areas and definate integrals

**gixxer** · December 12th 2008, 07:13 PM

I'm just curious how to solve something like S dx/4x+6 from x=1 to x=6
or S du/8u^1/2. I understand how to do the differentiation but I dont know what happens to the dx or du during the process. I'm trying to study for a final and this is driving me crazy. Thank you so much.

Greg

**Chris L T521** · December 12th 2008, 07:28 PM

Originally Posted by gixxer

I'm just curious how to solve something like S dx/4x+6 from x=1 to x=6
or S du/8u^1/2. I understand how to do the differentiation but I dont know what happens to the dx or du during the process. I'm trying to study for a final and this is driving me crazy. Thank you so much.

Greg

For the first one, $\int_1^6\frac{\,dx}{4x+6}$ , use a substitution $z=4x+6\implies \,dz=4\,dx\implies\tfrac{1}{4}\,dz=\,dx$ . You can also convert the limits of integration as well. $z(1)=10$ and $z(6)=30$ . Thus, $\int_1^6\frac{\,dx}{4x+6}$ is the same thing as $\tfrac{1}{4}\int_{10}^{30}\frac{\,dx}{z}$ . The second integral is a bit easier to evalute. Can you take this one from here?

For the second one, $\int\frac{\,du}{8u^{\frac{1}{2}}}=\tfrac{1}{8}\int u^{-\frac{1}{2}}\,du$ . From here, apply the power rule of integration: $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ (since this is indefinite integration). Can you take it from here?

Does this make sense?

**gixxer** · December 12th 2008, 08:38 PM

Thank you for all your help it made me understand this unit a lot better. You got me thinking correctly again.

One more question...

Would this be solved by substitution?

**Chop Suey** · December 12th 2008, 08:48 PM

Yes. Try $u = e^{2x}-2x$

**Pi314** · December 12th 2008, 09:15 PM

Yeah it uses substitution as previously stated, I actually was doing a problem like that earlier today.

Apparently any anti-deriv. problem that requires multiplication and/or division needs to be solved via subs. So far the examples I've come across follow this idea.

And at times you're not left with the "proper" dx so you might have to "break" one of the functions.

for instance, for the anti. derivative of (8x+8^(2x))(x^2+e^(2x)) dx you have to take out the 8 after making the second portion "u" and then you'll be left with (x^2+e^(2x)) dx and you can continue to solve the problem normally.

**Prove It** · December 13th 2008, 01:53 AM

Originally Posted by Pi314

Yeah it uses substitution as previously stated, I actually was doing a problem like that earlier today.

Apparently any anti-deriv. problem that requires multiplication and/or division needs to be solved via subs. So far the examples I've come across follow this idea.

And at times you're not left with the "proper" dx so you might have to "break" one of the functions.

for instance, for the anti. derivative of (8x+8^(2x))(x^2+e^(2x)) dx you have to take out the 8 after making the second portion "u" and then you'll be left with (x^2+e^(2x)) dx and you can continue to solve the problem normally.

Yes that's right...

So in the one above...

$\int_0^4{(e^{2x}-2x)^5 (e^{2x}-1)\,dx}$

as stated earlier, you'd have to use the substitution $u = e^{2x} - 2x$ . This would mean

$\frac{du}{dx} = 2e^{2x} - 2 = 2(e^{2x} - 1)$ .

So you would need to "break" the original function so that you get $\frac{du}{dx}$ as a factor...

$\int_0^4{(e^{2x}-2x)^5 (e^{2x}-1)\,dx}=\frac{1}{2}\int_0^4{(e^{2x}-2x)^5 2(e^{2x}-1)\,dx}$

$= \frac{1}{2}\int_{x=0}^{x=4}{u^5 \frac{du}{dx}\,dx}$

$= \frac{1}{2}\int_{x=0}^{x=4}{u^5 \, du}$

$= \frac{1}{2}\left[\frac{1}{6}u^6\right]_{x=0}^{x=4}$

$= \frac{1}{12}[(e^{2x}-2x)^6]_0^4$

$= \frac{1}{12}[(e^8 - 8)^6 - (e^0 - 0)^6]$

$= \frac{1}{12}[(e^8 - 8)^6 - 1]$ .

**GaloisTheory1** · December 13th 2008, 11:38 AM

Originally Posted by gixxer

I'm just curious how to solve something like S dx/4x+6 from x=1 to x=6
or S du/8u^1/2. I understand how to do the differentiation but I dont know what happens to the dx or du during the process. I'm trying to study for a final and this is driving me crazy. Thank you so much.

Greg

1. $\int_1^6 \frac{dx}{4x}+6 = \int_1^6 \frac{1}{4}\cdot\frac{dx}{x}+6=[\frac{1}{4}\cdot\ln{x}+6x]|_1^6$

2. $\int 8 \cdot u^{1/2}=8 \cdot \frac{2}{3}u^{\frac{3}{2}}+C$

**mr fantastic** · December 13th 2008, 01:04 PM

Originally Posted by GaloisTheory1

1. $\int_1^6 \frac{dx}{4x}+6 = \int_1^6 \frac{1}{4}\cdot\frac{dx}{x}+6=[\frac{1}{4}\cdot\ln{\color{red} |} x {\color{red} |} +6x]|_1^6$

2. $\int 8 \cdot u^{1/2} \, {\color{red} du}=8 \cdot \frac{2}{3}u^{\frac{3}{2}}+C$

Some minor corrections (in red).

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