How do you intergrate this function

Deleted member 2597

Deleted member 2597

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idk but meeks has advice for you
 
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keep grinding bro, shoot for your dreams
 
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first step: multiply it out

(root of t) * (64 - 16 (root t) + t) = 64(root of t) - 16t + t^(1.5)

integrate that and you get:

ANSWER:
(128/3)(t^1.5) - (8)t^2 + (2/5)t^2.5

i'm pretty sure
 
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first step: multiply it out

(root of t) * (64 - 16 (root t) + t) = 64(root of t) - 16t + t^(1.5)

integrate that and you get:

ANSWER:
(128/3)(t^1.5) - (8)t^2 + (2/5)t^2.5

i'm pretty sure
Thanks man
 
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Jfl @Syobevoli nigga aernt you physics cel major and you dont know
 
Confirmed site is full of retards they taught me how to do this at 13/14
 
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I REALLY DONT KNOW HOW TO DO THAT SHIT,here is a gif of a cute girl
E898D25D 905B 4563 AB0C 3924EAF3EB9B
 
expand it and reverse power rule
integral(x^ndx) = (x^n+1) / n+1
 
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first step: multiply it out

(root of t) * (64 - 16 (root t) + t) = 64(root of t) - 16t + t^(1.5)

integrate that and you get:

ANSWER:
(128/3)(t^1.5) - (8)t^2 + (2/5)t^2.5

i'm pretty sure
Wrong
Answer is 128/3 (t^3/2)-8t^2 +2/3t^3/2
 
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How do you do this
 

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Wrong
Answer is 128/3 (t^3/2)-8t^2 +2/3t^3/2
no the last part by you is wrong. we agree on the first two parts, but the last part would be (2/5)t^(2.5)
 
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How do you do this
int(2x-5)^-7dx
u = 2x-5
du = 2dx
du/2 = dx

(1/2) * int(u^-7)du = (1/2) * (u^-6)/-6 = 1/(-12(2x-5)^6)
just use symbolab or wolframalpha jfl stop asking us
 
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