1. ## Iterated Integrals

1. The problem statement, all variables and given/known data

Evaluate the iterated integral

2. Relevant equations

Fundamental Theorem of Calculus & Fubini's Theorem

3. The attempt at a solution

I have been working on this problem for the last hour and haven't been able to solve it thus far. I integrated the inside of the integral with respect to y from 4 to 3, and then integrated the result from the first integral with respect to x from 2 to 1. In the process, I used substitution to solve the integrals.

The answer I keep on getting is (1/(4(3x+4)^4)) - (1/(4(3x+3)^4)) and solve from x= 2 to 1 which results in a small fractional answer of 7.5674005*10^(-5). But my online program keeps on saying its incorrect.

Can anyone give me a heads up as to where I am going wrong? This is one of the few iterated integrals that is giving me problems right now

2. ## suggestion

Originally Posted by skaterbasist
1. The problem statement, all variables and given/known data

Evaluate the iterated integral

2. Relevant equations

Fundamental Theorem of Calculus & Fubini's Theorem

3. The attempt at a solution

I have been working on this problem for the last hour and haven't been able to solve it thus far. I integrated the inside of the integral with respect to y from 4 to 3, and then integrated the result from the first integral with respect to x from 2 to 1. In the process, I used substitution to solve the integrals.

The answer I keep on getting is (1/(4(3x+4)^4)) - (1/(4(3x+3)^4)) and solve from x= 2 to 1 which results in a small fractional answer of 7.5674005*10^(-5). But my online program keeps on saying its incorrect.

Can anyone give me a heads up as to where I am going wrong? This is one of the few iterated integrals that is giving me problems right now
what is the correct answer???i got 1/12 although i am not sure whether that is correct.

3. $\int_1^2\int_3^4 (3x+ y)^{-2}dydx$
Let u= 3x+y so that dy= du. When y= 3, u= 3x+3 and when y= 4, u= 3x+4. The integral becomes:
$\int_{x=1}^2\int_{u= 3x+3}^{3x+4} u^{-2}du= -\int_1^2\left(\frac{1}{3x+4}- \frac{1}{3x+ 3}\right)dx$
$= -\int_1^2 \frac{1}{3x+4}dx+ \int_1^2\frac{1}{3x+3}dx$

In the first integral, let v= 3x+ 4 so that dx=(1/3)dv. When x= 1, v= 7 and when x= 2, v= 10. That integral becomes $\frac{1}{3}\int_7^10 \frac{1}{v} dv= \frac{1}{3}(ln(10)- ln(7))= \frac{1}{3}ln(10/7)$

In the second integral, let w= 3x+ 3 so that dx= (1/3)dw. When x= 1, w= 6 and when x= 2, w= 9. That integral becomes $\frac{1}{3}\int_6^9 \frac{1}{w}dw= \frac{1}{3}(ln(9)- ln(6))= \frac{1}{3}ln(3/2)$.

The original integral is $-\frac{1}{3}\left(ln(10/7)- ln(3/2)\right)= -\frac{1}{3}ln(20/21)$