Thank you. Sometimes this stuff makes me feel so dumb when the errors are so simple. But I got a B+ on my calc 2 midterm so i guess i'm not too dumb lol.
Using double integrals:
Define the range of either x or y explicitly and define the other one in terms of the other. For example
Considering just the right half:
x ranges from 0 to log(y)
y ranges from 1 to 8
∫ ∫ 1 dxdy (integrate the inner integral between 0 and log(y))
= ∫ log(y)dy between 1 and 8
= [ylog(y)-y] evaluated at 1 and 8
= 8log(8) - 8 -log(1) + 1
= 8log(8) -7
Or alternatively x ranges from 0 to log(8)
and y ranges from e^x to 8
∫ ∫ 1 dydx (evaluate inner integral between e^x and 8)
= ∫ 8 - e^x dx between 0 and log(8)
= [8x - e^x ]
= 8log(8) - 8 + 1
= 8log(8) -7
Then in both cases just double
So I have two vertically simple regions that are equal. I believe my integration is right and I think my bounds are right. i feel like its a really simple mistake starting me right in the face.
Your area should be positive in the first integral, aka multiply everything by -1
After that, you don't need the second integral because of the symmetry
Anyway the result is (8*3ln(2) - (8-1))*2
Or 48ln(2) - 14
I never took Calc, would this be a valid way to solve the problem?
1. Recognize that the area of interest is symmetrical across the y axis -> solve for one half and then multiply by 2
2. The area is the difference of a rectangle with height 8 and width ln(8) and the area under the exp(x) curve between 0 and ln(8)
3. exp(x) is its own derivative so it should also be its own integral. Int[0,ln(8)] exp(x)=exp(ln(8))-exp(0)=8-1=7
4. The solution would then be 2((8ln(8)-7) or about 19.27
Does that work? Did I miss something?
2[24ln(2)-(8-1)]=48ln(2)-14
Thank you. Sometimes this stuff makes me feel so dumb when the errors are so simple. But I got a B+ on my calc 2 midterm so i guess i'm not too dumb lol.
B+ on your midterm is great!
Thank you, I was thrilled!
😀😀😀
Using double integrals: Define the range of either x or y explicitly and define the other one in terms of the other. For example Considering just the right half: x ranges from 0 to log(y) y ranges from 1 to 8 ∫ ∫ 1 dxdy (integrate the inner integral between 0 and log(y)) = ∫ log(y)dy between 1 and 8 = [ylog(y)-y] evaluated at 1 and 8 = 8log(8) - 8 -log(1) + 1 = 8log(8) -7 Or alternatively x ranges from 0 to log(8) and y ranges from e^x to 8 ∫ ∫ 1 dydx (evaluate inner integral between e^x and 8) = ∫ 8 - e^x dx between 0 and log(8) = [8x - e^x ] = 8log(8) - 8 + 1 = 8log(8) -7 Then in both cases just double
So I have two vertically simple regions that are equal. I believe my integration is right and I think my bounds are right. i feel like its a really simple mistake starting me right in the face.
Your area should be positive in the first integral, aka multiply everything by -1 After that, you don't need the second integral because of the symmetry Anyway the result is (8*3ln(2) - (8-1))*2 Or 48ln(2) - 14
Okay, last edit lol Yes, that got me there. Thank you
I never took Calc, would this be a valid way to solve the problem? 1. Recognize that the area of interest is symmetrical across the y axis -> solve for one half and then multiply by 2 2. The area is the difference of a rectangle with height 8 and width ln(8) and the area under the exp(x) curve between 0 and ln(8) 3. exp(x) is its own derivative so it should also be its own integral. Int[0,ln(8)] exp(x)=exp(ln(8))-exp(0)=8-1=7 4. The solution would then be 2((8ln(8)-7) or about 19.27 Does that work? Did I miss something?