integration question~~!! (1 Viewer)

[pLuvia]

Suppose g(x) is a continuous even function. Given the ∫(5 top, 2 bottom) g(x)dx = -6 and ∫(5 top, 0 bottom) g(x)dx = 10, evaluate

a) ∫(0 top, -5 bottom) g(x)dx

b) ∫(2 top, 0 bottom) g(x)dx

c)∫(2 top, -5 bottom) g(x) dx

can someone help me do this?

and also this one

If ∫("a" top, 1 bottom) (³√x² + 1/³√x^4) dx = 12/5, where "a" is a positive constant, find the value of "a"



and how do you do make those proper 1/4 stuff.. anyways

thanks in advanced

 

[haboozin]

Suppose g(x) is a continuous even function. Given the ∫(5 top, 2 bottom) g(x)dx = -6 and ∫(5 top, 0 bottom) g(x)dx = 10, evaluate

a) ∫(0 top, -5 bottom) g(x)dx

b) ∫(2 top, 0 bottom) g(x)dx

c)∫(2 top, -5 bottom) g(x) dx

can someone help me do this?

and also this one

If ∫("a" top, 1 bottom) (³√x² + 1/³√x^4) dx = 12/5, where "a" is a positive constant, find the value of "a"



and how do you do make those proper 1/4 stuff.. anyways

thanks in advanced

a: 10
b. 16
c. 26

??

 

[pLuvia]

a: 10
b. 16
c. 26

??

err.. can you show me how you got those please??

 

[acmilan]

1. A continuous even function is symmetrical about the y-axis. So the integration from 0 to 5 is the same as the one from -5 to 0.

2. int from 0 to 2 = int from 5 to 0 - int from 5 to 2

3. int from -5 to 2 = int from -5 to 0 + int from 0 to 2, which is the sum of the first two questions

 

[haboozin]

your second question is too annoying...

i dont know how it could be solved other than trial and error.....

 

[Trev]

a) Since it's an even function, area underneath curve from 0 to 5 (as in original thing) should be the same as integral from -5 to 0; so 10.
b) From 0 to 2, is the same as taking away area from 0 to 5, from 2 to 5. So 10-(-6) = 16.
c) Same as adding answer from a) to area from answer b) which is 26.

 

[rama_v]

Suppose g(x) is a continuous even function. Given the ∫(5 top, 2 bottom) g(x)dx = -6 and ∫(5 top, 0 bottom) g(x)dx = 10, evaluate

a) ∫(0 top, -5 bottom) g(x)dx

b) ∫(2 top, 0 bottom) g(x)dx

c)∫(2 top, -5 bottom) g(x) dx

can someone help me do this?

and also this one

If ∫("a" top, 1 bottom) (³√x² + 1/³√x^4) dx = 12/5, where "a" is a positive constant, find the value of "a"




and how do you do make those proper 1/4 stuff.. anyways

thanks in advanced

For the last one just integrate and find a, i got a = √5 ..
Integrate[From 1 to a] (x^(2/3) + x^(-4/3) ) dx

= x^5/3 / (5/3) - x^-1/3/(-1/3)
= 3a^5/3/5 - 3/a^1/3 +(12/5) = (12/5)
so 3a^5/3 * a^(1/3) = 15
3a²=15
so a = √5

 

[pLuvia]

For the last one just integrate and find a, i got a = √5 ..
Integrate[From 1 to a] (x^(2/3) + x^(-4/3) ) dx

= x^5/3 / (5/3) - x^-1/3/(-1/3)
= 3a^5/3/5 - 3/a^1/3 +(12/5) = (12/5)
so 3a^5/3 * a^(1/3) = 15
3a²=15
so a = √5

i dun get wat you did here

= 3a5/3/5 - 3/a1/3 +(12/5) = (12/5)
so 3a^5/3 * a^(1/3) = 15

... lol.. sorry

 

[rama_v]

i dun get wat you did here

= 3a5/3/5 - 3/a1/3 +(12/5) = (12/5)
so 3a^5/3 * a^(1/3) = 15

... lol.. sorry

Just bringing the three up because its x^5/3 /(5/3) = 3x^(5/3)/5 then replace with a because thats the limit ur integrating with the next one i worked out the integration and then cross multiplied...

 

[Trev]

i dun get wat you did here

= 3a5/3/5 - 3/a1/3 +(12/5) = (12/5)
so 3a^5/3 * a^(1/3) = 15

... lol.. sorry

I'd do it like this:
= 3a5/3/5 - 3/a1/3 +(12/5) = (12/5)
3/5*a^5/3 = 3a^-1/3
3a^{5/3 + 1/3} = 15a^{-1/3 + 1/3}
a²=5
a=√5