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June 17th, 2017, 12:32 AM   #1
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AlgeBRAAAAA! PLEASE HELP

please help, I have attached the problem !
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File Type: jpg Screen Shot 2017-06-17 at 17.34.33.jpg (19.2 KB, 21 views)
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June 17th, 2017, 03:01 AM   #2
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A clearer version . . .
Bridge.PNG
26. The cross-section of a platform is shown above.
All measurements are in metres.

(a) Find the height of the platform.
(b) Find the cross-sectional area of the platform.
(c) Find the volume of the concrete required to
$\ \ \ \,\,$ build this platform if it is 20 metres long.

Answers:

(a) The height of the platform is (1 - (-2)) metres = 3 metres.

(b) Cross-sectional area = $\displaystyle 3e^{-2} + \int_{e^{-2}}^e (1 - \ln(x))dx = 3e^{-2} + {\large[}2x - x*\ln(x){\large]}_{e^{-2}}^e$ metres² = $\displaystyle \left(e - e^{-2}\right)$ metres².

(c) Required volume = $\displaystyle 20\left(e - e^{-2}\right)$ metres³.
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June 17th, 2017, 04:40 AM   #3
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But this is definitely not an "algebra", not even an "AlgeBRAAAAA" problem!
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June 21st, 2017, 06:18 PM   #4
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Where you should start.

Calculus was invented as a tool for solving problems. Prior to the development of calculus, there were a variety of different problems that could not be addressed using the mathematics that was available. For example, scientists did not know how to measure the speed of an object when that speed was changing over time. Also, a more effective method was desired for finding the area of a region that did not have straight edges. Geometry, algebra, and trigonometry, which were well understood, did not provide the necessary tools to adequately address these problems.

Last edited by skipjack; August 7th, 2017 at 08:44 PM.
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June 21st, 2017, 06:39 PM   #5
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thanks for cutting and pasting that Quora article (without tribute) on the history of calculus. I'm sure that it helped the OP no end.
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June 21st, 2017, 06:41 PM   #6
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Quote:
Originally Posted by John Farman View Post
Calculus was invented as a tool for solving problems.
I thought it was a plot to make students miserable.

Skipjack, where did $e^{-2}$ come from? That's not on OP's diagram, and of course without some clue we don't know what the negative $y$-value is for the bottom of the shape.
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