Can u please answer question 7 Could you please show your st

Calculus 1 Given f (x) Test M3 Review X +1 derivatives x find value s) of c that satisfy the conditions of the Mean Value Theorem (of the on the interval .2 2. Given the following functions and associated intervals determine if Rolle\'s Theorem is satisfied on the interval a) 2x if 0 if exs3 [0,3] For 2b, apply Rolle\'s 4. Utilize Theorem and all values of e where f (x) 0. find Hopital\'s Rule to compute the following limits lim b) X-5x2 s. X-4 Solve the initial value x-2 problem f (3) 2e\' sinx with f(0) 7 e\"-1 6. Given the acceleration (r) e\" find the velocity function if the initial velocity is 2.5 per second. meters 7. Let R be the region bounded by the graph of f (J)--s and the x-axis on the interval 12.51 using n-6 subintervals, calculate the left and right Riemann sums. Round to 2 decimal places. 8. Find the value(s) of c at which the function f(x) -x 7 equals its average value on I-6.-31. 9. Evaluate the following integrals. Use substitution where needed. 2x 1 cos (sin x)cos xdx (x2 +x+ 10. Given f (x) x 3x 4x +12 on the interval [-2,3], find the following: a) The net area under the curve. b) The total area bounded by f (x) and thexaxis. Remember that you must find the zeros here to determine where positive and negative areas are located. increased from 150 to 200 units. (Hint: This is a net change!)

Solution

Given that

f(x) = 1/x² on the interval [ 2 , 5 ] and n = 6

Left end point = 2

Left riemann sum : ( incuding left end point )

( from a to b ) f(x)dx x( f(x₀) + f(x₁) + f(x₂) +...+ f(x_n2) + f(x_n1) ) , where x = ( ba ) /n.

we have a = 2 , b = 5 , n = 6

x = ( 5 - 2 ) / 6

    x = 3 / 6

  x = 1 / 2

The six subintervals are,

x₀ = 2

x₁ = 2 + x =    2 + 1 / 2 = 5/2

x₂ = x₁ + 1 / 2 = 5/2 + 1/2 = 6/2 = 3

x₃ = x₂ + 1/2 = 3 + 1/2 = 7/2

x₄ = x₃ + 1/2 = 7/2 + 1/2 = 4

x₅ = x₄ + 1/2 = 4 + 1/2 = 9/2

f(x) = 1/x²

f(x₀) = f(2) = 1/2² = 1/4

f(x₁) = f(5/2) = 1/(5/2)² = 4/25

f(x₂) = f(3) = 1/(3)² = 1/9

f(x₃) = f(7/2) = 1/(7/2)² = 4/49

f(x₄) = f( 4) = 1 /(4)² = 1/16

f(x₅) = f( 9/2) = 1/(9/2)² = 4/81

Hence,

   ( from 2 to 5 ) 1/x² dx x ( f(x₀) + f(x₁) + f(x₂) +...+ f(x_n2) + f(x_n1) ) , where x = ( ba ) /n

   1/2 [ (1/4) + ( 4/25) + ( 1/9) + ( 4/49) + (1/16) + (4/81) ]

   0.3573132401

Therefore,

Left reimann sum = 0.36

Right  riemann sum : ( including right end point )

Right end point = 5

( from a to b ) f(x)dx x( f(x₁) + f(x₂) +...+ f(x_n2) + f(x_n1) ) , where x = ( ba ) /n.

we have a = 2 , b = 5 , n = 6

x = ( 5 - 2 ) / 6

    x = 3 / 6

  x = 1 / 2

The six subintervals are

x₁ = 2 + x =    2 + 1 / 2 = 5/2

x₂ = x₁ + 1 / 2 = 5/2 + 1/2 = 6/2 = 3

x₃ = x₂ + 1/2 = 3 + 1/2 = 7/2

x₄ = x₃ + 1/2 = 7/2 + 1/2 = 4

x₅ = x₄ + 1/2 = 4 + 1/2 = 9/2

x₆ = 5

f(x) = 1/x²

f(x₁) = f(5/2) = 1/(5/2)² = 4/25

f(x₂) = f(3) = 1/(3)² = 1/9

f(x₃) = f(7/2) = 1/(7/2)² = 4/49

f(x₄) = f( 4) = 1 /(4)² = 1/16

f(x₅) = f( 9/2) = 1/(9/2)² = 4/81

f(x₆) = f(5) = 1/(5)2 = 1/25

Hence,

   ( from 2 to 5 ) 1/x² dx x ( f(x₁) + f(x₂) +...+ f(x_n2) + f(x_n1) ) , where x = ( ba ) /n

   1/2 [ ( 4/25) + ( 1/9) + ( 4/49) + (1/16) + (4/81) + (1/25) ]

   0.2523132401

Therefore,

Right reimann sum = 0.25