∫ l3t-5ldt from 0 to 3
assuming it means absolute value, then
= | [ (3/2t^2 - 5t ] | from 0 to 3
= | (3/2)(9) - 5(3) - 0 |
= 3/2
Find the distance from 0 to 3 of the ∫ l3t-5ldt. I don't understand how the answer is 41/6.
2 answers
not sure what you mean by the "distance"
If you mean the value of the integral, note that the vertex of the graph is at t = 5/3, where 3t-5=0
So, the integral is
∫[0,5/3] -(3t-5) dt + ∫[5/3,3] (3t-5)dt
The area is just the sum of the two triangles.
see
www.wolframalpha.com/input/?i=integral%5B0..3%5D+%7C3x-5%7C+dx
If you mean the value of the integral, note that the vertex of the graph is at t = 5/3, where 3t-5=0
So, the integral is
∫[0,5/3] -(3t-5) dt + ∫[5/3,3] (3t-5)dt
The area is just the sum of the two triangles.
see
www.wolframalpha.com/input/?i=integral%5B0..3%5D+%7C3x-5%7C+dx
3 answers