Question

Evaluate the following definite integral using the Fundamental Theorem on Calculus:
\[\int\limits_{1}^{4}1divx ^{2}dx\]

Answer 1

\div

Answer 2

or simply: \frac{1}{x^2}

Answer 3

it seems easier to just convert to exponents:

1/x^2 = x^-2

Answer 4

Let
\[ F(x) = \int \frac{1}{x^2} dx \]
By the F.T.Calculus, \( dF/dx = 1/x^2 = x^{-2} \).

Hence the question is: what function F has that property?

Answer 5

So ... what's your answer to the first part here? What's the function F(x)?

Answer 6

is that the antiderivative of f(x)?

Answer 7

If f(x) = 1/x^2, then F(x) is indeed the anti-derivative.

Answer 8

James, i don't think I can do this one :(

Answer 9

What is the derivative of 1/x ?

Answer 10

That is x^-1 (i think)

Answer 11

yes. Hence it's derivative is what?

Answer 12

1/x?

Answer 13

If g(x) = 1/x = x^(-1), then dg/dx = ... what?

Hint: what is the derivative of h(x) = x^n, for an arbitrary number n?

Answer 14

If h(x) = x^n, then
dh/dx = nx^(n-1).

Therefore what is dg/dx?

Answer 15

is it -x^-2

Answer 16

Correct. Now we want a function F(x) such that dF/dx = +x^(-2).

What then must that function be?

Answer 17

the function must be: -1/3 x^-3??

Answer 18

No. You know that (d/dx) 1/x = -1/x^2.

You want a function F so that (d/dx) F = +1/x^2.
But +1/x^2 = (-1)(-1/x^2). Therefore F must equal ... what?

Answer 19

Ok. let's try this one other way.

What is the derivative of g(x) = A/x, where A is a real number, a constant?

Answer 20

dg/dx = ...what?

Answer 21

that is Ax^-1?

Answer 22

yes.

Answer 23

Quickly now. You've already done this for x^(-1)

Answer 24

If g(x) = A/x = Ax^(-1), then
dg/dx = -A/x^2 = -Ax^(-2)

Answer 25

dg/gx= -Ax^-2

Answer 26

That being the case, for what value of A is dg/dx = 1/x^2 ?

Answer 27

A=1?

Answer 28

No. If A = 1, then dg/dx = -1/x^2

Answer 29

A= -1

Answer 30

Yes. That being the case, the anti-derivative of 1/x^2 is what?

Answer 31

The anti-derivative is the function F such that
dF/dx = 1/x^2

Answer 32

the anti-derivative of 1/x^2=> F(x)=1/2x^3??

Answer 33

i think I just dis all wrong

Answer 34

No. dF/dx = x^(-2).

What is the derivative of 1/x?

Answer 35

x^-1

Answer 36

1/x = x^(-1). What is the derivative of that function wrt x?

Answer 37

it is -x^-2?

Answer 38

(d/dx)(1/x) = -1/x^2 = -x^(-2).
Because writing 1/x = x^(-1), it's derivative is
(-1).x^(-1-1) = -x^(-2)

Answer 39

So what again is the derivative of g(x) = A/x ?

Answer 40

Ax^-1

Answer 41

that is NOT the derivative.

Answer 42

If g(x) = A/x
dg/dx = ...

Answer 43

oops... -Ax^-2?

Answer 44

dg/dx = -A/x^2, yes.

Hence for what value of A is dg/dx = 1/x^2 ?

Answer 45

A= -1

Answer 46

Correct. Hence what is the function F such that
dF/dx = 1/x^2
?

Answer 47

F=1?

Answer 48

If F = 1, dF/dx = 0.

With our A = -1, the function g is g(x) = -1/x. What is the derivative dg/dx in this case?

Answer 49

the derivative of dg/dx=1^-2

Answer 50

1x^(-2) = 1/x^2

I.e., if g(x) = -1/x, then dg/dx = 1/x^2.

Hence what is the antiderivative function F that has the property
dF/dx = 1/x^2
?

Answer 51

is it -1?

Answer 52

Kewnisha: have a coffee or a glass of coke or something.

Didn't we just show that
if g(x) = -1/x,
then dg/dx = 1/x^2?

So mustn't a function F which has the property dF/dx = 1/x^2 be exactly the same?

Answer 53

That is, F(x) = -1/x.

Answer 54

OHHH... now I get it... ouch... that was really bad :(

Answer 55

Ok. Returning now to your integral,
\[ \int_1^4 \frac{1}{x^2} dx = F(4) - F(1) \]
where dF/dx = 1/x^2

Answer 56

Therefore the integral is equal to what?
F(4) - F(1) = ...

Answer 57

-1/4 - -1/1 => -1/4+1= 3/4?

Answer 58

Correct.

Answer 59

So tell me again, what is the derivative of 1/x?

Answer 60

that is x^-1

Answer 61

1/x = x^(-1), yes. What is its derivative?

Answer 62

its derivative if -x^(-2)

Answer 63

The derivative of 1/x is -1/x^2, yes.

What is the derivative of -1/x?

Answer 64

the derivative of -1/x is x^-2

Answer 65

1/x^2 yes.

What is the derivative of A/x ?

Answer 66

..where A is a constant?

Answer 67

Ax^-1

Answer 68

the derivative of A/x is ...

Answer 69

it is Ax^-1

Answer 70

(d/dx)(A/x) = -A/x^2

Answer 71

What's the derivative of 2/x?

Answer 72

2x^-1

Answer 73

No. What is (d/dx)(1/x)?

Answer 74

(d/dx)(1/x)
= (d/dx)(x^-1)
= -x^-2
= -1/x^2

Answer 75

Hence what is (d/dx)(2/x)?

Answer 76

-2/x^2

Answer 77

correct. What is
(d/dx)(A/x) ?

Answer 78

that is: -A/x^2

Answer 79

Yes. What is the (d/dx)(-1/x) ?

Answer 80

1/x^2

Answer 81

yes. What is the anti-derivative of 1/x^2 ?

Answer 82

-1/x

Answer 83

Yes. What is the anti-derivative of -2/x^2 ?

Answer 84

2/x

Answer 85

Yes. What is (d/dx)(x^2) ?

Answer 86

-2/x^3

Answer 87

x^2, not 1/x^2

Answer 88

ohh... 2x

Answer 89

What's that antiderivative of 1/x^3 ?

Answer 90

that is: -3x^-4

Answer 91

No. What's (d/dx)(1/x^2) ?

Answer 92

-1/x

Answer 93

No, you wrote it down a moment ago by accident.
(d/dx)(1/x^2) = ...

Answer 94

-2/x^3

Answer 95

Yes. Hence the anti-derivative of (-2/x^3) is what?

Answer 96

1/x^2

Answer 97

Yes, good. What then is the anti-derivative of 1/x^3 ?

Answer 98

1/4 x^-4