1. Area under a curve region bounded by the given function, vertical lines and the x axis.


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1 Ares y Integrtion. Are uner urve region oune y the given funtion, vertil lines n the is.. Are uner urve region oune y the given funtion, horizontl lines n the y is.. Are etween urves efine y two given funtions.. Are uner urve region oune y the given funtion, vertil lines n the is. If f() is ontinuous n nonnegtive funtion of on the lose intervl [, ], then the re of the region oune y the grph of f, the is n the vertil lines = n = is given y: Are f ( ) A When lulting the re uner urve f(), follow the steps elow:. Sketh the re.. Determine the ounries n,. Set up the efinite integrl,. Integrte. E.. Fin the re in the first qurnt oune y Grph: f ( ) n the is. To fin the ounries, etermine the interepts: f ( ) ( ) or ( ) so Therefore the ounries re n n Pge of 9
2 Set up the integrl: Solve: A f ( ) ( ) ( ) 6 6 The re in the first qurnt uner the urve f ( ) E.. Fin the re oune y the following urves:, Grph: is equl to y y,, squre units A Fining the ounries: y, n y implies so or From the grph we see tht is our ounry t. The vlue is solution to the eqution ove ut it is not ouning the re. (Here s why the grph is n importnt tool to help us etermine orret results. Don t skip this step!) The other ounry vlue is given y the eqution of the vertil line, Bounries re:, n, Set up the integrl: Solve: A f ( ) ( ) ( ) The re oune y the urves y, y,, 8 is equl to squre units. Pge of 9
3 . Are uner urve given funtion, region oune y the horizontl lines n the y is. In ertin prolems it is esier to rewrite the funtion in terms of y n lulte the re using horizontl elements. In this se the formul for the re woul e: Are g y y When lulting the re uner urve, or in this se to the left of the urve g(y), follow the steps elow:. Sketh the re.. Determine the ounries n,. Set up the efinite integrl,. Integrte. E.. Fin the first qurnt re oune y the following urves:. y, y vertil elements n horizontl elements A Typil retngle (with with of ) Typil retngle (with with of y) There re two wys to solve this prolem: we n lulte the re etween two funtions y n y using the vertil elements n integrte with respet to, or we n use the horizontl elements n lulte the re etween the yis n the funtion y integrting the funtions with respet to y. We will solve it using the seon pproh y onsiering horizontl elements n the funtion in terms of y. The formul we will use is: Are gyy, so we nee to etermine the ounry vlues n first. The ounry vlue orrespons to the horizontl line y. To lulte, we nee to lote the yinterepts. Therefore set ounries re n solve the eqution y for y; hene y. So the Pge of 9
4 Now we nee to fin our g(y). Tht is esily one y solving y for. y y y We will ignore the negtive ril, sine our re is in the first qurnt. Now let s set up the integrl: A y y Solve the integrl using simple u sustitution: A units. y y y The first qurnt re oune y the following urves: y, y n squre units. squre is equl to E.. Fin the first qurnt re oune y the following urves:. y A y rtn, y n tn y thn y rtn Sine it is muh esier to integrte, we will rewrite the given funtion in terms of y, n integrte using the horizontl elements n the formul: A gyy to fin the re. The funtion y rtn implies tn y. So gy tn y The lower ounry = is esily otine from the grph or y solving the eqution rtn. The upper ounry is given y the eqution of the line: y. So the re we re looking for is given y the following integrl: A tn yy. Pge of 9
5 Solving the integrl yiels: tn yy ln os y ln ln ln ln os ln ln So the first qurnt re oune y the following urves: equl to ln squre units. ln os ln ln y rtn, y n is. Are etween two urves. This n e onsiere s more generl pproh to fining res. Thus eh of the previous emples oul hve een solve using suh n pproh y onsiering the  n y es s funtions with equtions y= n =, respetively. Mny res n e viewe s eing oune y two or more urves. When re is enlose y just two urves, it n e lulte using vertil elements y sutrting the lower funtion from the upper funtion n evluting the integrl. Anlogously, to lulte the re etween two urves using horizontl elements, sutrt the left funtion from the right funtion. As lwys, sketh of the grph n e very importnt tool in etermining the preise setup of the integrl. If you sutrt in the wrong orer, your result will e negtive. Tht mistke n e voie y tking the solute vlue of the ifferene of the funtions. Here is the universl formul for fining the re etween two urves: Using the vertil elements: Are y y Using the horizontl elements: Are y where y n y re funtions of where n re funtions of y. E.5. Fin the re of the region enlose y the following urves: n. As lwys, we will first rw sketh. y e y,, y e y Pge 5 of 9
6 In this se it is firly esy to integrte the funtions s given with respet to. So the ounries re: n. Notie tht in the region tht we re intereste in, the funtion y Solving it:, thus the integrl set up shoul look s follows: A e y e is ove the funtion e e e e e e e e e e. e e e e So the re of the region oune y y e y, n e e e squre units., is equl to, n y E.6. Fin the re of the region enlose y the following urves: y Sine the first funtion is etter efine s funtion of y, we will lulte the integrl with respet to y. As usul rw the piture first: y. y In this se the ounries re etermine y the points of intersetion of oth funtions. Rememer tht we wnt the yvlues sine we will e integrting with respet to y. We nee. This implies y y y y y y y or y n So n. The left funtion is y n the right funtion is y. y y y y y y A squre units. So the re of the region enlose y the urves: y units. 9, n y is equl to squre Pge 6 of 9
7 E.7. Fin the re of the region enlose y the following urves: n 5 As usul sketh rough grph first: y, 6 y, y y 6 5 In this se it is very importnt to rw the grph, sine the funtions interset etween the ounries. This mens tht we will hve to tully lulte two seprte integrls n then the results. Otherwise we woul en up sutrting the two piees from eh other. First we nee the mile intersetion point so we will solve the eqution: 5 5 The intersetion point t mile ounry vlue. or is outsie our re. We re intereste in In this se the integrl setup will look s follows:, this is our A squre units. 6 So the re of the region enlose y the urves: : y, y 6, n 5 57 is equl to squre units Pge 7 of 9
8 Prtie prolems: Fin the re of the region oune y the given urves. Deie whether to integrte with respet to or y.. y, y, n. y, y, n. y e, y e, n. y n y 5. y os, y sin, n Pge 8 of 9
9 Answers:. A 7. 7 A 6 A e A A Pge 9 of 9
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