# Math RE - Calculus II Area Page 1 of 12

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1 Mth --RE - Clculus II re Pge of re nd the Riemnn Sum Let f) be continuous function nd = f) f) > on closed intervl,b] s shown on the grph. The Riemnn Sum theor shows tht the re of R the region R hs re= f)d The re is evluted b the definite integrl. The Riemnn Sum theor lso shows tht b b if f) < on,b], the definite integrl will give negtive nswer negtive re??) R To void this sitution, lws sketch the region of the re; if the region is bove the is, it s fine; = f) if the region is below the is, tke the bsolute vlue of the definite integrl. Emple : Find the re of the region bounded b f) = 4, the is from = to = 6. The grph shows tht the region is bove the is. = f) Then the re: 6 4 d = ) 4 6 ] ] 4 4 6) ) = 88 6 = 5 squre units Note re is lws in squre units. Definite integrls hve no units. 6

2 Mth --RE - Clculus II re Pge of Emple : Find the re of the region bounded b f) = 4, the is from = to = 4. The grph shows tht the region is bove the is. Then the re: 4 44) ] 4) 4) ] ) 8 6 = squre units 4 )d = 4 ) 4 4 Emple : Find the re of the region bounded b f) =, the is from = to =. The grph shows tht the region is below the is; must insert negtive sign in the definite integrl. Then the re: ] ] 4 )4 4 )4 )d = ) = 4 squre units Note: the nswer of the re is lws positive number

3 Mth --RE - Clculus II re Pge of Emple 4: Find the re of the region bounded b f) =, the is from = to =. The grph shows tht the region is in prts. region from = to =, the region is below the is; must insert negtive sign in the definite integrl region from = to =, the region is bove the is; Therefore: = )d = ) = 4 )4 + ] ) 4 )4 + ] ) = 4 = = )d = 4 4 ) 4 )4 ] ) 4 )4 ] ) = 9 4 totl re: + = = 5 =.5 squre units

4 Mth --RE - Clculus II re Pge 4 of re of region between two grphs Let two continuous functions f) nd g) To find the points of intersection: mke f) = g) nd solve for. Let them be = nd = b, then the points of intersection re:,f) = g) ) nd b,fb) = gb) ). Find the re of the region between the two grphs see grph) re= f) g)] d Visul proof is shown below. In generl, the region R is evluted between = nd = b. Integrte the difference between highest function nd lowest function. Note: Remember to grph the region between curves if requested. f) R g) b f) f) f) R g) = g) g) R R b b b re = f) g)] d = f)d g) d re of R = f) g)] d = re of R re of R re of R = f) g)] d Note: Remember to grph the region between curves if requested.

5 Mth --RE - Clculus II re Pge 5 of Emple 5: Find the re of the region bounded b f) = + 4, g) = + from = to =. The grph shows no point of intersection since + 4 = + + = no solution. 5 f) re of the region: f) g)] d 4 + 4) + ) ] d = + ] d g) = ) + ) ] ) + ) ) ] ) + ) 7 6 = squre units Emple 6: Find the re of the region bounded b f) = +, g) =. The grph shows two points of intersection: = + = ) + ) = = = f ) = g ) = = = f) = g) = re of the region: = f) g)] d + ) ) ] d = + ) ] d ) + ] ) + ) ) + ] ) + ) f) g) 7 6 = 9 = 4.5 squre units

6 Mth --RE - Clculus II re Pge 6 of Emple 7: Find the re of the region bounded b f) = 5, the is from = to =. The grph shows no point of intersection between = nd = re of the region: g) f)] d where f) = 5 nd g) = is) ) 5) ] d = + 5 ] d ) ] 4 )4 + 5 ) ] 4 )4 + 5) = squre units Emple 8: Find the re of the region bounded b f) = +, g) = +. 5 The grph shows tht three points of intersection: + = + = ) + ) = = = f ) = g ) = = = f) = g) = ; = = f) = g) = 5 re of the region: + = = f) g)] d + g) f)] d + ) + ) ] d = 4 4 ] = 5 ] d = + ) + ) ] d = + + ] d = ] + = 8 totl re: + = = 7.8 squre units

7 Mth --RE - Clculus II re Pge 7 of Emple 9: Find the re of the region bounded b f) =, g) = from = to =. The grph shows tht two points of intersection: = = ) = = = f) = g) = = = f) = g) = ; = f) = ; g) = re of the region: + = g) f)] d + ) ) ] d = = + ] = 6 f) g)] d + ] d f) g) = ) ) ] d = ] d = ] = totl re: + = 6 + =.67 squre unit

8 Mth --RE - Clculus II re Pge 8 of Consumer Surplus Given the demnd D) = p) nd the suppl S) = p) To find the equilibrium point: mke D) = S) = p) nd solve for. Let e nd e = p e be the coordintes of equilibrium point; then the consumer surplus is evluted: e = p e C.S. D) C.S. = e D) e ] d S) C.S. is the region s shown on the grph. Note: Remember to grph the region between curves if requested. e Producer Surplus Given the demnd D) = p) nd the suppl S) = p) To find the equilibrium point: mke D) = S) = p) nd solve for. D) Let e nd e = p e be the coordintes of equilibrium point; e = p e then the producer surplus is evluted: P.S. e e S) ] d S) P.S. is the region s shown on the grph. Note: Remember to grph the region between curves if requested. e

9 Mth --RE - Clculus II re Pge 9 of Emple : Given the demnd p = + t the equilibrium quntit of 7 units, sketch nd identif the C.S. region, then evlute the C.S. The grph shows the C.S. region. To find the equilibrium price: replce e = 7 p) p = + in the demnd eqution: p e = 7 + = 5 C.S. C.S. = e demnd p e ] d 5 C.S. = C.S. = 7 + ) 5 ] d = ] d ) = 686 = 686 \$ Emple : Given the suppl p = t the equilibrium price of \$/per unit, sketch nd identif the P.S. region, then evlute the P.S. The grph shows the P.S. region. p) To find the equilibrium quntit: replce p e = in the demnd eqution: = P.S. + 9 = + ) 9) = e = 9 e p e suppl ] d 4 p = ) ] d + 9 ] d = ) = 5 \$56.5

10 Mth --RE - Clculus II re Pge of Emple : Given the demnd p = + 64 nd the suppl p = + 9, sketch nd identif the C.S. nd the P.S. regions, then evlute the P.S. To find the equilibrium price: mke demnd = suppl p) 64 p = = = 5) + 7) = The equilibrium point t 5,9) e p e suppl ] d 9 9 P.S. C.S. p = ) ] d = 5 ) 5 + = 5 = \$5. + ] d 5 re of region between two curves one or two is/re functions) of f) ) If one of the two curves is function of like = f) To find the points of intersection: mke f) = g) nd solve for. Let them be = nd = b, then the points of intersection re: ) ) f) = g), nd fb) = gb),b. = g) = b = f) Find the re of the region between the two grphs b integrting with respect to see grph) re = f) g)] d R Visul proof is shown below. = In generl, the region R is evluted between = nd = b. Integrte the difference between right function nd left function. Note: Remember to grph the region between curves if requested.

11 Mth --RE - Clculus II re Pge of Emple : Find the re of the region bounded b f) = + 4 nd g) = The grph shows the shded region hs two points of intersections: g) = f) = = = ) + ) = = = f) = g) = 5 = = f ) = g ) = = g) = f) re: f) g)] d + 4) ) ] d ] d + ) = squre units 5 4 Emple 4: Find the re of the region bounded b f) = + nd g) = The grph shows the shded region hs two points of intersections: g) = f) = = + = g) = f) = ) = = = f) = g) = = = f) = g) = re: f) g)] d + ) ) ] d + ] d + ) 6 =.7 squre unit 6

12 Mth --RE - Clculus II re Pge of Emple 5: Find the re of the region bounded b f) = 4 8, g) = 8 between = nd = The grph shows three points of intersections: f) = g) = 8 = = ) + ) = = = f ) = g ) = 6 = = f) = g) = 8 = f) = 4 ; g) = 7 = g) = f) re: f) g)] d 4 8) 8) ] d 4 ] d 4 ) = 9 =.5 squre units 4 6 8

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