Evaluate the following integral.

[flaren5]

I believe that I'm on the right track to evaluating the integral, however I'm getting stumped on what to do with the definite integral from 0 to 1. I would greatly appreciate a step-by-step, if possible from the original integral, to the final evaluation. Thank you.

 

[stapel]

\(\displaystyle \int_0^1\, \frac{1}{1\, +\, \sqrt{x}}\, dx\)

\(\displaystyle u\, =\, \sqrt{x}\, \mbox{ and }\, du\, =\, \frac{1}{2\sqrt{x}}\, dx\)

\(\displaystyle =\, 2\, \int\, \frac{u}{1\, +\, u}\, du\)

How did you get the last line above? Using your substitution... well, it isn't possible, is it? It almost looks like you "separated" the denominator into two fractions or something, and then maybe got rid of part of it...?? :shock:

To confirm, differentiate:

. . . . .\(\displaystyle \dfrac{d}{du}\, \dfrac{2u}{1\, +\, u}\, =\, \dfrac{2(1\, +\, u)\, -\, u(0\, +\, 1)}{(1\, +\, u)^2}\, =\, \dfrac{2\, +\, u}{1\, +\, 2u\, +\, u^2}\)

Now plug back in for u, and you won't get what you'd started with.

 

[mathmari]

View attachment 3220

I believe that I'm on the right track to evaluating the integral, however I'm getting stumped on what to do with the definite integral from 0 to 1. I would greatly appreciate a step-by-step, if possible from the original integral, to the final evaluation. Thank you.

You subtituted \(\displaystyle u=\sqrt{x} \), so when \(\displaystyle x=0 \Rightarrow u=0 \) and when \(\displaystyle x=1 \Rightarrow u=1 \).

\(\displaystyle \int_{0}^{1} \frac{1}{1+\sqrt{x}}dx=\int_{0}^{1} \frac{2 \sqrt{x}}{1+\sqrt{x}}\frac{dx}{2 \sqrt{x}}=\int_{0}^{1} \frac{2 u}{1+u}du=2\int_{0}^{1} \frac{u+1-1}{1+u}du=2\int_{0}^{1} (1-\frac{1}{1+u})du=2\int_{0}^{1}1 du-2\int_{0}^{1} \frac{1}{1+u}du=2-2 \cdot ln2 \)

 

[Deleted member 4993]

How did you get the last line above? Using your substitution... well, it isn't possible, is it? It almost looks like you "separated" the denominator into two fractions or something, and then maybe got rid of part of it...?? :shock:

To confirm, differentiate:

. . . . .\(\displaystyle \dfrac{d}{du}\, \dfrac{2u}{1\, +\, u}\, =\, \dfrac{2(1\, +\, u)\, -\, u(0\, +\, 1)}{(1\, +\, u)^2}\, =\, \dfrac{2\, +\, u}{1\, +\, 2u\, +\, u^2}\)

Now plug back in for u, and you won't get what you'd started with.

Liz, you need a cup of coffee!!

 

[flaren5]

How did you get the last line above? Using your substitution... well, it isn't possible, is it? It almost looks like you "separated" the denominator into two fractions or something, and then maybe got rid of part of it...?? :shock:

To confirm, differentiate:

. . . . .\(\displaystyle \dfrac{d}{du}\, \dfrac{2u}{1\, +\, u}\, =\, \dfrac{2(1\, +\, u)\, -\, u(0\, +\, 1)}{(1\, +\, u)^2}\, =\, \dfrac{2\, +\, u}{1\, +\, 2u\, +\, u^2}\)

Now plug back in for u, and you won't get what you'd started with.

Thank you for you response. I got the last line above by doing long division (wan't sure if there was a shorter way). ;-)

 

[flaren5]

You subtituted \(\displaystyle u=\sqrt{x} \), so when \(\displaystyle x=0 \Rightarrow u=0 \) and when \(\displaystyle x=1 \Rightarrow u=1 \).

\(\displaystyle \int_{0}^{1} \frac{1}{1+\sqrt{x}}dx=\int_{0}^{1} \frac{2 \sqrt{x}}{1+\sqrt{x}}\frac{dx}{2 \sqrt{x}}=\int_{0}^{1} \frac{2 u}{1+u}du=2\int_{0}^{1} \frac{u+1-1}{1+u}du=2\int_{0}^{1} (1-\frac{1}{1+u})du=2\int_{0}^{1}1 du-2\int_{0}^{1} \frac{1}{1+u}du=2-2 \cdot ln2 \)

Thank you for your response, it was straight forward, and I was able to understand it, for the most part, however there is a part that I don't completely follow, which is highlighted below, with a red box, in particular the +1-1. If you're able to give a quick explanation, I'd greatly appreciate it.

 

[JeffM]

View attachment 3220

I believe that I'm on the right track to evaluating the integral, however I'm getting stumped on what to do with the definite integral from 0 to 1. I would greatly appreciate a step-by-step, if possible from the original integral, to the final evaluation. Thank you.

\(\displaystyle u = \sqrt{x} \implies \dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}} = \dfrac{1}{2u} \implies 2u * du = dx.\)

\(\displaystyle u = \sqrt{x} \implies \dfrac{1}{1 + \sqrt{x}} * dx = \dfrac{1}{1 + u} * 2u * du = 2 * \dfrac{u}{u + 1} * du = 2 \left(1 - \dfrac{1}{u + 1}\right) * du.\) You were fine to here.

\(\displaystyle \int2 \left(1 - \dfrac{1}{u + 1}\right) * du = 2intdu - 2\int\dfrac{du}{u + 1} = 2u - 2ln(u + 1) + C.\) Do you follow this?

Now you must change the limits of integration from x to u. IN THIS CASE, you get lucky, but it is a crucial step.

\(\displaystyle x = 1 \implies \sqrt{x} = 1 \implies u = 1.\)

\(\displaystyle x = 0 \implies \sqrt{x} = 0 \implies u = 0.\)

\(\displaystyle 2 * 1 - 2ln(1 + 1) = 2 - 2ln(2) = 2 - ln(4).\)

\(\displaystyle 2 * 0 - 2ln(0 + 1) = 0 - 2ln(1) = 0 - 2 * 0 = 0.\)

So \(\displaystyle \int_0^1\dfrac{dx}{1 + sqrt{x}} = 2 - ln(4) - 0 = 2 - ln(4).\)

The question that you asked involves a "trick" that saves you the work of doing long division. u can be turned into u + 1 - 1 so

\(\displaystyle \dfrac{u}{u + 1} = \dfrac{u + 0}{u + 1} =\dfrac{u + (1 - 1)}{u + 1} = \dfrac{u + 1}{u + 1} - \dfrac{1}{u + 1} = 1 - \dfrac{1}{u + 1}.\)

 

[mathmari]

Thank you for your response, it was straight forward, and I was able to understand it, for the most part, however there is a part that I don't completely follow, which is highlighted below, with a red box, in particular the +1-1. If you're able to give a quick explanation, I'd greatly appreciate it. View attachment 3221

Since you cannot calculate the integral \(\displaystyle \int \frac{u}{u+1}du \) , you have to change the form of the integral.
You can appear the number of the denominator in the numerator by adding and subtracting 1.
\(\displaystyle u=u+1-1 \)
\(\displaystyle \frac{u}{u+1}=\frac{u+1-1}{u+1}=\frac{u+1}{u+1}-\frac{1}{u+1}=1-\frac{1}{u+1} \) .

An other way is that you can do the subtitution \(\displaystyle w=1+u \) .
\(\displaystyle dw=du \)
\(\displaystyle u=0 \Rightarrow w=1 \)
\(\displaystyle u=1 \Rightarrow w=2 \)
\(\displaystyle 2 \int_{0}^{1} \frac{u}{1+u}du =2 \int_{1}^{2} \frac{w-1}{w}dw=2 \int_{1}^{2}(1-\frac{1}{w})dw=2-2ln2 \)

 

[flaren5]

\(\displaystyle u = \sqrt{x} \implies \dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}} = \dfrac{1}{2u} \implies 2u * du = dx.\)

\(\displaystyle u = \sqrt{x} \implies \dfrac{1}{1 + \sqrt{x}} * dx = \dfrac{1}{1 + u} * 2u * du = 2 * \dfrac{u}{u + 1} * du = 2 \left(1 - \dfrac{1}{u + 1}\right) * du.\) You were fine to here.

\(\displaystyle \int2 \left(1 - \dfrac{1}{u + 1}\right) * du = 2intdu - 2\int\dfrac{du}{u + 1} = 2u - 2ln(u + 1) + C.\) Do you follow this?

Now you must change the limits of integration from x to u. IN THIS CASE, you get lucky, but it is a crucial step.

\(\displaystyle x = 1 \implies \sqrt{x} = 1 \implies u = 1.\)

\(\displaystyle x = 0 \implies \sqrt{x} = 0 \implies u = 0.\)

\(\displaystyle 2 * 1 - 2ln(1 + 1) = 2 - 2ln(2) = 2 - ln(4).\)

\(\displaystyle 2 * 0 - 2ln(0 + 1) = 0 - 2ln(1) = 0 - 2 * 0 = 0.\)

So \(\displaystyle \int_0^1\dfrac{dx}{1 + sqrt{x}} = 2 - ln(4) - 0 = 2 - ln(4).\)

The question that you asked involves a "trick" that saves you the work of doing long division. u can be turned into u + 1 - 1 so

\(\displaystyle \dfrac{u}{u + 1} = \dfrac{u + 0}{u + 1} =\dfrac{u + (1 - 1)}{u + 1} = \dfrac{u + 1}{u + 1} - \dfrac{1}{u + 1} = 1 - \dfrac{1}{u + 1}.\)

Awesome, thank you. I absolutely follow your response. I am a very visual learner, and greatly appreciate how you broke it down.

 

[flaren5]

Since you cannot calculate the integral \(\displaystyle \int \frac{u}{u+1}du \) , you have to change the form of the integral.
You can appear the number of the denominator in the numerator by adding and subtracting 1.
\(\displaystyle u=u+1-1 \)
\(\displaystyle \frac{u}{u+1}=\frac{u+1-1}{u+1}=\frac{u+1}{u+1}-\frac{1}{u+1}=1-\frac{1}{u+1} \) .

An other way is that you can do the subtitution \(\displaystyle w=1+u \) .
\(\displaystyle dw=du \)
\(\displaystyle u=0 \Rightarrow w=1 \)
\(\displaystyle u=1 \Rightarrow w=2 \)
\(\displaystyle 2 \int_{0}^{1} \frac{u}{1+u}du =2 \int_{1}^{2} \frac{w-1}{w}dw=2 \int_{1}^{2}(1-\frac{1}{w})dw=2-2ln2 \)

Thank for an explanation, and your time, it helped me understand this problem in a new light.;-)