Area between two curves: y=(x-1)^3 and y=x-1

[Whutever42]

The question is to find the area between the curves of the equations:
y=(x-1)³ and y=x-1

I have set the two equal to each other and got x = 0,1,2

Then I used integrals to solve for the areas:

. . .\(\displaystyle \displaystyle \int_0^1\, \left[(x\, -\, 1)^3\, -\, (x\, -\, 1)\right]\, dx\, +\, \int_1^2\, \left[(x\, -\, 1)\, -\, (x\, -\, 1)^3\right]\, dx\)

I solve and I get 0 for an area but another classmate put a minus between the integrals and got .5 for an answer. I want to know which one is correct, or if I solved wrong, and why? Thank you!

 

[ksdhart2]

Actually, you're both wrong, but you each have a piece of the puzzle. Your approach, as far as adding the integrals, is correct (and your friend is wrong to subtract them); but your friend's answer of 0.5 is correct. I think I know where you went wrong, but it's hard to say for sure. What did you get for each of the separate integrals? That is to say:

\(\displaystyle \displaystyle \int_{0}^{1} \left[ (x-1)^3 - (x-1) dx \right] = \text{???}\)

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 dx \right] = \text{???}\)

 

[Whutever42]

Actually, you're both wrong, but you each have a piece of the puzzle. Your approach, as far as adding the integrals, is correct (and your friend is wrong to subtract them); but your friend's answer of 0.5 is correct. I think I know where you went wrong, but it's hard to say for sure. What did you get for each of the separate integrals? That is to say:

\(\displaystyle \displaystyle \int_{0}^{1} \left[ (x-1)^3 - (x-1) dx \right] = \text{???}\)

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 dx \right] = \text{???}\)

For the first integral I got 1/4
and the second integral i got -1/4

 

[greg1313]

What did you get for each of the separate integrals? That is to say:

The closing brackets go in different places:

\(\displaystyle \displaystyle \int_{0}^{1} \left[ (x-1)^3 - (x-1) \right] dx= \text{???}\)

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 \right]dx = \text{???}\)

 

[Harry_the_cat]

For the first integral I got 1/4
and the second integral i got -1/4

You need to find the absolute value of both and then add.
(subtracting a negative would achieve this, but it's not the correct way to think about it)

 

[ksdhart2]

For the first integral I got 1/4
and the second integral i got -1/4

Almost. The first integral is correct, but the second one is not. You should get 1/4 for both. My suspicion is that you mixed up the order of the terms. In the first interval, from 0 to 1, (x - 1)³ is the greater of the two functions, so it comes first. However, in the second interval, from 1 to 2, (x - 1)³ is the lesser of the two functions so it comes second.

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 \right] dx \ne \int_{1}^{2} \left[ (x-1)^3 - (x-1) \right] dx\)

In fact:

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 \right] dx = -\int_{1}^{2} \left[ (x-1)^3 - (x-1) \right] dx\)

 

[Whutever42]

Almost. The first integral is correct, but the second one is not. You should get 1/4 for both. My suspicion is that you mixed up the order of the terms. In the first interval, from 0 to 1, (x - 1)³ is the greater of the two functions, so it comes first. However, in the second interval, from 1 to 2, (x - 1)³ is the lesser of the two functions so it comes second.

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 \right] dx \ne \int_{1}^{2} \left[ (x-1)^3 - (x-1) \right] dx\)

In fact:

\(\displaystyle \displaystyle \int_{1}^{2} \left[ (x-1) - (x-1)^3 \right] dx = -\int_{1}^{2} \left[ (x-1)^3 - (x-1) \right] dx\)

Thank you! Now I understand where I went wrong. With the negative integral I got .5 for a final answer.
Again thank you very much!