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#1 2018-03-18 04:34:43

Monox D. I-Fly
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From: Indonesia
Registered: 2015-12-02
Posts: 2,000

[ASK] Integral

The area of the region limited by the curve [MATH]y=x^2+2x-3[/MATH], X-axis, Y-axis, and the line x = 2 is ....
A. 4 area unit
B. 9 area unit
C. 11 area unit
D. 13 area unit
E. 27 area unit

My attempt so far:


(x + 3)(x - 1) = 0
x = -3 or x = 1
X-intercept is at x = -3 and x = 1.
After drawing the graph, the restriction is x = 1 to x = 2.










Where did I do wrong? I know from the graph that the answer is A since the area covered is less large than a [MATH]1\times5[/MATH] rectangle and the option A is the only one with a value less than 5, but what did I do wrong algebraically?


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#2 2018-03-18 04:54:32

zetafunc
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Registered: 2014-05-21
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Re: [ASK] Integral

The integral should evaluate to
. Your arithmetic slips take place in these three lines:



In particular, the second pair of brackets in the second line and the first pair of brackets in the third line are not correct.

However, the question appears to be badly worded. The area 'bounded' by the curve, the axes and the line x = 2 is not bounded at all -- it is infinite.

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#3 2018-03-18 05:24:17

Monox D. I-Fly
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From: Indonesia
Registered: 2015-12-02
Posts: 2,000

Re: [ASK] Integral

Then, do you think it has no answer?


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#4 2018-03-18 05:42:18

zetafunc
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Re: [ASK] Integral

The question as it stands has no answer (the area would be infinite). There is probably a typo in the question.

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#5 2018-03-18 06:54:03

Alg Num Theory
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Registered: 2017-11-24
Posts: 693
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Re: [ASK] Integral

zetafunc wrote:

The question as it stands has no answer (the area would be infinite). There is probably a typo in the question.

The question makes perfect sense and there is a perfectly good answer.

Graph of the function.

So, noting that the function is negative between 0 and 1 and positive between 1 and 2, the area should be

[list=*]
[*]

[/*]
[/list]

Last edited by Alg Num Theory (2018-03-18 06:58:10)


Me, or the ugly man, whatever (3,3,6)

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#6 2018-03-18 11:50:58

Monox D. I-Fly
Member
From: Indonesia
Registered: 2015-12-02
Posts: 2,000

Re: [ASK] Integral

darn, I forgot that it doesn't have to be in the first quadrant. Thanks Alg Num Theory.


Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.

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#7 2018-03-18 21:31:39

Alg Num Theory
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Registered: 2017-11-24
Posts: 693
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Re: [ASK] Integral

Any time. smile


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