Question

Evaluate the following integrals?

Answer 1

know the standard integral
\(\Large \int x^n dx =...?\)

Answer 2

x+1/n+1

Answer 3

you mean
\(\Large \dfrac{x^{n+1}}{n+1}+c
right ?\)

Answer 4

now we just need some algebra,
\(\dfrac{1}{x^2}\)
can you bring this in the form of x^n
?
and tell me what would be 'n'
?

Answer 5

x-2

Answer 6

yes, so n =-2

just plug in n=-2 in the general formula
what do u get ?

Answer 7

so x

Answer 8

just x ? no...

Answer 9

\(\dfrac{x^{-2+1}}{-2+1} = ...?\)

Answer 10

x^-1

Answer 11

thats the numerator
what about denominator ?
-2+1 = ... ?

Answer 12

x^-1/-1

Answer 13

correct, now we plug in the limits

fist plug in upper limit
put x = 1 in
\(-x^{-1}\)
what do u get

Answer 14

-x

Answer 15

no, plug in x = 1

Answer 16

how can we plug x=1 in -x^-1

Answer 17

and from where did we get -x^-1

Answer 18

x=1 in -x^-1
means
\(-(1)^{-1} = -1\)

Answer 19

x^-1/-1 = -x^-1

Answer 20

if we are moving -1 in numerators place shouldnt it be +1

Answer 21

thats not correct.

a/(-1) = -a

Answer 22

negative in numerator or denominator does not make any difference, answer would still be negative

Answer 23

ok

Answer 24

after plugging in upper limit, plug in lower limit

put x=-1 in
-x^-1

Answer 25

-1^-1=-1

Answer 26

there is negative in the beginning too
\(\Large - (-1)^{-1} = - (-1)=+1\)
got this ?

Answer 27

can you do all the steps in one reply so that what ever im missing or havent understood i can ask you

Answer 28

\(\Large \int \dfrac{1}{x^2}dx = \int x^{-2}dx = \dfrac{x^{-1}}{-1}+c = -x^{-1 }+c\)

\(\Large \int \limits_{-1}^1 \dfrac{1}{x^2}dx = [-x^{-1}]_{-1}^1 = -(1)^{-1} - [-(-1)^{-1}] \\ \Large = -1- [1] = -2 \)

Answer 29

let me know about any doubts in any step

Answer 30

from where did we get c

Answer 31

'c' is just a constant when we solve indefinite integration

while solving definite integration, since we plug in upper and lower limits, we need not add 'c'

Answer 32

and the 7th step

Answer 33

which ?
whats specifically your doubt in that step ?

Answer 34

how did we get so many negatives

Answer 35

don't get confused
take it term by term

plug in upper limit x=1 in -x^-1

we get -(1^-1) = -(1) = -1

Answer 36

plug in lower limit , x= -1 in -x^-1

we get
-(-1^-1) = - (-1) = 1

Answer 37

got those ?^

Answer 38

so we have to go to ways

Answer 39

becuase of the upper and lower bound of the integral

Answer 40

we plug in upper limit, then lower limit and then subtract them

here, after plugging in upper limit, we got -1
for lower limit , we got, +1
so,
\(-1 - (1) = -2\)

Answer 41

and power will be +1

Answer 42

what power?

Answer 43

[(-1^-1]-[-(-1^-1)]

Answer 44

[-1^-1]-1^-1

Answer 45

i am not getting what you are confused with ??
or you're not confused?

try it on your own first, then see my solution.
if there is any difference, then ask

Answer 46

so minus 2 will be the answer

Answer 47

i think i got it

Answer 48

if you're confused with negatives,
forget that there is a negative in the beginning, solve it and then add a negative in the end.



and then adding the negative in the end gives
\(\Large -2\)
yes.