Integral Example - # 3

Jason76 · Nov 13, 2013

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{3} + 2x - 6x - 6$

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{3} -4x - 6$

$\displaystyle \rightarrow \dfrac{2x^{4}}{4} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{x^{4}}{2} -2x^{2} - 6x$

$\displaystyle [\dfrac{(4)^{4}}{2} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{4}}{2} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{256}{2} -2(16) - 6(4)] - [0]$

$\displaystyle [128 -32 -24] - [0] = 72$ - ?? Computer says this is wrong.

Deleted member 4993 · Nov 13, 2013

Jason76 said:
$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{3} + 2x - 6x - 6$ ...............Incorrect .... it should be .....2x² - 4x - 6

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{3} -4x - 6$

$\displaystyle \rightarrow \dfrac{2x^{4}}{4} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{x^{4}}{2} -2x^{2} - 6x$

$\displaystyle [\dfrac{(4)^{4}}{2} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{4}}{2} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{256}{2} -2(16) - 6(4)] - [0]$

$\displaystyle [128 -32 -24] - [0] = 72$ - ?? Computer says this is wrong.

.

stapel · Nov 13, 2013

Jason76 said:
$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

The lingo (they'll cover this when you take a calculus class) is "the integral, from zero to four, of...". The LaTeX coding for the "limits" of the integral is:

Code:

\int_{lower limit}^{upper limit}

...which yields:

. . . . .$\displaystyle \int_{\mbox{lower limit}}^{\mbox{upper limit}}$

You can use "displaystyle" coding to enlarge the integration symbol:

Code:

\displaystyle{ \int_{lower limit}^{upper limit} }

...to get:

. . . . .$\displaystyle \displaystyle{ \int_{\mbox{lower limit}}^{\mbox{upper limit}} }$

You can also insert spacing inside the LaTeX formatting by using a "slash" character followed by a "comma" character: " \, ". So your integral can be formatted as:

. . . . .$\displaystyle \displaystyle{\int_0^4\, (x\, -\, 3)(2x\, +\, 2)\, dx}$

Have fun!

Jason76 · Nov 13, 2013

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} + 2x - 6x - 6$

correction made

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} -4x - 6$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} -2x^{2} - 6x$

$\displaystyle [\dfrac{2(4)^{3}}{3} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{3}}{3} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{64}{3} -2(16) - 6(4)] - [0]$

$\displaystyle [\dfrac{64}{3} -32 -24] - [0]$

$\displaystyle [\dfrac{64}{3} - 56] - [0] = \dfrac{64}{3} - 56$

stapel · Nov 13, 2013

Jason76 said:
$\displaystyle \int$ evaluated at top 4 and bottom 0

No; as explained previously, the expression reads as "the integral, from zero to four, of...".

Jason76 said:
$\displaystyle [\dfrac{64}{3} - 56] - [0] = \dfrac{64}{3} - 56$

What is your question?

lookagain · Nov 13, 2013

Jason76 said:
$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} + 2x - 6x - 6$ correction made

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} -4x - 6$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} -2x^{2} - 6x$

$\displaystyle [\dfrac{2(4)^{3}}{3} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{3}}{3} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{64}{3} -2(16) - 6(4)] - [0] \ \ \ \ \ $No, it should be 128/3 - ....

$\displaystyle [\dfrac{64}{3} -32 -24] - [0] \ \ \ \ \ $ No, see above.

$\displaystyle [\dfrac{64}{3} - 56] - [0] = \dfrac{64}{3} - 56 \ \ \ \ \ \ $No, it should be 128/3 - 56 = ?

.

Jason76 · Nov 13, 2013

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} + 2x - 6x - 6$ :wink: correction made

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} -4x - 6$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} -2x^{2} - 6x$

$\displaystyle [\dfrac{2(4)^{3}}{3} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{3}}{3} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{128}{3} -2(16) - 6(4)] - [0]$

$\displaystyle [\dfrac{128}{3} - 8 $

$\displaystyle [\dfrac{128}{3} - \dfrac{24}{3} = \dfrac{104}{3}$

Computer says answer is wrong

DrPhil · Nov 13, 2013

Jason76 said:
$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} + 2x - 6x - 6$ :wink: correction made

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} -4x - 6$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} -2x^{2} - 6x$

$\displaystyle [\dfrac{2(4)^{3}}{3} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{3}}{3} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{128}{3} -2(16) - 6(4)] - [0]$....OK here

$\displaystyle [\dfrac{128}{3} - 8 $....X

$\displaystyle [\dfrac{128}{3} - \dfrac{24}{3} = \dfrac{104}{3}$ Computer says answer is wrong

.

Jason76 · Nov 14, 2013

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle (x - 3)(2x + 2) dx$

$\displaystyle \int$ evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} + 2x - 6x - 6 dx$

$\displaystyle \int$evaluated at top 4 and bottom 0

$\displaystyle 2x^{2} -4x - 6 dx$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} - \dfrac{4x^{2}}{2} - 6x$

$\displaystyle \rightarrow \dfrac{2x^{3}}{3} -2x^{2} - 6x$

$\displaystyle [\dfrac{2(4)^{3}}{3} -2(4)^{2} - 6(4)] - [\dfrac{(0)^{3}}{3} -2(0)^{2} - 6(0)]$

$\displaystyle [\dfrac{128}{3} -2(16) - 6(4)] - [0]$

$\displaystyle \dfrac{128}{3} - 56 $

Correction made

$\displaystyle \dfrac{128}{3} - \dfrac{168}{3} = -\dfrac{40}{3}$

Also wrong, and also wrong when I divide the number

srmichael · Nov 14, 2013

-40/3 is correct. What does the computer say is the correct answer?

Jason76 · Nov 14, 2013

srmichael said:
-40/3 is correct. What does the computer say is the correct answer?

I don't know, cause we can't find out the correct answer (via the computer) until after the due date. I guess I'll just have to ask the professor. But I agree, this seems like the right one.

srmichael · Nov 14, 2013

Jason76 said:
I don't know, cause we can't find out the correct answer (via the computer) until after the due date. I guess I'll just have to ask the professor. But I agree, this seems like the right one.

Do the instructions say to leave answers in improper fraction form (-40/3)? Or maybe mixed number form (-13 1/3)? Or maybe round to one decimal place (-13.3)? Or round to two decimal places (-13.33)? OR ANYTHING?!?!?!

The instructions have to have said SOMETHING in terms of how to express your answers! -40/3 IS the answer. Now how do you display it in a format that the computer wants? That's the $64,000 question.

Deleted member 4993 · Nov 14, 2013

Jason76 said:
I don't know, cause we can't find out the correct answer (via the computer) until after the due date. I guess I'll just have to ask the professor. But I agree, this seems like the right one.

That is true if you have posted the problem correctly.......

Integral Example - # 3

Jason76

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Deleted member 4993

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stapel

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Jason76

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stapel

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lookagain

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Jason76

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DrPhil

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Jason76

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srmichael

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Jason76

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srmichael

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