Question

can someone please help me with double integration

Answer 1

\[\int\limits_{0}^{1 } \int\limits_{5y}^{5} e^(x^2)\]

that should be e^(x^2)

Answer 2

Ive tried this several times and keep getting the wrong answer

Answer 3

tell me how you did it please

Answer 4

we need to change the order of integration i hope you understand it

Answer 5

i tried to do it dx dy

Answer 6

but you can solve
e^(x^2) dx

Answer 7

yes i got 2xe^(x^2)

Answer 8

it is incorrect..

Answer 9

and then with the limit 10e^10 - 10ye^10y dy

Answer 10

take the derivative of this and see that it is not e^(x^2)

Answer 11

[2xe^(x^2)]' = 2(e^(x^2) + x* (2x) * e^(x^2))
and this is not
e^(x^2)

Answer 12

you cant integrate e^(x^2) normally ..

Answer 13

we have to change limits and hope that then we will be able to integrate..

Answer 14

ok so how do we do that?

Answer 15

wait i did a mistake in my sketch

Answer 16

i deleted and drawing again :

Answer 17

now if we want to change the order of integration
we have to change the limits
so if we want dydx we get:

\[\int\limits_{0}^{5} \int\limits_{0}^{\frac{ x }{ 5 }} e^{x^2} dydx\]

Answer 18

tell me if you understand how i changed the limits

Answer 19

solving this integral gives

\[\int\limits_{0}^{5} \frac{ x }{ 5 } \times e^{x^2} - 0 \times e^{x^2} dx = \int\limits_{0}^{5} \frac{ x }{ 5 } \times e^{x^2} dx\]

Answer 20

now we can do it!! since we may call u = x^2
du = 2xdx
so du/2 = xdx

\[\int\limits_{0}^{25} \frac{ e^{u} }{ 10 }du\] =
(e^25) / 10 - 1/10 = (e^(25) - 1)/10

Answer 21

Let me know when you read this

Answer 22

ok i read it

Answer 23

understand what i did?

Answer 24

kind of

Answer 25

im not really sure how the limits changed i guess i thought i got it but i guess i dont

Answer 26

the first limits was according to dxdy so :