Multivariable Double Integration Problem

[methstudent]

1. The problem statement
Fill in the blanks ∫ [0,1] ∫ [2x^2,x+1] f(y) dy dx = ∫ [0,1] ( ) dy + ∫ [1,2] ( ) dy
The expressions you
obtain for the ( ) should not contain integral signs.

The brackets are the bounds of integration, and the open parenthesis are the blanks.

The Attempt at a Solution

I graphed the region and figured that the oder of integration has to be changed. I see that 2x^2 runs from (0,0) to (1,2) and that x+1 runs from (0,1) to (1,2) with these two creating the section we are integrating. It's unclear to me which how the integration goes from in terms of dy and dx to just being an integration in terms of dy.

 

[STEMucator]

1. The problem statement
Fill in the blanks ∫ [0,1] ∫ [2x^2,x+1] f(y) dy dx = ∫ [0,1] ( ) dy + ∫ [1,2] ( ) dy
The expressions you
obtain for the ( ) should not contain integral signs.

The brackets are the bounds of integration, and the open parenthesis are the blanks.

The Attempt at a Solution

I graphed the region and figured that the oder of integration has to be changed. I see that 2x^2 runs from (0,0) to (1,2) and that x+1 runs from (0,1) to (1,2) with these two creating the section we are integrating. It's unclear to me which how the integration goes from in terms of dy and dx to just being an integration in terms of dy.

Originally you have $0 ≤ x ≤ 1$ and $2x^2 ≤ y ≤ x+1$.

Graphing the region yields $0 ≤ y ≤ 2$ by inspection while $\sqrt{\frac{y}{2}} ≤ x ≤ y - 1$.

 

[methstudent]

How does inspection help when the new integrations are both in terms of dy?

 

[STEMucator]

You have a function $f(y)$ with a region that's already been nicely setup. The problem is, you don't know what $f(y)$ is directly, so you can't just integrate with respect to $y$ outright.

If you switch the limits first though, you are able to pull the $f(y)$ further out and then integrate in terms of $x$ first.

How does inspection help when the new integrations are both in terms of dy?

Simply breaking up the resulting integral into two integrals, you have found the answer.

 

[methstudent]

oh ok thanks i see how to do it now, thank you very much