Is this correct?

**gammaman** · February 21st 2009, 10:09 AM

$\int(2x+7)\sqrt{4x-5}$
$\frac{1}{4}\int(2x+7)\sqrt{4x-5}du$
$\frac{1}{4}\int\frac{2u+12}{4}u^{1/2}du$
$\frac{1}{16}[2(\frac{2}{5})u^{5/2}+12(\frac{2}{3})u^{3/2}]du$
$\frac{1}{16}[\frac{4}{5}(4x-5)u^{5/2}+8(4x-5)^{3/2}]+c$

**Moo** · February 21st 2009, 10:18 AM

Hello,

Originally Posted by gammaman

$\int(2x+7)\sqrt{4x-5}$
$\frac{1}{4}\int(2x+7)\sqrt{4x-5}du$
$\frac{1}{4}\int\frac{2u+12}{4}u^{1/2}du$
$\frac{1}{16}[2(\frac{2}{5})u^{5/2}+12(\frac{2}{3})u^{3/2}]du$
$\frac{1}{16}[\frac{4}{5}(4x-5)u^{5/2}+8(4x-5)^{3/2}]+c$

$u=4x-5 \Rightarrow x=\frac{u+5}{4}$

So $2x+7=\frac{u+5}{2}+7=\frac{u+5+14}{2}=\frac{u+19}{ 2} \neq \frac{2u+12}{4}$

**gammaman** · February 21st 2009, 10:21 AM

ok but how did the u+5/4 become u+5/2? did the 4 reduce because of the 2x?

**Moo** · February 21st 2009, 10:24 AM

Originally Posted by gammaman

ok but how did the u+5/4 become u+5/2? did the 4 reduce because of the 2x?

Yes

Otherwise, your method is correct, though :

$\frac{1}{16}[2(\frac{2}{5})u^{5/2}+12(\frac{2}{3})u^{3/2}]du$

this line could be skipped (it makes no sense to write this du)

**gammaman** · February 21st 2009, 10:41 AM

what do I do from here

$\frac{1}{4}[(\frac{u+19}{2})u^{1/2}]du$

**Moo** · February 21st 2009, 10:45 AM

Originally Posted by gammaman

what do I do from here

$\frac{1}{4}[(\frac{u+19}{2})u^{1/2}]du$

It's $\frac 14\int \frac{u+19}{2} u^{1/2} ~du$
$=\frac 14 \int \frac 12 u^{3/2} ~du+\frac 14 \int \frac{19}{2} u^{1/2} ~du$

just do what you did in the first time !

**gammaman** · February 21st 2009, 10:54 AM

Is this the answer?

$\frac{1}{20}[(4x-5)^{5/2}+\frac{19}{3}(4x-5)^{3/2}]+c$

**Krizalid** · February 21st 2009, 11:15 AM

Why don't you differentiate the result? That should yield the integrand.

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