You can put this solution on YOUR website! What is the area under the graph of the function f(t)= t^8+t between t=0 and t=1?
---
I assumne you mean area under the graph and above the t-axis.
------------------
Ans:
int from 0 to 1 of t^8+t
(1/9)t^9 + (1/2)t^2 evaluated at t = 1 minus the evaluation when t = 0.
-----
= (1/9)(1^9)+(1/2)(1^2) - [(1/9)*0^9 + (1/2)*0^2]
---
= (1/9) + (1/2)
--
= 11/18 sq. units
=======================
Cheers,
Stan H.