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Intregal definite, The area of [a,b] where a=-2.5 and b=2.5, by rectangles I have a similar area, is the arch of a bridge(model) . $$ y=-\frac{4}{25}x^2+1 $$ $$ \int_{-2.5}^{2.5}(-\frac{4}{25}x^2+1)dx$$ $$-\int_{-2.5}^{2.5}\frac{4}{25}x^2dx+\int_{-2.5}^{2.5}dx$$ $$-\frac{4}{25}\frac{x^{2+1}}{2+1}+x$$ $$-\frac{4}{25}\frac{x^{3}}{3}+x$$ $$-\frac{4}{75}x^{3}|_{-2.5}^{2.5}+x|_{-2.5}^{2.5}$$ $$[-\frac{4}{75}(2.5)^{3}+(2.5)]-[-\frac{4}{75}(-2.5)^{3}+(-2.5)]$$ $$[-\frac{4}{75}(15.625)+2.5]-[-\frac{4}{75}(-15.625)-2.5]$$ $$[-\frac{62.5}{75}+2.5]-[\frac{62.5}{75}-2.5]$$ $$[-\frac{62.5}{75}+\frac{187.5}{75}]-[\frac{62.5}{75}-\frac{187.5}{75}]$$ $$[\frac{125}{75}]-[-\frac{125}{75}]$$ $$\frac{125}{75}+\frac{125}{75}$$ $$\frac{250}{75}u^2\approx 3.3333 u^2$$ scale: $$1u=\frac{110}{9}cm$$ $$1u^2=\frac{12100}{81}cm^2$$ real scale: $$1cm^2=4225m^2$$ I'm doing this for a school project, I need to show the use of integrals in real life. So I made a scale model of a bridge, so I have this function: -(4/25x^2 + 1) for the quadratic function. the only I need to know if my scale and my maths are right so I can continue this problem.

Si pueedes en español mejor jaja

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What's your doubt then? –  user1620696 Jun 9 '13 at 3:27
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What do you mean by $1cm^2=4225cm^2$? –  Erdos Yi Jun 9 '13 at 3:32
    
ok... thanks for you attention it's m^2 –  JgMc Jun 9 '13 at 3:36
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2 Answers 2

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This is Evariste's answer translated into Spanish.

La integral que has computado es correcta, pero tal vez te saldria mejor si incluyes los limites de integracion antes de evaluar la integral, es decir, no dejarla en forma indefinida asi: $$\frac{4}{25}\frac{x^{2+1}}{2+1}+x$$

Si dudaste que todo lo que has hecho este bien, puedes aprovecharte de WolframAlpha. Es un sitio web muy util que puedes usar para verifiar tus computaciones.

Desde el punto de vista de estrategio, visto que has separado las integrales ya en la suma de dos integrales (las que son mas facil de evaluar), ?porque las agregaste otra vez cuando llegaste a la etapa de evaluacion? Sugiero que trabajes mas en computar integrales y que adoptes una manera sistematica para ello. Eso te va a ahorrar muchisimo tiempo.

La cifra final debe ser 51122500/81$m^2$, o aproximadamente 631141.98$m^2$.

Un saludo.

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Gracias... ya lo había leido. –  JgMc Jun 10 '13 at 3:43
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The integral you've done is absolutely right, although you might always want to include the upper and lower limits before you evaluate the integral(i.e. never leave it in the indefinite form like this: $$-\frac{4}{25}\frac{x^{2+1}}{2+1}+x$$ ).

You were probably wondering if you did everything correctly. I think it was correct. Next time you might want to put your stuff into Wolfram Alpha(www.wolframalpha.com). It's a useful website that will check your integrals.

Also in terms of your strategy, since you have already split your integrals into the sum of two integrals(which is easier to evaluate and less nasty), why did you put them back together when you actually got the the evaluation step? I suggest that you practice more on solving integrals and set up a systematical way of doing it. This will save you tons of time.

The final answer should be $51122500/81 m^2$ which is approximately $631141.98m^2$

Hope this helps.

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all is step by step for any people –  JgMc Jun 9 '13 at 3:51
    
@JgMc I don't understand what you're talking about. What do you mean by "all is step by step for any people"? –  Evariste Jun 9 '13 at 4:34
    
Oops.. "Que cualquier persona entienda, todo debe ser paso a paso" I don't know how to tell you my idea. My english is not perfect. :/ –  JgMc Jun 10 '13 at 3:42
    
@JgMc That's alright. I don't know how to express myself in Spanish either :p But I reckon what you did was correct(although you can prob follow a better and clearer approach). If you wonder what a better and clearer approach is, I suggest you check out Calculus by Stewart. Look at those examples that he did. –  Evariste Jun 10 '13 at 6:29
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