx where u= -5x
Step 1: apply change of variable. Let -5x=t
After that you will integrate and obtain : -1/125 * int ( e^t * t^2) dt
Step2: Now apply integration by parts formula
Let u= t^2 ( then du=2t dt) and dv= e^t ( then v is anti derivative of dv which is e^t )
Step3: now plug in everything into:
uv- int( v du) formula integrate and simply.
Without u sub directly apply integration by parts formula.
Let u=x^2 then du: 2x dx and let dv= e^-5x then v is the anti derivative : -1/5 * e^-5x the best use tabular method. You can check integration by part video I made that also shows practical method. If you check other IBP videos I made you will be fluent in these kind of integrals.
What if power of x is negative
Apply substitution first let t=-x
Then dt=-dx
Rewrite the integral in terms of t
Int(e^t*t^3 dt
Now start integration by parts :(uv-int ( v du)
Let u=t^3 and dv=e^t
Then du: 3t^2 and v=e^t
@@mathbyleo what we have to do if there is x^-3 in place of x^3
X^2E^3dx=? What values
It is (e^3*x^3)/3 + C
e^3 is a constant so you technically integrating x^2 dx which is x^3/3
thanks 🌹🌹🌹
Thanks Zainab. :)
Thanks
You are welcome Hijfur.
No entendí nada pero me ayudo ver como lo hiciste, genial!
Gracias por el bonito comentario. ¿Necesitas algún otro ejemplo paso a paso? se trata solo de la fórmula y el enchufe en sus valores u y dv primero, luego derivan u para obtener du e integran v para obtener v y enchufar en la fórmula: uv-integral of v du
Muchas gracias! Me fue excelente en mi escuela gracias este video, pase al frente a explicarlo jaja
@@fernandtt1282 De nada. Me alegro de que te beneficies.
Which language you talk to each other can you tell me
It is Spanish but I use translate. :)
W vid
Thanks bro
You are very welcome.
Fast reply Anna please