Question

In calculus when taking the integral of a constant while using u-sub does it become x or u?

Answer 1

well it's constant * u, but you have to substitute back in whatever you set to u in terms of x

Answer 2

so the integral of 1+u would be u+u^2/2?

Answer 3

yeah

Answer 4

yup !

\(\large \int c ~dx = cx + C\)

\(\large \int c ~du = cu + C\)

\(\large \int c ~dt = ct + C\)

\(\large \int c ~dw = cw + C\)

Answer 5

dont forget the integration constant^

Answer 6

thank you the actual problem is a definite integral so i dont have to worry about that

Answer 7

ohk... cool :)

Answer 8

If any of you could help me double check my answer of the integral from 0 to pi/2 of cos^5 c dx it would be helpful.

Answer 9

What do you get for \[\int_0^{\frac{\pi}{2}}\cos^5 x\,dx\]

Answer 10

\[\int\limits_{0}^{\pi/2} (1-\sin^2 x)^2 \cos x dx\]

Answer 11

then i used u-sub for sinx

Answer 12

Wouldn't it be easier to just start out with \(u=\cos x\)?

Answer 13

If it is i wouldn't know how to do it

Answer 14

oh, maybe not...

Answer 15

So after i used u-sub and distributed i got \[\int\limits_{0}^{\pi/2} 1-2u^2+u^4 \]

Answer 16

Now, if you do change of variable, you have to also change the limits of integration, unless you're going to put the variable back before evaluating the definite integral

Answer 17

yeah put it back into the equation before evaluating

Answer 18

but strictly speaking, you shouldn't be writing it with the old limits...

Okay, and if you integrate \(1-2u^2 + u^4\)?

Answer 19

oh ok yeah so i would have to multiply the old limits by cosx correct?

Answer 20

well, you apply the u substitution equation to the limits to find the new limits

Answer 21

\[\int\limits_{0}^{1}\]

Answer 22

if you did \(u=\sin x\), then your new limits would be \(u_1 = \sin 0\) and \(u_2 = \sin \pi/2\)

yes, \[\int_0^1 1-2u^2+u^4\,du\]

Answer 23

you're getting close :-)

Answer 24

and what i got for the integral was \[u-2u^3/3+u^5/5 \]
and then i substituted sinx for u

Answer 25

okay, but you could just use 1 and 0 and stay with u now...

in any case, what is your final answer? I have to go in a minute or two...

Answer 26

oh really? \[\sin(1)-2/3\sin^3(1)+1/5\sin^5(1)\]

Answer 27

No, \[u-\frac{2}{3}u^3+\frac{1}{5}u^5\]Evaluate that at \(u=1\) and subtract the value at
\(u=0\)

Answer 28

but doesnt the F(0) evaluate to 0?

Answer 29

It does, but what's your point?

Answer 30

that it isn't needed in the final answer i would only need to evaluate F(1)

Answer 31

Okay...

Answer 32

So I guess im wondering if \[\sin(1)-2/3\sin^3(1)+1/5\sin(1)^5 \] is correct

Answer 33

No, it's not. Look, either you use the original equation with the original limits, or you use the substituted equation with the substituted limits.

Answer 34

They give the same answer if you do it correctly.

Answer 35

Either you evaluate

\[\left.u-\frac{2}{3}u^3+\frac{1}{5}u^5\right|_0^1\]

or
\[\left.\sin x - \frac{2}{3}\sin^3x+\frac{\sin^5x}5\right|_0^{\frac{\pi}2}\]