MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 1 of 12 MARK_________/54 =______________% EXAMPLE 1: Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = x, x = 2, x = 5 Method I: Geometry Method II: Calculus Area Rectangle = Area Formula = antiderivative of f(x) 1 2x3=6 π΄ = 2 π₯2 Area of triangle = π π π (π)π β (π)π = ππ. π (3)(3) = 4.5 π¨ = π π π Total area = 6 + 4.5 = 10.5 Method III: Calculator nd 2 CALC 7:β«f(x) dx Lower limit 2. Upper limit 5. 5 Area = 10.5 β«2 (π₯)ππ₯ = ππ. π EXAMPLE 2: Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = β 2x β 4, x = β2 0 x=0 Method I: Geometry Area Triangle = π (2)(4) = 4 π Method II: Calculus Area Formula = antiderivative of f(x) A = β x2 β 4x A = | [ β(0)2 β 4(0)] β [β(β 2 )2 β 4(β 2)] | A=|0β4|=|β4|=4 β2 β4 Method III: Calculator 2nd CALC 7:β«f(x) dx Lower limit β 2 Upper limit 0 0 β«β2(β2π₯ β 4)ππ₯ = β π BUT Area = 4 Areas below the x-axis work out to be negative but because we are thinking of them as areas we take the absolute value of the answer. U7 L3 ANS Areas under Curves MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 2 of 12 EXAMPLE 3: Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = x β 2 , x=β1 x=4 Determine the x-intercept algebraically. Use geometry to find the areas of the 2 triangles. π¨ = (π)(π) (π)(π) + = π. π π π If the x -intercept lies between the lower limit and the upper limit then you must determine the negative part of the area take the absolute value then find the positive part of the area and add the two answers. Find the antiderivative and use this to find the areas of the 2 triangles. π A = x2 β 2x π π π A = | [π (2)2 β 2(2)] β [π (β1)2 β 2(-1)] | = | β 4.5| = 4.5 π π A = | [π (4)2 β 2(4)] β [π (2)2 β 2(2)] | = |2| = 2 Total area = 6.5 4 Compare to the definite integral β« (π₯ β 2)ππ₯ β1 π π π β ππ π π π [ (π)π β π(π)] β [ (βπ)π β π(βπ)] π π πβ π π = β = βπ. π π π The definite integral is not equal to the area if part of the area is below the x-axis. U7 L3 ANS Areas under Curves MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 3 of 12 EXAMPLE 4: Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = x2 β 4 , x=0 x=4 Determine the x-intercepts and label. x2 β 4 = 0 x=+2 y-axis . Shade the areas (β 2, 0) ( 2, 0) x=0 Find the antiderivative and use this to find the areas. π π¨ = ππ β ππ π π π ππ ππ |[ (π)π β π(π)] β [ (π)π β π(π)]| = | | = π π π π π π ππ ππ |[ (π)π β π(π)] β [ (π)π β π(π)]| = |β | = π π π π Total Area ππ ππ ππ + = = ππ π π π U7 L3 ANS Areas under Curves x-axis x=4 MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 4 of 12 EXAMPLE 5: Find the area between the graphs of y = x2 + 2 and y = x and the lines x = 1 and x = 4. WINDOW [-1, 5, 1] [-2, 20, 1] Area under y = x2 + 2 π¨= π π π + ππ π π π |[ (π) + π(π)] β [ (π) + π(π)]| = ππ Antiderivative of y =πx is π Area under y = x π¨= π π π π π π ππ |[ (π)] β [ (π)]| = π π π U7 L3 ANS Areas under Curves Area between the curves is ππ ππ ππ β = π π MATH 31 LESSON #3 AREA UNDER THE CURVE /3 NAME ANSWERS Page 5 of 12 QUESTIONS 1. Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = x + 2, x = - 2, x = 2 Method I: Geometry Method II: Calculus Area Formula = π Area of triangle = π (π)(π) antiderivative of f(x) π A = π x2 + 2x Total area = 8 π π π¨ = |[π (π)π + π(π)] β [π (βπ)π + π(βπ)]| π¨ = |π β (βπ)| = |π| = π /4 2. Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = 6 β 2x, x = - 1, x = 3 Method I: Geometry Method II: Calculus Area Formula = π Area of triangle = π (4)(8) antiderivative of f(x) Total area = 16 A = 6x β x 2 A = [6(3) β (3)2] β [6(-1) β (-1)2] A = (18 β 9) β ( -6 β 1 ) = 16 π¨ = |[π(π) β (π)π )] β [π(βπ) β (βπ)π ]| π¨ = |π β (βπ)| = |ππ| = ππ Use the antiderivative of the function to determine the following areas 3. Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = 2x2 + 4, x=β2 x=2 /3 π¨= π π π + ππ π π π π¨ = |[π (π)π + π(π)] β [π (βπ)π + π(βπ)]| π¨ = |[ ππ ππ ππ ππ ] β [β ]| = | | = π π π π U7 L3 ANS Areas under Curves MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS 4. f(x) = 4x3 x = 0 /3 Page 6 of 12 x=2 A= x4 A = [ (2)4 ] β [ (0)4 ] A = 16 5. f ( x) ο½ /3 3 x2 x=1 x=3 βπ π¨ = βππβπ = π βπ βπ π¨=[ ]β[ ] π π π¨ = βπ + π = π 6. a) Find the area between the graph of y = x, the x-axis and x = β 3 and x = 2 π¨= x-intercept is (0, 0) s π π π π Area between β 3 and 0 π π π π π¨ = | (π)π β (βπ)π | = |β | = π π π π Area between 0and 2 π π π¨ = | (π)π β (π)π | = |π| = π π π 2 b) Evaluate the following definite integral. β« π₯ ππ₯ = β3 π π π π π π (π)π β (βπ)π π π /7 πβ U7 L3 ANS Areas under Curves π π =β π π Total Area is π ππ +π= π π MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 7 of 12 7. Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = x2 β 1 , x=β2 x=0 a) Determine the x-intercepts. x2β1=0 x=+1 b) Shade the required areas. x-axis /6 y-axis x = -2 x=0 c) Find the antiderivative and use it to find the required areas. π¨= π π π βπ π π π π π π π¨ = |[ (βπ)π β (βπ)] β [ (βπ)π β (βπ)]| = | β (β )| = π π π π π π π π π π π¨ = |[ (π)π β (π)] β [ (βπ)π β (βπ)]| = |π β ( )| = |β | = π π π π π Total Area is π π + =π π π U7 L3 ANS Areas under Curves MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 8 of 12 8. Find the area between the graph of y = f(x), the x-axis and the given lines. f(x) = x3 , x=β1 x=1 a) Determine the x-intercepts. x=0 b) Shade the required areas. x = -1 x=1 /6 c) Find the antiderivative and use it to find the required areas. π¨= π π π π π π π π π¨ = |[ (π)π β (βπ)π ]| = |β | = π π π π π π π π π¨ = |[ (π)π β (π)π ]| = | | = π π π π Total Area is π π π + = π π π U7 L3 ANS Areas under Curves MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS 9. Find the area enclosed between the graphs of y = x and y = x2. WINDOW [-1, 3, 1] [-1, 2, 1] Page 9 of 12 Algebraically find the points where the two curves intersect. x2 = x x2 β x = 0 x(x β 1) = 0 x = 0 or x = 1 (0, 0) (1, 1) Area under y = x between the points of intersection. π¨= π π π π π π π π π¨ = |[ (π)π β (π)π ]| = | | = π π π π /6 Area under y = x2 between the points of intersection. π¨= π π π π π π π π π¨ = |[ (π)π β (π)π ]| = | | = π π π π Area between the curves is π π π β = π π π 10. Find the area in the first quadrant, between the graphs of U7 L3 ANS Areas under Curves π¦= 1 π₯2 and y = x2 MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS and the line x = 2. WINDOW [0, 5, 1] [ β1, 5, 1] Page 10 of 12 (1, 1) Algebraically find the point in the first quadrant where the two curves intersect. π ππ = ππ π ππ = π π = ±π (π, π) ππ ππ πππππ ππππ ππππ Area under y = x2 between the point of intersection and x = 2. π¨= /6 π π π π π π π -1 π π ο2 π¨ = |[ (π)π βof y(π) ]| is= | -x |= Antiderivative ο½ x π π π π Area under y ο½ x ο2 between the point of intersection and x = 2. πβπ+π π π¨= = βπβπ = β βπ + π π π βπ π π π¨ = |[β β ]| = | | = π π π π Area between the curves is π π ππ β = π π π U7 L3 ANS Areas under Curves MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS 11. Find the area enclosed between the graphs of y = x2 and y = x + 6. WINDOW [-5, 5, 1] [-1, 15, 1] Algebraically find the points where the two curves intersect. ππ = π + π ππ β π β π = π (π β π)(π + π) = π π = π ππ π = βπ Area under y = x + 6 between the points of intersection. /6 π¨= π π π + ππ π π π ππ ππ π¨ = |[ (π)π + π(π)] β [ (βπ)π + π(βπ)]| = | | = π π π π Area under y = x2 between the points of intersection. π π¨ = ππ π π π ππ ππ π¨ = |[ (π)π β (βπ)π ]| = | | = π π π π Area between the curves is ππ ππ πππ β = π π π U7 L3 ANS Areas under Curves Page 11 of 12 MATH 31 LESSON #3 AREA UNDER THE CURVE NAME ANSWERS Page 12 of 12 BONUS CHALLENGE QUESTION π 12. y = sin (x) and y = cos (x), between the y-axis and 4 radians along the positive x-axis. π = π¬π’π§ π π βπ ( , ) π π π 4 π = ππ¨π¬ π Algebraically find the points where the two curves intersect. sin x = cos x π¬π’π§ π ππ¨π¬ π = ππ¨π¬ π ππ¨π¬ π π = πππ§βπ π = π π π βπ ( , ) π π tan x = 1 /6 Area under y = cos x between the y βaxis and the point of intersection. π¨ = π¬π’π§ π π βπ βπ [π¬π’π§ ] β [π¬π’π§ π] = βπ= π π π Area under y = sin x between the y βaxis and the point of intersection. π¨ = β ππ¨π¬ π π βπ [βππ¨π¬ ] β [β ππ¨π¬ π] = β +π π π Area between the curves is πβπ βπ βπ βπ βπ β (β + π) = + βπ= βπ π π π π π βπ β π U7 L3 ANS Areas under Curves
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