int_0^8 int_(y^1/3)^2 y/(x^7+1)dxdy

Lily D

Lily D

Answered question

2022-06-03

Answer & Explanation

xleb123

xleb123

Skilled2023-05-19Added 181 answers

To evaluate the double integral
08y1/32yx7+1dxdy
we will first integrate with respect to x and then with respect to y.
Let's begin by evaluating the integral with respect to x:
y1/32yx7+1dx
To integrate this expression, we can use the substitution u=x7+1. This gives us du=7x6dx, which implies dx=17x6du. Substituting these values into the integral, we have:
y1/32yx7+1dx=y1/32yu·17x6du=17y1/32yu·1x6du
Now, we can evaluate the integral with respect to u:
17y1/32yu·1x6du=17[yu·15x5]y1/32du
Simplifying further, we obtain:
17[y5u·1x5]y1/32du
Next, we can evaluate the integral with respect to y:
0817[y5u·1x5]y1/32du
To do this, we substitute the limits of integration into the expression:
0817[25u·1x5y1/35u·1x5]du
Now, let's integrate with respect to y:
0817[25u·1x5y1/35u·1x5]du=17[25u·1x5·y15u·1x5·34y4/3]08du
Simplifying further, we get:
17[25u·1x5·y15u·1x5·34y4/3]08du
Now, let's substitute the limits of integration:
17[25u·1x5·815u·1x5·34(8)4/3(25u·1x5·015u·1x5·34(0)4/3)]du
Simplifying further, we have:
17[165u·1x534·84/35u·1x5]du
Now, we can simplify the expression:
17[165u·1x534·84/35u·1x5]du=17[165u·1x534·84/35u·1x5]du
Finally, we can integrate with respect to u:
17[165u·1x534·84/35u·1x5]du=17[165·1x5·ln|u|34·84/35·1x5·ln|u|]du
Now, we substitute u=x7+1 back into the expression:
17[165·1x5·ln|(x7+1)|34·84/35·1x5·ln|(x7+1)|]du
To complete the evaluation of the double integral, we need to substitute the limits of integration:
17[165·1x5·ln|(x7+1)|34·84/35·1x5·ln|(x7+1)|]08
Now, let's evaluate the expression at the upper limit:
17[165·185·ln|(87+1)|34·84/35·185·ln|(87+1)|]
And evaluate the expression at the lower limit:
17[165·105·ln|(07+1)|34·84/35·105·ln|(07+1)|]
Since 07+1=1, we have:
17[165·105·ln|1|34·84/35·105·ln|1|]
Since ln|1|=0, we can simplify further:
17[165·105·034·84/35·105·0]
Both terms in the square brackets evaluate to 0, so the final result is:
17·0=0
Therefore, the value of the double integral
08y1/32yx7+1dxdy
is 0.

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