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Challenging problems Area between curves
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Starter: solve the following simultaneous equations
Areas between curves KUS objectives BAT use integration to find the area between a curve and the x-axis BAT use integration to find the area between a line and curve or between two curves Starter: solve the following simultaneous equations π¦= π₯ 2 +5π₯+4 π¦=π₯+1 π₯ 2 + π¦ 2 =13 π¦=π₯+1
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To calculate the Area between a Curve and a Straight Line
Introduction To calculate the Area between a Curve and a Straight Line To work out the Region between 2 lines, you work out the region below the βhigherβ line, and subtract the region below the βlowerβ line y Region R y2 y1 x a b ο Sometimes you will need to work out the values of a and b ο Sometimes a and b will be different for each part ο MAKE SURE you put y1 and y2 the correct way around!
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Expand and rearrange (higher equation β lower equation)
WB13 Below is a diagram showing the equation y = x, as well as the curve y = x(4 β x). Find the Area bounded by the line and the curve. y 1) Find where the lines cross (set the equations equal) y = x Expand the bracket R Subtract x Factorise x 3 2) Integrate to find the Area y = x(4 β x) Expand and rearrange (higher equation β lower equation) Integrate Split and Substitute
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WB14 The diagram shows a sketch of the curve with equation y = x(x β 3), and the line with Equation 2x. Calculate the Area of region R. 1) Work out the coordinates of the major points.. x y y = x(x β 3) y = 2x R O A B C As the curve is y = x(x β 3), the x-coordinate at C = 3 ο Set the equations equal to find the x-coordinates where they crossβ¦ Expand Bracket Subtract 2x Factorise 2) Area of the Triangleβ¦ The Area we want will be The Area of Triangle OAB β The Area ACB, under the curve. Substitute values in Work it out!
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Area of Triangle OAB β The Area ACB
WB14 continued The diagram shows a sketch of the curve with equation y = x(x β 3), and the line with Equation 2x. Calculate the Area of region R. 3) Area under the curve Expand Bracket y = x(x β 3) y y = 2x (5,10) Integrate Split and Substitute 16 1/3 R x 3 5 Area of Triangle OAB β The Area ACB 25 - 26/3
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1) Find the area enclosed between π¦= π₯ 2 +2π₯+3 and π¦=π₯+5
Practice 1 1) Find the area enclosed between π¦= π₯ 2 +2π₯ and π¦=π₯+5 2) Find the area enclosed between π¦= π₯ 2 β2π₯ and π¦=5βπ₯ 3) Find the area enclosed between π¦= π₯ 2 β3π₯ and π¦=5βπ₯ Solutions Intersection points (-2, 3) (1, 6) Area = 4.5 Intersection points (-1, 6) (2, 3) Area = 9 2 Intersection points (0, 5) (2, 3) Area = 4 3
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Practice 1 1) Find the area enclosed between π¦= π₯ 2 +2π₯ and π¦=π₯+5 πππ‘πππ πππ‘πππ πππππ‘π β2, 3 πππ (1, 6) π₯+5 β π₯ 2 +2π₯+3 =2βπ₯β π₯ 2 β2 1 2βπ₯β π₯ 2 = 2π₯β 1 2 π₯ 2 β 1 3 π₯ β2 = 2(1)β β 1 3 (1) 3 β 2(β2)β β2 2 β 1 3 (β2) 3 = 2β 1 2 β 1 3 β β4β2+ 8 3 = β β = 9 2
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Practice 2 2) Find the area enclosed between π¦= π₯ 2 β2π₯ and π¦=5βπ₯ πππ‘πππ πππ‘πππ πππππ‘π β1, 6 πππ (2, 3) 5βπ₯ β π₯ 2 β2π₯+3 =2+π₯β π₯ 2 β π₯β π₯ 2 = 2π₯ π₯ 2 β 1 3 π₯ β1 = 4+2β 8 3 β β = β β = 9 2
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= 4 3 3) Find the area enclosed between π¦= π₯ 2 β3π₯+5 and π¦=5βπ₯
Practice 3 3) Find the area enclosed between π¦= π₯ 2 β3π₯ and π¦=5βπ₯ πππ‘πππ πππ‘πππ πππππ‘π 0, 5 πππ (2, 3) 5βπ₯ β π₯ 2 β3π₯+5 =2π₯β π₯ 2 0 2 2π₯β π₯ 2 = π₯ 2 β 1 3 π₯ = 4β 8 3 β 0 = 4 3
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One thing to improve is β
KUS objectives BAT use integration to find the area between a curve and the x-axis BAT use integration to find the area between a line and curve or between two curves self-assess One thing learned is β One thing to improve is β
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