Definite Integration Problem - # 4

[Jason76]

Does this question sound right? Also, what is the latex notation for "evaluated at upper and lower bounds" for after you began to actually take the integral?

What is the area of a curve with a derivative of $\displaystyle x(5x^{2} - 3)^{1/2} dx$ on the closed interval $\displaystyle [-2,1]$?

$\displaystyle \int_{-2}^{1} x(5x^{2} - 3)^{1/2} dx$.

$\displaystyle \int_{-2}^{1} x(u)^{1/2} dx$.

$\displaystyle u = 5x - 3$.

$\displaystyle du = 10x dx$.

$\displaystyle \dfrac{1}{10}du = x dx$.

$\displaystyle \int_{-2}^{1} (u)^{1/2} \dfrac{1}{10}du$.

$\displaystyle \dfrac{1}{10} \int_{-2}^{1} (u)^{1/2} \dfrac{1}{10}du$.

$\displaystyle \rightarrow (\dfrac{1}{10}) \dfrac{(u)^{3/2}}{\dfrac{3}{2}}$.

$\displaystyle \rightarrow (\dfrac{1}{10})(\dfrac{2}{3}) (u)^{3/2}$.

$\displaystyle \rightarrow (\dfrac{2}{30})(u)^{3/2}$.

$\displaystyle \rightarrow (\dfrac{1}{15})(u)^{3/2}$. Some missing latex symbols here.

$\displaystyle \rightarrow \dfrac{1}{15}(5x - 3)^{3/2}$.

$\displaystyle [\dfrac{1}{15}(5(-2) - 3)^{3/2}] - [\dfrac{1}{15}(5(1) - 3)^{3/2}]$.

 

[stapel]

what is the latex notation for "evaluated at upper and lower bounds" for after you began to actually take the integral?

Do a "left" tag that you leave blank, by typing a full-stop (a "period") after the "left", as: \left.

Then, to a "right" tag at the other end, and use the "pipe" character (probably above the "Enter" key on your keyboard). Then use the underscore and carat notations to indicate the limits, as: \right_{lower limit}^{upper limit}

To see any coding you're wanting to imitate, try a "reply with quote". You don't actually have to reply, but the quoted reply will display the LaTeX coding.

 

[stapel]

Does this question sound right?

Do you mean "Does this solution look right?"? If so, then...

$\displaystyle \int_{-2}^{1} x(5x^{2} - 3)^{1/2} dx$.

$\displaystyle \int_{-2}^{1} x(u)^{1/2} dx$.

As has been mentioned so many times before, no, you cannot do this. The integral needs to be in terms of $\displaystyle x$ or else in terms of $\displaystyle u$, but not both at once.

$\displaystyle \int_{-2}^{1} (u)^{1/2} \dfrac{1}{10}du$.

As has been mentioned so many times before, no, you cannot do this. Either work with the indefinite integral when doing the computations with $\displaystyle u$, or else properly change the limits of integration.

While you may have ended up with the correct final value (I haven't checked), many ("most"?) instructors would stop at one of these earlier lines, considering anything "right" that happened later to be mere luck. It really would be a good idea to start reading the responses you receive and then consider incorporating the advice into your work.

 

[Jason76]

Sorry, but the Latex code: \right_{-2}^{1} doesn't produce the right result on either a forum or blog.

Back to the problem:

$\displaystyle \int_{-2}^{1} x(5x^{2} - 3)^{1/2} dx$.

:grin: The troublesome step omitted. So is it right, or do I need to add something here?

$\displaystyle u = 5x - 3$.

$\displaystyle du = 10x dx$.

$\displaystyle \dfrac{1}{10}du = x dx$.

Another troublesome step omitted. So is it right, or do I need to add something here?

$\displaystyle \dfrac{1}{10} \int_{-2}^{1} (u)^{1/2} du$.

$\displaystyle \rightarrow (\dfrac{1}{10}) \dfrac{(u)^{3/2}}{\dfrac{3}{2}}$.

$\displaystyle \rightarrow (\dfrac{1}{10})(\dfrac{2}{3}) (u)^{3/2}$.

$\displaystyle \rightarrow (\dfrac{2}{30})(u)^{3/2}$

$\displaystyle \rightarrow (\dfrac{1}{15})(u)^{3/2}$

$\displaystyle \rightarrow \dfrac{1}{15}(5x - 3)^{3/2}$.

$\displaystyle [\dfrac{1}{15}(5(-2) - 3)^{3/2}] - [\dfrac{1}{15}(5(1) - 3)^{3/2}]$.

$\displaystyle 3.124811105 - 0.188561808 = 2.936249297$ Answer

 

[stapel]

Sorry, but the Latex code: \right_{-2}^{1} doesn't produce the right result on either a forum or blog.

Really?

. . . . .$\displaystyle \displaystyle{\int_{-2}^1\, 2x\, dx\, =\, \left. x^2 \right|_{-2}^1}$

 

[Deleted member 4993]

Sorry, but the Latex code: \right_{-2}^{1} doesn't produce the right result on either a forum or blog.

Back to the problem:

$\displaystyle \int_{-2}^{1} x(5x^{2} - 3)^{1/2} dx$.

Did you investigate what happens to your function, when:

$\displaystyle \displaystyle{(5x^2 - 3) < 0 → -\sqrt{\frac{3}{5}} \ < \ x \ < \ \sqrt{\frac{3}{5}}}$

 

[pka]

The code \mathop {\left. {F(x)} \right|}\nolimits_a^b gives $\displaystyle \displaystyle\mathop {\left. {F(x)} \right|}\nolimits_a^b$.

But mainly your mistake is in not changing the limits of integration after making the u-substitution.
\(\displaystyle \begin{array}{*{20}{c}}
x&|& { - 2}&{}&1\\
\hline
u&|& { 20}&{}&2
\end{array}\)

So it becomes $\displaystyle \displaystyle\int_{ 20}^2 {f(u)du}$

That saves you from going back to the original.

 

[Deleted member 4993]

Sorry, but the Latex code: \right_{-2}^{1} doesn't produce the right result on either a forum or blog.

Back to the problem:

$\displaystyle \int_{-2}^{1} x(5x^{2} - 3)^{1/2} dx$.

:grin: The troublesome step omitted. So is it right, or do I need to add something here?

$\displaystyle u = 5x - 3$..... Or is it u = 5x² - 3

$\displaystyle du = 10x dx$.

$\displaystyle \dfrac{1}{10}du = x dx$.

Another troublesome step omitted. So is it right, or do I need to add something here?

$\displaystyle \dfrac{1}{10} \int_{-2}^{1} (u)^{1/2} du$.

$\displaystyle \rightarrow (\dfrac{1}{10}) \dfrac{(u)^{3/2}}{\dfrac{3}{2}}$.

$\displaystyle \rightarrow (\dfrac{1}{10})(\dfrac{2}{3}) (u)^{3/2}$.

$\displaystyle \rightarrow (\dfrac{2}{30})(u)^{3/2}$

$\displaystyle \rightarrow (\dfrac{1}{15})(u)^{3/2}$

$\displaystyle \rightarrow \dfrac{1}{15}(5x - 3)^{3/2}$.

$\displaystyle [\dfrac{1}{15}(5(-2) - 3)^{3/2}] - [\dfrac{1}{15}(5(1) - 3)^{3/2}]$.

$\displaystyle 3.124811105 - 0.188561808 = 2.936249297$ Answer

.

 

[Jason76]

\(\displaystyle \right_{-2}^{1}\)

Still not right rendering

 

[JeffM]

Is this what you are trying to do?

$\displaystyle \mathop{\left.{F(x)}\right|}\nolimits_{-2}^1$

It's this code (as pka told you)

\mathop{\left.{F(x)}\right|}\nolimits_{-2}^1

Actually you do not even need the mathop as stapel told you.

$\displaystyle \left.F(x)\right|_a^b.$

 

[Deleted member 4993]

\(\displaystyle \right_{-2}^{1}\)

Still not right rendering

Because

you did not start with \left.

and

you do not have | after \right

when you fix those, you will get the following:

$\displaystyle \left . \right |_{-2}^{1}$

 

[Jason76]

$\displaystyle \mathop{\left.{}\right|}\nolimits_{-2}^1$

$\displaystyle \mathop{\left.{F(x)}\right|}\nolimits_{-2}^1$

 

[JeffM]

$\displaystyle \mathop{\left.{}\right|}\nolimits_{-2}^1$

$\displaystyle \mathop{\left.{F(x)}\right|}\nolimits_{-2}^1$

You do not need all that as stapel, subhotosh, and I have already explained.

\left. F(x) \right |_a^b gives $\displaystyle \left. F(x) \right |_a^b$

 

[daon2]

You do not need all that as stapel, subhotosh, and I have already explained.

\left. F(x) \right |_a^b gives $\displaystyle \left. F(x) \right |_a^b$

Or: F(x) \Big|_a^b

$\displaystyle F(x) \Big|_a^b$