integration by substitution

[bold belfry]

Where am I going wrong here everytime I sub the numbers into this I get the wrong answer

[glossy tinsel]

When you make a substitution you also have to change the limits

[bold belfry]

Yeah I did that I got 25 and 0

[glossy tinsel]

Well you should then have

$$\int_{0}^{25} \frac{u^{\frac{1}{2}}(u - 1)}{16} ,du$$

[acoustic ice]

AmbivalentChamp

[glossy tinsel]

which it looks like you already have

so the problem must've come when actually integrating it

should be

so

$$\int_{0}^{25} \frac{u^{\frac{3}{2}} - u^{\frac{1}{2}}}{16} ,du$$

[acoustic ice]

AmbivalentChamp

[glossy tinsel]

what did u get when integrating this?

[bold belfry]

I integrated just the u subbed the numbers in them multiplied by the 1/16 would that work?

[glossy tinsel]

could u show me if u don't mind?

Cause I get

$$[\frac{1}{40}u^{\frac{5}{2}} - \frac{1}{24}u^{\frac{3}{2}}]_{0}^{25}$$

[acoustic ice]

AmbivalentChamp

[glossy tinsel]

And subbing 25 gives u 875/12

[glossy tinsel]

You forgot the 1/16

And if u sub back in the original u don't use the 25

Use the original limits of 1 and 6

[bold belfry]

So the 1/16 has to he in the equation when I'm integrating

[glossy tinsel]

yeah

[bold belfry]

Subbing these in I seem to get the wrong answer. I must be working it out wrong

[glossy tinsel]

what are you subbing in?

[bold belfry]

25 and 0

[glossy tinsel]

that's the thing

u don't sub in 25 and 0

If you revert it back to x

you use 1 and 6 again

[bold belfry]

Ohh I sub 25 in for u. Not into u=4x+1

Or 1 and 6 into the 4x+1? Because they are in terms of x?

[glossy tinsel]

Well if u sub 25 = 4x + 1

[bold belfry]

I was going to do 1/40(25)^5/2....

[glossy tinsel]

ohhhh

[glossy tinsel]

U just subbing it into this?

Then u should get the answer

[bold belfry]

Yes
Where u is 25 and 0

Ah ok that's where I was going wrong

That's for the help

Thanks*