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The FUNDAMENTAL Theorem of Calculus (yay!) 11/11/11 (also yay!) - PDF document

The FUNDAMENTAL Theorem of Calculus (yay!) 11/11/11 (also yay!) Warm-up Suppose a particle is traveling at velocity v ( t ) = t 2 from t = 1 to t = 2. if the particle starts at y (0) = y 0 , 1. what is the function y ( t ) which gives the


  1. The FUNDAMENTAL Theorem of Calculus (yay!) 11/11/11 (also yay!) Warm-up Suppose a particle is traveling at velocity v ( t ) = t 2 from t = 1 to t = 2. if the particle starts at y (0) = y 0 , 1. what is the function y ( t ) which gives the particles position as a function of time (will have a y 0 in it)? 2. how far does the particle travel from t = 1 to t = 2? Compare your answer to the upper and lower estimates of the area under the curve f ( x ) = x 2 from x = 1 to x = 2: Upper U ( f , P ) Lower L ( f , P ) n � 1 n ✓ ◆ ✓ 1 ◆ ✓ ◆ ✓ 1 ◆ 1 + i 1 + i X X ∗ ∗ n n n n i =1 i =0 n U ( f , P ) L ( f , P ) 10 2.485 2.185 100 2.34835 2.31835 1000 2.33483 2.33183

  2. The Fundamental Theorem of Calculus Theorem (the baby case) d If F ( x ) is any function satisfying dx F ( x ) = f ( x ) , then Z b b � f ( x ) dx = F ( x ) x = a = F ( b ) − F ( a ) � � a Z 2 x 2 dx ? Q. What is 1 A. F ( x ) = x 3 3 So 2 Z 2 � x 2 dx = x 3 ✓ 2 3 ✓ 1 3 ◆ ◆ � = � − 3 3 3 � 1 � x =1 = 8 3 − 1 3 = 7 3 ≈ 2 . 333 (same answer!) Examples Use the fundamental theorem of calculus, Z b f ( x ) dx = F ( b ) − F ( a ) a to calculate Z 3 1. 3 x dx 2 Z 1 x 3 dx 2. � 1 Z π 3. sin( x ) dx 0 Z 0 4. sin( x ) dx π

  3. The Fundamental Theorem of Calculus Theorem (the big case) If F ( x ) is any function satisfying d dt F ( t ) = f ( t ) , then Z b ( x ) � ( x ) b ( x ) f ( t ) dt = F ( t ) t = a ( x ) = F ( b ( x )) − F ( a ( x )) � � a Z ln( x ) t 2 dt ? Q. What is sin( x ) A. F ( t ) = 1 3 t 3 . So ln( x ) Z ln( x ) � t 2 dt = 1 ✓ 1 ◆ ✓ 1 ◆ � 3 t 3 3(ln( x )) 3 3(sin( x )) 3 = . � − � sin( x ) � t =sin( x ) Examples Use the fundamental theorem of calculus, Z b ( x ) f ( t ) dt = F ( b ( x )) − F ( a ( x )) a ( x ) to calculate Z cos( x ) 1. 3 t dt sin( x ) Z 5 x 2 � 3 t 3 dt 2. x +1 Z 0 3. sin( t ) dt arccos( x )

  4. Z ln( x ) For reference, we calculated f ( t ) dt where a ( x ) f ( t ) = t 2 a ( x ) = sin( x ) b ( x ) = ln( x ) . Notice: ✓ 1 ◆ d 3(ln( x )) 3 − 1 = 1 3(sin( x )) 3 x (ln( x )) 2 − cos( x )(sin( x )) 2 dx = b 0 ( x ) f ( b ( x )) − a 0 ( x ) f ( a ( x )) . In general: Z b ( x ) d f ( t ) dt = b 0 ( x ) f ( b ( x )) − a 0 ( x ) f ( a ( x )) . dx a ( x ) (Don’t even have to know F ( t )!) Why? Z sin( x ) Example: Calculate d e t 2 dt . dx tan( x ) Z e t 2 dt ! Answer: We can’t even calculate (There is no elementary function F ( t ) which satisfies F 0 ( t ) = e t 2 ) e t 2 dt is a function. Call it F ( t ). Z But we know Z sin( x ) e t 2 dt = F (sin( x )) − F (tan( x )). So tan( x ) Z sin( x ) d e t 2 dt = d Therefore dx ( F (sin( x )) − F (tan( x ))) dx tan( x ) = cos( x ) F 0 (sin( x )) − sec 2 ( x ) F 0 (tan( x )) = cos( x ) f (sin( x )) − sec 2 ( x ) f (tan( x )) = cos( x ) e (sin( x )) 2 − sec 2 ( x ) e (tan( x )) 2

  5. Part 2: Integration by substitution Warmup Fill in the blank: 1. Since d dx cos( x 2 + 1) = , Z dx = cos( x 2 + 1) + C . so 2. Since d dx ln | cos( x ) | = , Z so dx = ln | cos( x ) | + C . 3 x 2 dx = x 3 + C .) dx x 3 dx = 3 x 2 , so d R (Example:

  6. Undoing chain rule In general: d dx f ( g ( x )) = f 0 ( g ( x )) ∗ g 0 ( x ) , Z f 0 ( g ( x )) ∗ g 0 ( x ) dx = f ( g ( x )) + C . so Example: Calculate the (extremely suggestively written) integral Z cos( x 3 + 5 x − 10) ∗ (3 x 2 + 5 ∗ 1 + 0) dx Less obvious chain rules. Look for a buried function g ( x ) and it’s derivative g 0 ( x ) which can be paired with dx : Examples: Z cos( x ) Z 1 sin( x ) + 1 dx = sin( x ) + 1 ∗ cos( x ) dx Z 1 Z p x 2 + 1 dx = p x 2 + 1 ∗ 2 x dx x 2 Z cos( √ x ) Z 1 cos( √ x ) ∗ dx = 2 √ x dx 2 √ x

  7. Method of Substitution Look for a buried function g ( x ) and it’s derivative g 0 ( x ) which can be paired with dx . Z cos( x ) Example: sin( x ) + 1 dx Let u = g ( x ). Let u = sin( x ) + 1 du Calculate du . dx = cos( x ) so du = cos( x ) dx Z 1 Clear out all of the x ’s, u du replacing them with u ’s. R 1 Calculate the new integral. u du = ln | u | + C Substitute back into x ’s. ln | u | + C = ln | sin( x ) + 1 | + C 1 d Check dx ln | sin( x ) + 1 | + C = sin( x )+1 ∗ cos( x ) X Method of Substitution Look for a buried function g ( x ) and it’s derivative g 0 ( x ) which can be paired with dx . Z p x 2 + 1 dx Example: x Let u = x 2 + 1 Let u = g ( x ). dx = 2 x so 1 du Calculate c ∗ du . 2 du = x dx Z √ u ∗ 1 Clear out all of the x ’s, 2 du replacing them with u ’s. � 2 1 u 1 / 2 du = 1 3 u 3 / 2 � R Calculate the new integral. + C 2 2 3 ( x 2 + 1) 3 / 2 + C = 1 Substitute back into x ’s. 3 ( x 2 + 1) 3 / 2 + C = 1 2 ( x 2 + 1) 1 / 2 ∗ 2 x X 1 3 d Check dx 3

  8. Give it a try: Look for a buried function g ( x ) and it’s derivative g 0 ( x ) which can be paired with dx . Z e p x Example: √ x dx Let u = g ( x ). Calculate c ∗ du . Clear out all of the x ’s, replacing them with u ’s. Calculate the new integral. Substitute back into x ’s.

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