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Article #12Re: Area under a graph !help!
From: Robert Simon Martin
Date: Wed, 08 Mar 2000 00:00
Date: Wed, 08 Mar 2000 00:00
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1626 bytes
>Hi I've got my GCSE coursework due in on Monday, >and I have no idea what the formula for the area under >a curved graph is, eg. area when the x-axis ranges for 0-4, >in the graph of the function y=x^2 (y=x squared) First you need to *integrate* y=x^2, you do this by increasing the power by 1 and dividing by the new power, so I=(x^3)/3. Now you have to work out the value of this expression when x=4 and subtract the value where x=0, this gives the area. ie. A=(4^3)/3-(0^3)/3=4*4*4/3=21.3333... Rob * Sent from RemarQ http://www.remarq.com The Internet's Discussion Network * The fastest and easiest way to search and participate in Usenet - Free!
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X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: f458d,2bcea0a7b99fab7 X-Google-Attributes: gidf458d,public X-Google-Thread: 10d67e,2bcea0a7b99fab7 X-Google-Attributes: gid10d67e,public X-Google-Thread: 102524,2bcea0a7b99fab7 X-Google-Attributes: gid102524,public From: Robert Simon Martin <robNOroSPAM@rdale.force9.co.uk.invalid> Subject: Re: Area under a graph !help! Date: 2000/03/08 Message-ID: <14fbf8d8.dc379e3c@usw-ex0102-016.remarq.com> X-Deja-AN: 594881882 References: <8a6for$bcs$1@gxsn.com> Bytes: 670 X-Originating-Host: 193.113.185.167 X-Complaints-To: wrenabuse@remarq.com X-Trace: WReNphoon2 952551395 10.0.2.16 (Wed, 08 Mar 2000 13:36:35 PST) Organization: http://www.remarq.com: The World's Usenet/Discussions Start Here NNTP-Posting-Date: Wed, 08 Mar 2000 13:36:35 PST Newsgroups: uk.education.maths,alt.math,aus.mathematics X-Wren-Trace: eODF7ez1svizsvvqxbLl6P72vuPq9qPzvKLs7Kfh8uC+v/Ovv/GmoL25sau/9qQ=