Title | Webwork Section 1.3: fundamental theorem of calculus |
---|---|
Course | Biology for Science Majors II |
Institution | St. Philip's College |
Pages | 3 |
File Size | 70.7 KB |
File Type | |
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Webwork ch 1 problems 1-20, section 1.3: fundamental theorem of calculus...
Sarah Galvan MAT 1224 007 Fall 2021 Assignment HW Section 1.3 The Fundamental Theorem of Calculus due 08/30/2021 at 11:59pm CDT Problem 8. (1 point)
Problem 1. (1 point) Use the Fundamental Theorem of Calculus to find the derivative. If g(x) =
Z x
If f (x) = then f ′ (x) =
g′ (x) =
Problem 9. (1 point) If f (x) =
Problem 2. (1 point) Use the Fundamental Theorem of Calculus to find the derivative. If g(x) =
Z x
f ′ (x) =
If f (x) =
g′ (x) =
t 2 dt then
−2
Z x4
t 5 dt
x
then f ′ (x) =
Problem 3. (1 point) Use the Fundamental Theorem of Calculus to find the derivative. Z √x
Z x3
Problem 10. (1 point)
2
e−t dt then,
11
If g(x) =
t 3 dt
x
3 dt then, t3 − 5
−7
Z 6
Problem 11. (1 point) If f (x) =
tdt then,
Z x
(t 3 + 6t 2 + 5)dt
0
6
g′ (x) = Problem 4. (1 point) Use the Fundamental Theorem of Calculus to find the derivative.
Problem 12. (1 point) Part1.
Z d sin x p
1 − t 2 dt = dx 6 Problem 5. (1 point) Use the Fundamental Theorem of Calculus to find the derivative.
Find the most general antiderivative by evaluating the following indefinite integral: Z x2 − 11x dx =
Z x d e 2 ln u du = dx 10 Problem 6.Z(1 point) x
NOTE: The general antiderivative should contain an arbitrary constant.
t 3 dt. Evaluate the following.
Let f (x) =
−1
Part 2.
f ′ (x) = f ′ (4) = Problem 7. (1 point) If f (x) =
Z 19
Evaluate the given definite integral. Z −1
4
t dt then
−4
x
f ′ (x) = 1
x2 − 11x dx =
Problem 13. (1 point)
Problem 15. (1 point)
Part1.
Part1.
Find the most general antiderivative by evaluating the following indefinite integral:
Find the most general antiderivative by evaluating the following indefinite integral:
Z x2 − 1 9 − x2 dx =
Z
NOTE: The general antiderivative should contain an arbitrary constant.
NOTE: The general antiderivative should contain an arbitrary constant.
Part 2.
Part 2.
Evaluate the given definite integral.
Evaluate the given definite integral. Z 2 −5 − dt = 2t 1
Z −1
x2 − 1
−4
−
9 − x2 dx =
−5 2t
dt =
Problem 14. (1 point)
Problem 16. (1 point)
Part1.
Part1.
Find the most general antiderivative by evaluating the following indefinite integral:
Find the most general antiderivative by evaluating the following indefinite integral:
Z
Z
1 12t 2
+ 6t
2
dt =
sec2 (θ)dθ =
NOTE: The general antiderivative should contain an arbitrary constant.
NOTE: type ’theta’ for the variable θ NOTE: The general antiderivative should contain an arbitrary constant.
Part 2.
Part 2.
Evaluate the given definite integral.
Evaluate the given definite integral.
Z 16
Z 7π 4
9
1 12t 2 + 6t 2 dt =
2π
2
sec2 (θ)dθ =
Problem 18. (1 point) Evaluate the definite integral
Problem 17. (1 point)
Z 8
Part1.
4
Find the most general antiderivative by evaluating the following indefinite integral: Z
Problem 19. (1 point) Evaluate the definite integral Z 5
7 dx = 1 + x2
2
Z 7 2 2x + 2 4
Part 2.
√
3 3
3 3
√ dx x
Problem 21. (1 point) Evaluate the definite integral
Evaluate the given definite integral. √
(12x2 − 4x + 2)dx
Problem 20. (1 point) Evaluate the definite integral
NOTE: The general antiderivative should contain an arbitrary constant.
Z −
(2x + 6)dx
Z π
7 dx = 1 + x2
0
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3
2 sin(x)dx...