MATH 12002 - CALCULUS I 7.1: Area Between Curves Professor Donald - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 7.1: Area Between Curves Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 10

Area Between Curves We saw previously that the area of the region bounded by the graph of a function and the x -axis can be computed in term of definite integrals. We now generalize this to compute the area of a region bounded by the graphs of two functions, such as the area of the shaded region S in the figure below: § 7.1 FIGURE 1 D.L. White (Kent State University) 2 / 10

Area Between Curves Theorem If f ( x ) and g ( x ) are continuous functions such that f ( x ) � g ( x ) on the interval [ a , b ] , then the area of the region bounded by the graphs of f and g and the lines x = a and x = b is � b A = [ f ( x ) − g ( x )] dx . a The formula applies only when f ( x ) � g ( x ) for all x in [ a , b ]. We will see later how to adapt it to more general situations. If f and g are both positive on [ a , b ], then the area A between the � b graphs of f and g is the area under the graph of f , that is, a f ( x ) dx , � b minus the area under the graph of g , that is a g ( x ) dx , hence � b � b � b A = f ( x ) dx − g ( x ) dx = [ f ( x ) − g ( x )] dx . a a a The formula is proved in general using Riemann sums in the text. D.L. White (Kent State University) 3 / 10

Examples 1 Find the area of the region between the graphs of y = x 2 and y = x on the interval [0 , 1]. We need to find the area of the shaded region in the figure below: Since x � x 2 for 0 � x � 1, the area is � 1 1 � 1 2 x 2 − 1 �� = 1 2 − 1 3 = 1 x − x 2 dx = 3 x 3 � A = 6. � � 0 0 D.L. White (Kent State University) 4 / 10

Examples If two graphs intersect at two or more points, we can find the area of the region bounded by the two graphs between their first (left-most) intersection point and their last (right-most) intersection point. 2 Find the area of the region bounded by the graphs of y = x 2 − 3 and y = 5 − x 2 . We first need to determine where the graphs intersect. An intersection point occurs at an x -value where the y -values are equal, so we equate the two y -values and solve the equation for x : x 2 − 3 = 5 − x 2 , 2 x 2 − 8 = 0 , 0 = 2( x 2 − 4) = 2( x − 2)( x + 2) . Therefore, the two graphs intersect at x = − 2 and x = 2. D.L. White (Kent State University) 5 / 10

Examples [Example 2, continued] We therefore want to find the area of the shaded region in the figure below: Since 5 − x 2 � x 2 − 3 for − 2 � x � 2, the area is � 2 � 2 (5 − x 2 ) − ( x 2 − 3) dx = 8 − 2 x 2 dx . A = − 2 − 2 D.L. White (Kent State University) 6 / 10

Examples [Example 2, continued] Finally, we have � 2 8 − 2 x 2 dx A = − 2 2 � 8 x − 2 �� 3 x 3 � = � � − 2 � 8(2) − 2 � � 8( − 2) − 2 � 3(2 3 ) 3(( − 2) 3 ) = − � 16 − 16 � � − 16 + 16 � = − 3 3 16 − 16 3 + 16 − 16 = 3 32 − 32 3 = 64 = 3 . D.L. White (Kent State University) 7 / 10

Examples 3 Find the area of the region bounded by the graphs of y = x 2 − 2 x and y = x + 4. We first determine the points of intersection. We set x 2 − 2 x = x + 4 , so that 0 = x 2 − 3 x − 4 = ( x + 1)( x − 4) , and so the graphs intersect at x = − 1 and x = 4. Hence we want to find the area of the shaded region in the following figure: D.L. White (Kent State University) 8 / 10

Examples [Example 3, continued] Since x + 4 � x 2 − 2 x for − 1 � x � 4, the area is � 4 � 4 ( x + 4) − ( x 2 − 2 x ) dx = 4 + 3 x − x 2 dx . A = − 1 − 1 D.L. White (Kent State University) 9 / 10

Examples [Example 3, continued] Finally, we have � 4 4 + 3 x − x 2 dx A = − 1 4 �� � 4 x + 3 2 x 2 − 1 3 x 3 � = � � − 1 � � � � 4(4) + 3 2(4 2 ) − 1 4( − 1) + 3 2(( − 1) 2 ) − 1 3(4 3 ) 3(( − 1) 3 ) = − � � � � 16 + 24 − 64 − 4 + 3 2 + 1 = − 3 3 40 − 64 3 + 4 − 3 2 − 1 3 = 125 = 6 . D.L. White (Kent State University) 10 / 10