Question

I need help finding the indefinite integral (1 to 2) of 1/X^2 (dx).

Answer 1

hey im not sure how i should use the quotient rule with anti derivs.

Answer 2

So you want to find the anti-derivative of f(x) = 1/x^2. Call it F(x), where
F'(x) = f(x) = 1/x^2

Answer 3

Once you've found F(x), then your integral is
\[ \int_1^2 \frac{1}{x^2} dx = F(2) - F(1) \]

Answer 4

In other words, what function when differentiated gives 1/x^2 ?

Answer 5

oh i c mental block how do i find that

Answer 6

hint: what's the derivative of 1/x ?

Answer 7

derivative? why not antiderivative? (sorry i need help with this stuff too)

Answer 8

it is -1/x^2?

Answer 9

We are looking for the function F(x) with the property that
\[ F\ '(x) = 1/x^2. \]

Then by the Fundamental Theorem of Calculus
\[ \int_1^2 \frac{1}{x^2} dx = F(2) - F(1) \]


Now another way to write this
\[ F \ '(x) = x^{-2} \]

Let's take a step back. What is the derivative of
\[ g(x) = x^n \]

Answer 10

its nx^e?

Answer 11

nx^9n-1)

Answer 12

ya what she said

Answer 13

sorry nx^(n-1)

Answer 14

its (n+1)x^n+1/n+1 if that makes sense lol

Answer 15

(1/(n+1))x^(n+1)

Answer 16

Yes, if g(x) = x^n, then \[ g′(x)=nx^{n−1} \].

Hence, what is the anti-derivative of
h(x)=xn

I.e., what is the function H such that H' = x^n ?

Answer 17

yeah that one lol

Answer 18

Right ... the antiderivative of g(x) = x^n is
\[ G(x) = \frac{1}{n+1} x^{n+1} \]

Given that, what is the antiderivative of
\[ f(x) = 1/x^2 = x^{-2} \]

Answer 19

x^1?

Answer 20

oh!! now i get it!!!

Answer 21

oh sorry -1x
?

Answer 22

thank you thank you!

Answer 23

No, f(x) = x^-2. So n = -2. Use now the formula above to find the anti-derivative F(x)

Answer 24

hmm -1x^-1???

Answer 25

Right.

\[ F(x) = \frac{1}{n+1} x^{n+1} = \frac{1}{-2+1} x^{-2+1} = (-1) x^{-1} = -1/x \]

Answer 26

Therefore
\[ \int_1^2 \frac{dx}{x^2} \]
equals what?

Answer 27

lol hold on sec

Answer 28

\[ \int_1^2 \frac{dx}{x^2} = F(2) - F(1) \]
where F'(x) = 1/x^2

Answer 29

you explain soo well just new for me trying to get it visualy

Answer 30

Sketch the graph of f(x) = 1/x^2. We're calculating the tea under the curve between x=1 and x=2.

Answer 31

So whatever else the answer is, as the area is above the x-axis, the area must be a positive number.

Answer 32

ahh i c

Answer 33

so this -1(2)^-1 - -1(1)^-1

Answer 34

ok. Simplify that.

Answer 35

you'll find it easier if you think of F(x) as F(x) = -1/x.

Answer 36

oh i c

Answer 37

1/2!?

Answer 38

Yes
F(2) - F(1) = -1/2 - (-1/1)
= -1/2 + 1
= 1/2

Answer 39

yessssssssss finally thanks soo much

Answer 40

The derivative of 1/x is so important, you should learnt it cold.
The antiderivative of 1/x^2 is almost as important and you should learn that as well.

Answer 41

will do :)

Answer 42

For the record, what is the anti-derivative of -1/x^2 ?

Answer 43

is it x^2? the other was negative

Answer 44

1/x^2 = x^-2

Answer 45

no wait its possitive so -1/2?

Answer 46

what's the anti-derivative of 1/x^2 ? It's what we just used for your problem.

Answer 47

its x^-2

Answer 48

sorry confused it was -1X^-1

Answer 49

1/x^2 = x^-2
But what is its antiderivative? I.e., what is the function F such that
F'(x) = 1/x^2 ?

Answer 50

was it this -1x^-1?

Answer 51

It is -1/x, yes.
That is, if F(x) = -1/x, then F'(x) = 1/x^2.

Ok. Now, what is the antiderivative of f(x) = -1/x^2 ?

Answer 52

its 1/x?lol i know i have fail memory

Answer 53

Yes. If F(x) = 1/x, then F'(x) = -1/x^2.

ok. ttyl

Answer 54

Brilliant much thankks for giving me time man