Evaluate the following integral          ∫_0^4 ∫_(x^2)^16 x^3 e^(y^3 ) dy dx
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Evaluate the following integral          ∫_0^4 ∫_(x^2)^16 (x^3 e^(y^3)) dy dx

We need to change the order of integration.

The relationship between x and y is y = x^2 (from the 2nd, lower limit)

Now x varies from 0 to 4.

Since we have y = x^2, then we can say that x varies from 0 to sqrt(y) and y varies from 0 to 16.

Therefore we can rewrite our integral as,

∫_(0)^16 ∫_(0)^sqrt(y) (x^3 e^(y^3)) dx dy

Let I1 = ∫_(0)^sqrt(y) (e^(y^3).x^3) dx

Then, I1 = e^(y^3).[x^4/4][0 .. sqrt(y) ] = e^(y^3).{ y^2/4 – 0}

I1 = (1/4).e^(y^3).y^2

And, I2 = ∫_(0)^16 I1 dy

I2 = ∫_(0)^16 (1/4).e^(y^3).y^2 dy

I2 = (1/4). ∫_(0)^16 e^(y^3).y^2 dy

I2 = (1/4).(e^(y^3)/3)[0 .. 16] = (1/12).{e^(16^3) – e^(0)}

I2 = (1/12)(e^4096 – 1)

Answer: the integral evaluates to: (1/12)(e^4096 – 1)

by Level 11 User (81.5k points)

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