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

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

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)

Related questions

9 answers
1 answer
13 answers
1 answer
asked Apr 10, 2013 in Calculus Answers by anonymous | 817 views
1 answer
asked Jun 19, 2013 in Calculus Answers by anonymous | 235 views
1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
85,844 questions
91,601 answers
15,919 users