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Calc 1 5.4

 

<== Calculus 1
Number Question Answer
See the work
1. Let

f(x)  = \begin{cases} 
              0 if x < -4 \
              3  if  -4 < x < 0 \
              -5 if  0 < x < 5 \
              0 if  x > 5
and
g(x)  = integral from -4 to x f(t) dt

Determine the value of each of the following:
(a) g(-8) =
(b) g(-3) =
(c) g(1) =
(d) g(6) =
(e) The absolute maximum of g(x) occurs when x = and is the value
A) 0

B) 3

C) 7

D) -13

E) x = 0
value = 12
The work

Point Cost: 4
2. Use part I of the Fundamental Theorem of Calculus to find the derivative of

f(x)  = integral from 3 to x (1)/(1+t^(2)) dt

f'(x) =
See the Work
The work

Point Cost: 3
3. Use part I of the Fundamental Theorem of Calculus to find the derivative of

f(x) = integral from -2 to x sqrt(t^3+8) dt

f'(x) =
See the Work
The work

Point Cost: 3
4. If f(x) = integral from x to 8 t^3 dt then
f'(x) = =
See the Work
The work

Point Cost: 3
5. Use part I of the Fundamental Theorem of Calculus to find the derivative of

F(x) = integral from x to 2 sin(t^2) dt

F'(x) =
See the Work
The work

Point Cost: 3
6. Use part I of the Fundamental Theorem of Calculus to find the derivative of

h(x) = integral from 7 to (1/x) 7arctan(t) dt

h'(x) =
h(x) = integral from 7 to 1/x of 7arctan (t) dt
The work

Point Cost: 4
7. Use part I of the Fundamental Theorem of Calculus to find the derivative of

y = integral from -6 to sqrt(x) (cos (t))/(t^(9)) dt

dy/dx =
dy/dx
The work

Point Cost: 4
8. If \displaystyle f(x) = integral from 1 to x^2 t^2 dt then
f'(x) = =
2x^5
The work

Point Cost: 4
9. Use part I of the Fundamental Theorem of Calculus to find the derivative of

h(x) = integral from -2 to sin(x) (cos(t^5)+t) dt

h'(x) =
h(x) = integral from -2 to sin(x) of (cos(t^5)+t) dt
The work

Point Cost: 4
10. Find the derivative of the following function

F(x) = integral from 1 to sqrt(x) (s^2)/(1 + 7 s^4) ds

F'(x) =
F(x) = integral from 1 to sqrt(x) (s^2)/(1 + 7 s^4) ds
The work

Point Cost: 4
11. If \displaystyle f(x) = integral from (1-9 x) to 9 (sin(t))/(1+t^2) dt , then f'(x) = =. f'(x) =
The work

Point Cost: 4
12. Find the derivative of the following function

F(x) = integral from x^3 to x^4 (2t-1)^3 dt
using the Fundamental Theorem of Calculus.
F'(x) =
F(x) = \int_{x^3}^{x^4} (2t-1)^3\, dt
The work

Point Cost: 5
13. If \displaystyle f(x) = integral from x to x^2 t^(2) dt

then
f'(x) = =

f'(x) =
The work

Point Cost: 5
14. Find the derivative of

g(x)  = integral from 5x to 8x (u+2)/(u-8) du

g(x)  = integtral from 5x to 8x (u+2)(u-8) du
The work

Point Cost: 5
15. Find the derivative of the function:

y = integral from sqrt(x) to (x^6) sqrt(t)sin t^(2) dt

dy/dx =
\displaystyle{rac{dy}{dx}}
The work

Point Cost: 5
16. Use part I of the Fundamental Theorem of Calculus to find the derivative of

y = integral from cos x to 7x cos(u^7) du

dy/dx =
rac{dy}{dx}
The work

Point Cost: 5
17. Given

f(x) = integral from 0 to x (t^2 - 16)/(1+cos^2(t)) dt
At what value of x does the local max of f(x) occur?

x =

x = -4
The work

Point Cost: 3
18. If \displaystyle f(x) =  integral from 0 to x (t^3 + 4 t^2 + 7) dt
then
f''(x) =
The work

Point Cost: 3
19. Evaluate the definite integral

integral from 1 to 3 d/dt sqrt(5 + 4t^4) dt
using the Fundamental Theorem of Calculus.
integral from 1 to 3 d/dtsqrt(5 + 4t^4) dt =
See Work
The work

Point Cost: 3
20. Find the average value of f(x) = x^(2) on the interval [3,6].

21
The work

Point Cost: 3
21. Find the average value of : f(x) =  8sin(x) + 4cos(x)
on the interval [0,13pi/6  ]

Average value =
0.4513
The work

Point Cost: 3
22. Find the mean value of the function f(x) = |6 - x| on the closed interval [3, 9].
mean value =
1.5
The work

Point Cost: 3
23. (a) Find the average value of f(x) = 25-x^2 on the interval [0,2].



(b) Find a value c in the interval [0,2] such that f (c) is equal to the average value.

A) 71/3

B) 1.1547

The work

Point Cost: 3