evaluate the integral `int_0^2int_y^(2y)xydxdy`

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evaluate the integral `int_0^2int_y^(2y)xydxdy`

Postby barnamah » Fri 24 June, 2011 15:16

The question says evaluate the integral `int_0^2int_y^(2y)xydxdy`
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Re: evaluate the integral `int_0^2int_y^(2y)xydxdy`

Postby Integr88 » Fri 24 June, 2011 15:33

this is a double integral to be evaluated.`int_0^2int_y^(2y)xydxdy`
we can evaluate the inner integral `int_y^(2y)xydx` treating y constant and then evaluate the outer integral with the result of this integral

`int_y^(2y)xydx=1/2x^2y|_(x=y)^(x=2y)`
using fundamental theorem of calculus we can evaluate it as follow with replacing every occurance of x with the upper and lower values.
Remember that upper bound is x=2y and lower bound is x=y

:
`1/2[(2y)^2y-1/2(y)^2y]= 1/2[ 4y^3-y^3]`
now sub this result `1/2[ 4y^3-y^3]` into the outer integral and we have:
`int_0^(2)1/2[ 4y^3-y^3]dy`
`1/2int_0^(2)[ 4y^3-y^3]dy` 1/2 was constant and I took it out of the integral

Integrating ` 4y^3-y^3` we have `1/2[y^4-1/4y^4]_0^2`
`1/2[y^4-1/4y^4]_0^2=[1/2y^4-1/8y^4]_0^2`

Now this expression `[1/2y^4-1/8y^4]_0^2` means evaluate this from 0 to 2 meaning first substitute the upper bound 2 into the equation and subtract it from this result with the lower bound value.
`[1/2(2)^4-1/8(2)^4]-[1/2(0)^4-1/8(0)^4]=[1/2(2)^4-1/8(2)^4]-[0]`
`1/2(16) - 1/8(16)=8-2=6`

`:.` the value is 6


Evaluate the double integral `int_0^2int_y^(2y)xydxdy`
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