Calculus Integration Problem

[Jason76]

du bigger than dx in original problem

Here is a problem where the du value is bigger than the dx value in the top of the original problem. If anybody sees anything wrong with the problem or some mistake in my "converting \(\displaystyle du\) and \(\displaystyle dx\) into proper form" notation, then please say.

Formulas to be used:

Integration Power Rule \(\displaystyle \dfrac{x^{n + 1}}{n + 1}\) on the left fraction, and \(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + a^{2}}} = \ln(u + \sqrt{u^{2} = a^{2}}) + C\) on the right one.

\(\displaystyle \int \dfrac{2x - 1}{\sqrt{9x^{2} + 16}}dx\)

\(\displaystyle \int \dfrac{2x}{\sqrt{9x^{2} + 16}} dx - \int \dfrac{dx}{\sqrt{9x^{2} + 16}}\)

Looking at left fraction:

\(\displaystyle u = 9x^{2}\)

\(\displaystyle du = 18x^{2} dx\)

\(\displaystyle 18xdu = 2dx\)

\(\displaystyle du = \dfrac{2x}{18x}dx\)

\(\displaystyle du = \dfrac{1}{9}dx\)

\(\displaystyle \int(9x^{2} + 16)^{-1/2} 2x dx - \int \dfrac{dx}{\sqrt{9x^{2} + 16}}\) - Rewriting the left fraction.

Let's focus on the left fraction:

\(\displaystyle \dfrac{1}{9} \int \dfrac{(9x^{2} + 16)^{1/2}}{1/2} - 2x dx - \int \dfrac{dx}{\sqrt{9x^{2} + 16}}\)

\(\displaystyle \int \dfrac{1}{9}(2) (9x^{2} + 16)^{1/2} - 2x dx - \int \dfrac{dx}{\sqrt{9x^{2} + 16}}\)

\(\displaystyle \int \dfrac{2}{9}(9x^{2} + 16)^{1/2} - 2x dx - \int \dfrac{dx}{\sqrt{9x^{2} + 16}}\)

Now let's look at the right fraction.

\(\displaystyle \int \dfrac{2}{9}(9x^{2} + 16)^{1/2} - 2x dx - \int \dfrac{dx}{\sqrt{(3x)^{2} + (4)^{2}}}\)

\(\displaystyle u = 3x\)

\(\displaystyle du = 3 dx\)

\(\displaystyle 3du = dx\)

\(\displaystyle du = \dfrac{1}{3}dx\)

So now employing a \(\displaystyle \ln\) integral formula on the right we get:

\(\displaystyle \dfrac{2}{9}\sqrt{9x^{2} + 16} - \dfrac{1}{3}\ln(3x + \sqrt{9x^{2} + 16}) + C\) - Final answer

 

[stapel]

Here is a problem where the du value is bigger than the dx value in the top of the original problem.

What do you mean by "bigger"? And are you suggesting that there is something "wrong" with being "bigger"?

Integration Power Rule \(\displaystyle \dfrac{n^{n + 1}}{n + 1}\) on the left fraction...

Does the above actually mean the following?

. . . . .\(\displaystyle \int \, x^n\, dx\,=\,\frac{x^{n+1}}{n\, +\, 1}\, +\, C\)

\(\displaystyle u = 9x^{2}\)

\(\displaystyle du = 18x^{2}\)

Shouldn't the second line be "du = 18x dx"?

I haven't looked further than this.... ;-)

 

[Jason76]

What do you mean by "bigger"? And are you suggesting that there is something "wrong" with being "bigger"?


Does the above actually mean the following?

. . . . .\(\displaystyle \int \, x^n\, dx\,=\,\frac{x^{n+1}}{n\, +\, 1}\, +\, C\)


Shouldn't the second line be "du = 18x dx"?

I haven't looked further than this.... ;-)

Problem fixed, but still can't see logic behind notation (when doing a conversion in integration where the du value doesn't match what's in the "numerator of the original problem").

 

[Deleted member 4993]

du bigger than dx in original problem

Here is a problem where the du value is bigger than the dx value in the top of the original problem. If anybody sees anything wrong with the problem or some mistake in my "converting \(\displaystyle du\) and \(\displaystyle dx\) into proper form" notation, then please say.

Formulas to be used:

Integration Power Rule \(\displaystyle \dfrac{x^{n + 1}}{n + 1}\) on the left fraction, and \(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + a^{2}}} = \ln(u + \sqrt{u^{2} + a^{2}}) + C\) on the right one.

\(\displaystyle \int \dfrac{2x - 1}{\sqrt{9x^{2} + 16}}dx\)

This is how I would do it:

u = 3x

du = 3 dx

\(\displaystyle \dfrac{2}{3}\int \dfrac{3x - 1.5}{\sqrt{9x^{2} + 16}}dx\)

\(\displaystyle = \ \dfrac{1}{9}\int \dfrac{2u du}{\sqrt{u^{2} + 4^2}} \ - \ \dfrac{1}{3}\int \dfrac{du}{\sqrt{u^{2} + 4^2}}\)

\(\displaystyle = \ \dfrac{1}{9}*2(u^2 + 16)^{\frac{1}{2}} \ - \ \dfrac{1}{3}ln(u + \sqrt{u^2 + 16}) + C\)

\(\displaystyle = \ \dfrac{2}{9}\sqrt{9x^2 + 16} \ - \ \dfrac{1}{3}ln(3x + \sqrt{9x^2 + 16}) + C\)