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Worksheet: The Fundamental Theorem of Calculus

Q1:

Suppose that 𝑓 is a function on the interval [ π‘Ž , 𝑏 ] and we are able to define 𝐹 by 𝐹 ( π‘₯ ) = ο„Έ 𝑓 ( 𝑑 ) 𝑑 π‘₯ π‘Ž d and find that 𝐹 is NOT differentiable on ( π‘Ž , 𝑏 ) . What can we conclude?

  • A 𝑓 is not differentiable somewhere in the interval ( π‘Ž , 𝑏 ) .
  • B 𝑓 is discontinuous everywhere on ( π‘Ž , 𝑏 ) .
  • C 𝑓 is not differentiable everywhere on ( π‘Ž , 𝑏 ) .
  • D 𝑓 is discontinuous somewhere in the interval ( π‘Ž , 𝑏 ) .
  • EThere is a mistake because whenever we integrate a function, it must be differentiable and 𝐹 β€² ( π‘₯ ) = 𝑓 ( π‘₯ ) .

Q2:

Use the Fundamental Theorem of Calculus to find the derivative of the function β„Ž ( 𝑒 ) = ο„Έ √ 3 𝑑 4 𝑑 + 2 𝑑 𝑒 4 d .

  • A β„Ž β€² ( 𝑒 ) = βˆ’ 3 ( 4 𝑑 βˆ’ 2 ) 2 √ 3 𝑑 ( 4 𝑑 + 2 ) 2
  • B β„Ž β€² ( 𝑒 ) = √ 3 𝑑 4 𝑑 + 2
  • C β„Ž β€² ( 𝑒 ) = βˆ’ 3 ( 4 𝑒 βˆ’ 2 ) 2 √ 3 𝑒 ( 4 𝑒 + 2 ) 2
  • D β„Ž β€² ( 𝑒 ) = √ 3 𝑒 4 𝑒 + 2
  • E β„Ž β€² ( 𝑒 ) = βˆ’ 3 ( 4 𝑑 βˆ’ 2 ) 2 √ 3 𝑑 ( 4 𝑑 + 2 )

Q3:

Let 𝑦 = ο„Έ √ 2 + 5 𝑑 𝑑 2 2 π‘₯ 2 s i n d . Use the Fundamental Theorem of Calculus to find 𝑦 β€² .

  • A 𝑦 β€² = βˆ’  2 + 5 2 π‘₯ s i n 2
  • B 𝑦 β€² = 2 ( 2 π‘₯ )  2 + 5 2 π‘₯ c o s s i n 2
  • C 𝑦 β€² =  2 + 5 2 π‘₯ s i n 2
  • D 𝑦 β€² = βˆ’ 2 ( 2 π‘₯ )  2 + 5 2 π‘₯ c o s s i n 2
  • E 𝑦 β€² = √ 2 + 5 𝑑 2

Q4:

Given that 𝑓 ( π‘₯ ) = ο„Έ ο€Ή 8 π‘₯ βˆ’ 5 π‘₯ + 4  π‘₯ 2 d , find d d 𝑓 π‘₯ .

  • A16
  • B 8 3 π‘₯ βˆ’ 5 2 π‘₯ + 4 π‘₯ 3 2
  • C 1 6 π‘₯ βˆ’ 5
  • D 8 π‘₯ βˆ’ 5 π‘₯ + 4 2

Q5:

The figure shows the graph of the function ο„Έ 𝑓 ( 𝑑 ) 𝑑 . π‘₯ 0 d

Which of the following is the graph of 𝑦 = 𝑓 ( π‘₯ ) ?

  • A
  • B
  • C
  • D
  • Enone of the above

Q6:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑔 ( 𝑠 ) = ο„Έ ο€Ή 3 𝑑 βˆ’ 4 𝑑  𝑑 𝑠 1 3 5 4 d .

  • A 𝑔 β€² ( 𝑠 ) = 4 ο€Ή 9 𝑑 βˆ’ 2 0 𝑑  ο€Ή 3 𝑑 βˆ’ 4 𝑑  2 4 3 5 3
  • B 𝑔 β€² ( 𝑠 ) = ο€Ή 3 𝑑 βˆ’ 4 𝑑  3 5 4
  • C 𝑔 β€² ( 𝑠 ) = 4 ο€Ή 9 𝑠 βˆ’ 2 0 𝑠  ο€Ή 3 𝑠 βˆ’ 4 𝑠  2 4 3 5 3
  • D 𝑔 β€² ( 𝑠 ) = ο€Ή 3 𝑠 βˆ’ 4 𝑠  3 5 4
  • E 𝑔 β€² ( 𝑠 ) = 4 ο€Ή 9 𝑑 βˆ’ 2 0 𝑑  ο€Ή 3 𝑑 βˆ’ 4 𝑑  2 4 3 5 4

Q7:

Find the derivative of the function 𝑔 ( π‘₯ ) = ο„Έ 5 𝑑 𝑑 𝑑 1 + π‘₯ 1 βˆ’ 2 π‘₯ s i n d .

  • A 𝑔 β€² ( π‘₯ ) = ( 5 βˆ’ 2 π‘₯ ) ( 1 βˆ’ 2 π‘₯ ) + ( 5 + 5 π‘₯ ) ( 1 + π‘₯ ) s i n s i n
  • B 𝑔 β€² ( π‘₯ ) = βˆ’ ( 1 0 βˆ’ 2 0 π‘₯ ) ( 1 βˆ’ 2 π‘₯ ) + ( 5 + 5 π‘₯ ) ( 1 + π‘₯ ) s i n s i n
  • C 𝑔 β€² ( π‘₯ ) = βˆ’ ( 5 βˆ’ 1 0 π‘₯ ) ( 1 βˆ’ 2 π‘₯ ) + ( 5 + 5 π‘₯ ) ( 1 + π‘₯ ) s i n s i n
  • D 𝑔 β€² ( π‘₯ ) = ( 1 0 βˆ’ 2 0 π‘₯ ) ( 1 βˆ’ 2 π‘₯ ) + ( 5 + 5 π‘₯ ) ( 1 + π‘₯ ) s i n s i n
  • E 𝑔 β€² ( π‘₯ ) = βˆ’ ( 1 0 βˆ’ 2 0 π‘₯ ) ( 1 βˆ’ 2 π‘₯ ) βˆ’ ( 5 + 5 π‘₯ ) ( 1 + π‘₯ ) s i n s i n

Q8:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝐹 ( π‘₯ ) = ο„Έ √ 2 βˆ’ 3 𝑑 𝑑 4 π‘₯ s e c d .

  • A 𝐹 β€² ( π‘₯ ) = √ 2 βˆ’ 3 π‘₯ s e c
  • B 𝐹 β€² ( π‘₯ ) = √ 2 βˆ’ 3 𝑑 s e c
  • C 𝐹 β€² ( π‘₯ ) = βˆ’ 3 𝑑 𝑑 2 √ 2 βˆ’ 3 𝑑 s e c t a n s e c
  • D 𝐹 β€² ( π‘₯ ) = βˆ’ √ 2 βˆ’ 3 π‘₯ s e c
  • E 𝐹 β€² ( π‘₯ ) = 3 π‘₯ π‘₯ 2 √ 2 βˆ’ 3 π‘₯ s e c t a n s e c

Q9:

Find the derivative of the function 𝑦 ( π‘₯ ) = ο„Έ ( 1 βˆ’ 𝑣 ) 𝑣 4 π‘₯ 3 π‘₯ s i n c o s l n d .

  • A 𝑦 β€² ( π‘₯ ) = ( 1 βˆ’ 3 π‘₯ ) + ( 1 βˆ’ 4 π‘₯ ) l n c o s l n s i n
  • B 𝑦 β€² ( π‘₯ ) = βˆ’ 3 π‘₯ ( 1 βˆ’ 3 π‘₯ ) + 4 π‘₯ ( 1 βˆ’ 4 π‘₯ ) s i n l n c o s c o s l n s i n
  • C 𝑦 β€² ( π‘₯ ) = βˆ’ ( 1 βˆ’ 3 π‘₯ ) + ( 1 βˆ’ 4 π‘₯ ) l n c o s l n s i n
  • D 𝑦 β€² ( π‘₯ ) = 3 π‘₯ ( 1 βˆ’ 3 π‘₯ ) + 4 π‘₯ ( 1 βˆ’ 4 π‘₯ ) s i n l n c o s c o s l n s i n
  • E 𝑦 β€² ( π‘₯ ) = βˆ’ 3 π‘₯ ( 1 βˆ’ 3 π‘₯ ) βˆ’ 4 π‘₯ ( 1 βˆ’ 4 π‘₯ ) s i n l n c o s c o s l n s i n

Q10:

Use the Fundamental Theorem of Calculus to find the derivative of the function β„Ž ( π‘₯ ) = ο„Έ βˆ’ 𝑑 𝑑 𝑒 2 5 π‘₯ l n d .

  • A β„Ž β€² ( π‘₯ ) = βˆ’ 1 𝑑
  • B β„Ž β€² ( π‘₯ ) = βˆ’ 5 π‘₯
  • C β„Ž β€² ( π‘₯ ) = βˆ’ 𝑑 l n
  • D β„Ž β€² ( π‘₯ ) = βˆ’ 2 5 π‘₯ 𝑒 5 π‘₯
  • E β„Ž β€² ( π‘₯ ) = βˆ’ 5 𝑑

Q11:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑔 ( π‘₯ ) = ο„Έ ο€Ή 1 + 𝑑  𝑑 π‘₯ 3 5 l n d .

  • A 𝑔 β€² ( π‘₯ ) = 5 𝑑 1 + 𝑑 4 5
  • B 𝑔 β€² ( π‘₯ ) = ο€Ή 1 + 𝑑  l n 5
  • C 𝑔 β€² ( π‘₯ ) = 5 π‘₯ 1 + π‘₯ 4 5
  • D 𝑔 β€² ( π‘₯ ) = ο€Ή 1 + π‘₯  l n 5
  • E 𝑔 β€² ( π‘₯ ) = 1 1 + 𝑑 5

Q12:

Use the Fundamental Theorem of Calculus to find the derivative of the function β„Ž ( π‘₯ ) = ο„Έ 3 𝑧 𝑧 + 2 𝑧 √ π‘₯ 4 2 4 d .

  • A β„Ž β€² ( π‘₯ ) = 3 √ 𝑧 2 ( 𝑧 + 2 ) 2
  • B β„Ž β€² ( π‘₯ ) = 3 𝑧 𝑧 + 2 2 4
  • C β„Ž β€² ( π‘₯ ) = π‘₯ π‘₯ + 2 2
  • D β„Ž β€² ( π‘₯ ) = 3 √ π‘₯ 2 ( π‘₯ + 2 ) 2
  • E β„Ž β€² ( π‘₯ ) = 6 𝑧 + 1 2 𝑧 βˆ’ 1 2 𝑧 ( 𝑧 + 2 ) 4 5 4 2

Q13:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑦 = ο„Έ 3 πœƒ 5 πœƒ πœƒ πœ‹ 3 √ 5 π‘₯ t a n d .

  • A 𝑦 β€² = βˆ’ 3 √ 5 π‘₯ ο€» 5 √ 5 π‘₯  t a n
  • B 𝑦 β€² = 1 5 2 ο€» 5 √ 5 π‘₯  t a n
  • C 𝑦 β€² = 3 √ 5 π‘₯ ο€» 5 √ 5 π‘₯  t a n
  • D 𝑦 β€² = βˆ’ 1 5 2 ο€» 5 √ 5 π‘₯  t a n
  • E 𝑦 β€² = 3 πœƒ 5 πœƒ t a n

Q14:

Given that ο„Έ 𝑓 ( π‘₯ ) π‘₯ = π‘₯ βˆ’ 7 π‘₯ βˆ’ π‘₯ + 9 + d C 3 2 , find 𝑓 β€² ( 1 ) .

Q15:

Given that ο„Έ 𝑓 ( π‘₯ ) π‘₯ = 3 π‘₯ + π‘₯ βˆ’ 8 π‘₯ + 5 + d C 3 2 , find 𝑓 β€² ( βˆ’ 1 ) .

Q16:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑦 = ο„Έ 2 𝑑 2 + 𝑑 𝑑 5 π‘₯ + 3 4 5 d .

  • A 𝑦 β€² = 2 ( 5 𝑑 + 3 ) 2 + ( 5 𝑑 + 3 ) 5
  • B 𝑦 β€² = 2 ( 5 π‘₯ + 3 ) 2 + ( 5 π‘₯ + 3 ) 5
  • C 𝑦 β€² = 1 0 ( 5 𝑑 + 3 ) 2 + ( 5 𝑑 + 3 ) 5
  • D 𝑦 β€² = 1 0 ( 5 π‘₯ + 3 ) 2 + ( 5 π‘₯ + 3 ) 5
  • E 𝑦 β€² = 2 𝑑 2 + 𝑑 5

Q17:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑔 ( π‘₯ ) = ο„Έ βˆ’ 2 𝑑 𝑑 π‘₯ 2 4 d .

  • A 𝑔 β€² ( π‘₯ ) = βˆ’ 8 𝑑 3
  • B 𝑔 β€² ( π‘₯ ) = βˆ’ 2 𝑑 4
  • C 𝑔 β€² ( π‘₯ ) = βˆ’ 8 π‘₯ 3
  • D 𝑔 β€² ( π‘₯ ) = βˆ’ 2 π‘₯ 4
  • E 𝑔 β€² ( π‘₯ ) = βˆ’ 8 π‘₯ 4

Q18:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑦 = ο„Έ 5 ( 5 πœƒ ) πœƒ π‘₯ 2 2 4 c o s d .

  • A 𝑦 β€² = 5 ( 5 πœƒ ) c o s 2
  • B 𝑦 β€² = 5 ο€Ή 5 π‘₯  c o s 2 4
  • C 𝑦 β€² = βˆ’ 5 0 5 πœƒ 5 πœƒ s i n c o s
  • D 𝑦 β€² = 2 0 π‘₯ ο€Ή 5 π‘₯  3 2 4 c o s
  • E 𝑦 β€² = 5 0 5 πœƒ 5 πœƒ s i n c o s

Q19:

Find the derivative of the function 𝐹 ( π‘₯ ) = ο„Έ 2 𝑒 𝑑 2 π‘₯ 5 π‘₯ 𝑑 2 2 d .

  • A 𝐹 β€² ( π‘₯ ) = 2 π‘₯ 𝑒 βˆ’ 2 𝑒 4 π‘₯ 2 5 π‘₯ 4 2
  • B 𝐹 β€² ( π‘₯ ) = 8 π‘₯ 𝑒 + 1 0 𝑒 4 π‘₯ 2 5 π‘₯ 4 2
  • C 𝐹 β€² ( π‘₯ ) = 8 π‘₯ 𝑒 βˆ’ 1 0 𝑒 4 π‘₯ 2 5 π‘₯ 2 2
  • D 𝐹 β€² ( π‘₯ ) = 8 π‘₯ 𝑒 βˆ’ 1 0 𝑒 4 π‘₯ 2 5 π‘₯ 4 2
  • E 𝐹 β€² ( π‘₯ ) = βˆ’ 8 π‘₯ 𝑒 + 1 0 𝑒 4 π‘₯ 2 5 π‘₯ 4 2

Q20:

Find the derivative of the function 𝑔 ( π‘₯ ) = ο„Έ 𝑒 βˆ’ 3 𝑒 + 5 𝑒 4 π‘₯ 3 π‘₯ 2 2 d .

  • A 𝑔 β€² ( π‘₯ ) = βˆ’ 4 ο€Ή 1 6 π‘₯ βˆ’ 3  1 6 π‘₯ + 5 + 3 ο€Ή 9 π‘₯ βˆ’ 3  9 π‘₯ + 5 2 2 2 2
  • B 𝑔 β€² ( π‘₯ ) = 4 ο€Ή 1 6 π‘₯ βˆ’ 3  1 6 π‘₯ + 5 + 3 ο€Ή 9 π‘₯ βˆ’ 3  9 π‘₯ + 5 2 2 2 2
  • C 𝑔 β€² ( π‘₯ ) = 1 6 π‘₯ βˆ’ 3 1 6 π‘₯ + 5 βˆ’ 9 π‘₯ βˆ’ 3 9 π‘₯ + 5 2 2 2 2
  • D 𝑔 β€² ( π‘₯ ) = 4 ο€Ή 1 6 π‘₯ βˆ’ 3  1 6 π‘₯ + 5 βˆ’ 3 ο€Ή 9 π‘₯ βˆ’ 3  9 π‘₯ + 5 2 2 2 2
  • E 𝑔 β€² ( π‘₯ ) = 1 6 π‘₯ βˆ’ 3 1 6 π‘₯ + 5 + 9 π‘₯ βˆ’ 3 9 π‘₯ + 5 2 2 2 2

Q21:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑅 ( 𝑦 ) = ο„Έ 3 𝑑 2 𝑑 𝑑 5 𝑦 2 s i n d .

  • A 𝑅 β€² ( 𝑦 ) = 3 𝑦 2 𝑦 2 s i n
  • B 𝑅 β€² ( 𝑦 ) = βˆ’ 3 𝑑 2 𝑑 2 s i n
  • C 𝑅 β€² ( 𝑦 ) = 6 𝑑 2 𝑑 + 6 𝑑 2 𝑑 2 c o s s i n
  • D 𝑅 β€² ( 𝑦 ) = βˆ’ 3 𝑦 2 𝑦 2 s i n
  • E 𝑅 β€² ( 𝑦 ) = 6 𝑑 2 𝑑 βˆ’ 6 𝑑 2 𝑑 2 c o s s i n

Q22:

Use the Fundamental Theorem of Calculus to find the derivative of the function 𝑅 ( 𝑦 ) = ο„Έ βˆ’ 𝑑 3 𝑑 𝑑 1 𝑦 2 s i n d .

  • A 𝑅 β€² ( 𝑦 ) = βˆ’ 𝑦 3 𝑦 2 s i n
  • B 𝑅 β€² ( 𝑦 ) = 𝑑 3 𝑑 2 s i n
  • C 𝑅 β€² ( 𝑦 ) = βˆ’ 3 𝑑 3 𝑑 βˆ’ 2 𝑑 3 𝑑 2 c o s s i n
  • D 𝑅 β€² ( 𝑦 ) = 𝑦 3 𝑦 2 s i n
  • E 𝑅 β€² ( 𝑦 ) = βˆ’ 3 𝑑 3 𝑑 + 2 𝑑 3 𝑑 2 c o s s i n

Q23:

Given that 𝐹 ( π‘₯ ) = ο„Έ 𝑑 𝑑 4 π‘₯ √ π‘₯ βˆ’ 1 t a n d , find 𝐹 β€² ( π‘₯ ) .

  • A 𝐹 β€² ( π‘₯ ) = βˆ’ 4 4 π‘₯ + 1 2 √ π‘₯ √ π‘₯ t a n t a n βˆ’ 1 βˆ’ 1
  • B 𝐹 β€² ( π‘₯ ) = 4 4 π‘₯ + 1 2 √ π‘₯ √ π‘₯ t a n t a n βˆ’ 1 βˆ’ 1
  • C 𝐹 β€² ( π‘₯ ) = 4 π‘₯ βˆ’ √ π‘₯ t a n t a n βˆ’ 1 βˆ’ 1
  • D 𝐹 β€² ( π‘₯ ) = 4 4 π‘₯ βˆ’ 1 2 √ π‘₯ √ π‘₯ t a n t a n βˆ’ 1 βˆ’ 1
  • E 𝐹 β€² ( π‘₯ ) = 4 π‘₯ + √ π‘₯ t a n t a n βˆ’ 1 βˆ’ 1