Problem

On what interval is the curve $$ y = \displaystyl…

01:01

Problem 64 Hard Difficulty

If $ \displaystyle f(x) = \int^x_0 (1 - t^2) e^{t^2} \,dt $, on what interval is $ f $ increasing?

Answer

increasing when $|x|<1,$ decreasing otherwise

More Answers

00:57

Frank L.

Topics

Integrals

Integration

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Discussion

You must be signed in to discuss.

Top Calculus 1 / AB Educators

Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Joseph L.

Boston College

Watch More Solved Questions in Chapter 5

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Video Transcript

Okay, The first thing you know is the directive one minus X squared times E to the X squared. Now we know that we have one month's ex, Corde said. Equal to zero axes, Juan or negative one cause one are negative on squared, sort of this positive one know either the export always has to be positive. Therefore, we know that when access between negative 11 the function is increasing, so it's increasing. The absolute value of axe is less than one. That's essentially this simplified form in which we can write this, otherwise it's decreasing.

Amrita B.

Topics

Integrals

Integration

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Top Calculus 1 / AB Educators

Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Joseph L.

Boston College