Expert: Alon Mandes - 10/10/2009

Question
QUESTION: 2 questions - struggling for 2 hours now, need help!

1) Integrate:  ∫ 3t^2 + 4t + 8  dt

In the 2nd questions, a "1" goes at the bottom of the "∫" and a "4" goes at the top of the "∫," to indicate a "range" of numbers to be integrated over, but I don't know how to type it):
 
2) Integrate:  ∫ 3t^2 + 4t + 8  dt

Thank you so much for your help. If you could "spell out" the steps for me, I'd appreciate it. I do know that I have to separate each one first, like this: 5t^3 dt + 6t^2 dt + 3t dt + 1 dt. and I know that I have to add 1 to the "power" and then .... divide??  That's about as far as I'm getting, and not very good, either. thanks


ANSWER: Ok Felipe, let's do it step by step :
Due to the linearity of integration , we can claim that :
∫ (3t^2 + 4t + 8)  dt = ∫3t^2 dt + ∫4t dt + ∫8 dt = 3∫t^2 dt + 4∫t dt + 8∫ dt .
Now let's treat each one alone :
4
3∫t^2 dt = 3*(1/3)t^3 "{from 1 to 4}" = 3*(1/3)(4)^3 - 3*(1/3)*(1)^3 = 64 - 1 = 63 .
1

4
4∫t dt = (1/2)t^2 "{form 1 to 4}" = (1/2)(4)^2 - (1/2)(1)^2 = 8 - 1 = 7 .
1

4
8∫ dt = t "{form 1 to 4}" = 4 - 1 = 3 .
1

Therefore :
4
∫ 3t^2 + 4t + 8  dt = 63 + 7 + 3 = 73 .
1

Alon.

---------- FOLLOW-UP ----------

QUESTION: First of all, thank you for your help. Secondly, I messed up big time. I had 2 separate equations to figure out, and wrote the same numbers in for BOTH of them, when they should have been different. The first one was right - it was the ∫ 3t^2 + 4t + 8  dt. There was no "4" and "1" at the top and the bottom of the "∫" - it was just the "∫" by itself.

It was my 2nd equation that has the "4" and the "1" at the top and bottom of the "∫" but the numbers/exponents, etc. are different. The second equation with the "4" and the "1" at the top and bottom of the "∫" should have said: ∫ t^6 + 2t + 1 dt.   
Could you possibly help me with these again, as separate issues? I am going to try to do the first one according to the instructions as you already gave me ... and all I do is NOT give them a range of numbers to be integrated over, right?  I am SOOOO sorry ... I've just been trying to work on all my problems for HOURS today, and I think I'm getting tired. I really appreciate your help. Thanks!


Answer
Hello Felipe,
In general , the rule for polynoms integration is :
∫t^n dt = (1/n+1)t^(n+1) .
When the integral is attched with upper & lower limits , then
b
∫t^n dt = (1/n+1)t^(n+1) {from a to b} = (1/n+1)b^(n+1) - (1/n+1)a^(n+1) .
a

In our case,
4          4          4    4
∫ t^6 + 2t + 1 dt = ∫ t^6 dt + 2∫t dt + ∫1 dt .
1          1          1    1  

4
∫t^6 dt = (1/7)t^7 {from 1 to 4} = (1/7)4^7 - (1/7)1^7 =  2340.42
1

 4
2∫t dt = t^2 {from 1 to 4} = 4^2 - 1^2 = 16-1=15
1

4
∫1 dt = t {from 1 to 4} = 4 - 1 = 3
1

Therefore,
4
∫ t^6 + 2t + 1 dt = 2340.42 + 15 + 3 = 2358.42
1

Alon.